233 reviewexam1 print sp 2013
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Solutions to old Exam 1 problems
Hi students!I am putting this old version of my review for the first midtermreview, place and time to be announced. Check for updates onthe web site as to which sections of the book will actually becovered. Enjoy!!Best, Bill Meeks
PS. There are probably errors in some of the solutionspresented here and for a few problems you need to completethem or simplify the answers; some questions are left to youthe student. Also you might need to add more detailedexplanations or justifications on the actual similar problems onyour exam. I will keep updating these solutions with bettercorrected/improved versions.
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Problem 1(a) - Fall 2008
Find parametric equations for the line L which contains A(1, 2, 3)and B(4, 6, 5).
Solution:
To get the parametric equations of L you need a pointthrough which the line passes and a vector parallel to the line.
Take the point to be A and the vector to be theββAB.
The vector equation of L is
r(t) =ββOA + t
ββAB = γ1, 2, 3γ+ t γ3, 4, 2γ = γ1 + 3t, 2 + 4t, 3 + 2tγ ,
where O is the origin.The parametric equations are:
x = 1 + 3ty = 2 + 4t, t β R.z = 3 + 2t
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Problem 1(b) - Fall 2008
Find parametric equations for the line L of intersection of theplanes x β 2y + z = 10 and 2x + y β z = 0.
Solution:The vector part v of the line L of intersection is orthogonal tothe normal vectors γ1,β2, 1γ and γ2, 1,β1γ. Hence v can betaken to be:
v = γ1,β2, 1γ Γ γ2, 1,β1γ =
β£β£β£β£β£β£i j k1 β2 12 1 β1
β£β£β£β£β£β£ = 1i + 3j + 5k.
Choose P β L so the z-coordinate of P is zero. Setting z = 0,we obtain: x β 2y = 10
2x + y = 0.Solving, we find that x = 2 and y = β4. Hence,P = γ2,β4, 0γ lies on the line L.The parametric equations are:
x = 2 + ty = β4 + 3tz = 0 + 5t = 5t.
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Problem 2(a) - Fall 2008
Find an equation of the plane which contains the pointsP(β1, 0, 1), Q(1,β2, 1) and R(2, 0,β1).
Solution:
Method 1
Consider the vectorsββPQ = γ2,β2, 0γ and
ββPR = γ3, 0,β2γ
which lie parallel to the plane.
Then consider the normal vector:
n =ββPQ Γ
ββPR =
β£β£β£β£β£β£i j k2 β2 03 0 β2
β£β£β£β£β£β£ = 4i + 4j + 6k.
So the equation of the plane is given by:
γ4, 4, 6γ Β· γx + 1, y , z β 1γ = 4(x + 1) + 4y + 6(z β 1) = 0.
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Problem 2(a) - Fall 2008
Find an equation of the plane which contains the pointsP(β1, 0, 1), Q(1,β2, 1) and R(2, 0,β1).
Solution:
Method 2The plane consists of all the points S(x , y , z) β R3, such thatββPS ,
ββPQ and
ββPR are in the same plane (coplanar).
But this happens if and only if their box product is zero.
So the equation of the plane is:β£β£β£β£β£β£x + 1 y z β 1
2 β2 03 0 β2
β£β£β£β£β£β£ = 4(x + 1) + 4y + 6(z β 1) = 0.
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Problem 2(b) - Fall 2008
Find the distance D from the point (1, 6,β1) to the plane2x + y β 2z = 19.
Solution:
Recall the distance formula D = |ax1+by1+cz1+d |βa2+b2+c2
from a point
P = (x1, y1, z1) to a plane ax + by + cz + d = 0.In order to apply the formula, rewrite the equation of theplane in standard form: 2x + y β 2z β 19 = 0.So, the distance from (1, 2,β1) to the plane is:
D =|(2 Β· 1) + (1 Β· 6) + (β2 Β· β1)β 19|β
22 + 12 + (β2)2=| β 9|β
9= 3.
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Problem 2(c) - Fall 2008
Find the point Q in the plane 2x + y β 2z = 19 which is closest tothe point (1, 6,β1). (Hint: You can use part b) of this problem tohelp find Q or first find the equation of the line L passing throughQ and the point (1, 6,β1) and then solve for Q.)
Solution:
The line L in the Hint passes through (1, 6,β1) and is parallelto n = γ2, 1,β2γ.So, L has parametric equations:
x = 1 + 2ty = 6 + t, t β R.z = β1β 2t
L intersects the plane 2x + y β 2z = 19 if and only if
2(1 + 2t) + (6 + t)β 2(β1β 2t) = 19ββ 9t = 9ββ t = 1.
Substituting t = 1 in the parametric equations of L givesthe point Q = (3, 7,β3).
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Problem 3(a) - Fall 2008
Find the volume V of the parallelepiped such that the followingfour points A = (3, 4, 0), B = (3, 1,β2), C = (4, 5,β3),D = (1, 0,β1) are vertices and the vertices B,C ,D are alladjacent to the vertex A.
Solution:
The parallelepiped is determined by its edges
ββAB = γ0,β3,β2γ ,
ββAC = γ1, 1,β3γ ,
ββAD = γβ2,β4,β1γ .
Its volume can be computed as the absolute value of the box
productββAB Β· (
ββAC Γ
ββAD), i.e.,
V =
β£β£β£β£β£β£β£β£β£β£β£β£
0 β3 β21 1 β3β2 β4 β1
β£β£β£β£β£β£β£β£β£β£β£β£ = |3(β1β 6)β 2(β4 + 2)| = |β17| = 17.
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Problem 3(b) - Fall 2008
Find the center and radius of the spherex2 β 4x + y2 + 4y + z2 = 8.
Solution:
Completing the square we get
x2β4x+y2+4y+z2 = (x2β4x+4)β4+(y2+4y+4)β4+(z2)
= (x β 2)2 β 4 + (y + 2)2 β 4 + z2 = 8
ββ
(x β 2)2 + (y + 2)2 + z2 = 16.
This gives:
Center = (2,β2, 0) Radius = 4
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Problem 4(a) - Fall 2008
The position vector of a particle moving in space equalsr(t) = t2iβ t2j + 1
2 t2k at any time t β₯ 0.
a) Find an equation of the tangent line to the curve at the point(4,β4, 2).
Solution:The parametrized curve passes through the point (4,β4, 2) ifand only if
t2 = 4, βt2 = β4,1
2t2 = 2ββ t2 = 4ββ t = Β±2.
Since we have that t β₯ 0, we are left with the choice t0 = 2.The velocity vector field to the curve is given by
rβ²(t) = γ 2t,β2t, tγ hence rβ²(2) = γ4,β4, 2γ .The equation of the tangent line in question is:
x = 4 + 4ty = β4β 4t, t β₯ 0.z = 2 + 2t
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Problem 4(b) - Fall 2008
The position vector of a particle moving in space equalsr(t) = t2iβ t2j + 1
2 t2k at any time t β₯ 0.
(b) Find the length L of the arc traveled from time t = 1 to timet = 4.
Solution:
The velocity field is:
v(t) = rβ²(t) = γ2t,β2t, tγ .
Since t β₯ 0, the speed is:β£β£rβ²(t)β£β£ =β
9t2 =ββ£β£rβ²(t)
β£β£ = 3t.
Therefore, the length is:
L =
β« 4
13t dt =
3
2t2β£β£β£β£41
=3
2Β· 16β 3
2Β· 1 =
3
2Β· 15 =
45
2.
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Problem 4(c) - Fall 2008
Suppose a particle moving in space has velocity
v(t) = γsin t, 2 cos 2t, 3etγ
and initial position r(0) = γ1, 2, 0γ. Find the position vectorfunction r(t).
Solution:
One can recover the position, by integrating the velocity:
r(t) =
β« t
0v(Ο)dΟ + r(0)
Carrying out this integral yields:
r(t) =
β¨β cos Ο
β£β£β£t0, sin 2Ο
β£β£β£t0, 3et
β£β£β£t0
β©+ γ1, 2, 0γ
=β¨2β cos t, 2 + sin 2t, 3et β 3
β©.
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Problem 5(a) - Fall 2008
Consider the points A(2, 1, 0), B(3, 0, 2) and C (0, 2, 1). Find thearea of the triangle ABC . (Hint: If you know how to find the areaof a parallelogram spanned by 2 vectors, then you should be ableto solve this problem.)
Solution:
Recall that:
Area =1
2
β£β£β£β£ββAB Γ ββAC β£β£β£β£ .Since
ββAB = γ1,β1, 2γ and
ββAC = γβ2, 1, 1γ,
Area =1
2
β£β£β£β£β£β£β£β£β£β£β£β£
i j k1 β1 2β2 1 1
β£β£β£β£β£β£β£β£β£β£β£β£ =
1
2|γβ3,β5,β1γ| =
1
2
β35.
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Problem 5(b) - Fall 2008
Three of the four vertices of a parallelogram are P(0,β1, 1),Q(0, 1, 0) and R(2, 1, 1). Two of the sides are PQ and PR. Findthe coordinates of the fourth vertex.
Solution:
Denote the fourth vertex by S. Then
ββOS =
ββOQ +
ββPR = γ0, 1, 0γ+ γ2, 2, 0γ = γ2, 3, 0γ ,
where O is the origin. That is,
S = (2, 3, 0).
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Problem 6(a) - Spring 2008
Find an equation of the plane through the points A = (1, 2, 3),B = (0, 1, 3), and C = (2, 1, 4).
Solution:
Since a plane is determined by its normal vector n and a point onit, say the point A, it suffices to find n. Note that:
n =ββAB Γ
ββAC =
β£β£β£β£β£β£i j kβ1 β1 01 β1 1
β£β£β£β£β£β£ = γβ1, 1, 2γ.
So the equation of the plane is:
β(x β 1) + (y β 2) + 2(z β 3) = 0.
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Problem 6(b) - Spring 2008
Find the area of the triangle β with vertices at the pointsA = (1, 2, 3), B = (0, 1, 3), and C = (2, 1, 4).(Hint: the area of this triangle is related to the area of a certainparallelogram)
Solution:
Consider the points A = (1, 2, 3), B = (0, 1, 3) and C = (2, 1, 4).Then the area of the triangle β with these vertices can be found
by taking the area of the parallelogram spanned byββAB and
ββAC
and dividing by 2. Thus:
Area(β) =|ββAB Γ
ββAC |
2=
1
2
β£β£β£β£β£β£β£β£β£β£β£β£
i j kβ1 β1 01 β1 1
β£β£β£β£β£β£β£β£β£β£β£β£
=1
2|γβ1, 1, 2γ| =
1
2
β1 + 1 + 4 =
1
2
β6.
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Problem 7(a) - Spring 2008
Find the parametric equations of the line passing through thepoint (2, 4, 1) that is perpendicular to the plane 3x β y + 5z = 77.
Solution:
The vector part of the line L is the normal vectorn = γ3,β1, 5γ to the plane.
The vector equation of L is:
r(t) = γ2, 4, 1γ+ tn
= γ2, 4, 1γ+ tγ3,β1, 5γ = γ2 + 3t, 4β t, 1 + 5tγ.
The parametric equations are:
x = 2 + 3ty = 4β tz = 1 + 5t.
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Problem 7(b) - Spring 2008
Find the intersection point of the line L(t) = γ2 + 3t, 4β t, 1 + 5tγin part (a) and the plane 3x β y + 5z = 77.
Solution:
By part (a), we have L has parametric equations:x = 2 + 3ty = 4β tz = 1 + 5t.
Plug these t-values into equation of plane and solve for t:
3(2 + 3t)β (4β t) + 5(1 + 5t) = 77,
6 + 9t β 4 + t + 5 + 25t = 77,
35t = 70; =β t = 2.
So L intersects the plane at time t = 2.At t = 2, the parametric equations give the point:
γ2 + 3 Β· 2, 4β 2, 1 + 5 Β· 2γ = γ8, 2, 11γ.
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Problem 8(a) - Spring 2008
A plane curve is given by the graph of the vector function
u(t) = γ1 + cos t, sin tγ, 0 β€ t β€ 2Ο.
Find a single equation for the curve in terms of x and y byeliminating t.
Solution:
Rewriting u, we get:
u(t) = γ1 + cos t, sin tγ = γ1, 0γ+ γcos t, sin tγ.
Since γcos t, sin tγ is the parametrization of the circle ofradius 1 centered at the origin, then u is a circle of radiusr = 1 centered at (1, 0).
So the answer is:
(x β 1)2 + (y β 0)2 = 12
or (x β 1)2 + y2 = 1.
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Problem 8(b) - Spring 2008
Consider the space curve given by the graph of the vector function
r(t) = γ1 + cos t, sin t, tγ, 0 β€ t β€ 2Ο.
Sketch the curve and indicate the direction of increasing t in yourgraph.
Solution:
The sketch would be the following one translated 1 unit along thex-axis.
par
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Problem 8(c) - Spring 2008
Determine parametric equations for the line T tangent to thegraph of the space curve for r(t) = γ1 + cos t, sin t, tγ at t = Ο/3,and sketch T in the graph obtained in part (b).
Solution:First find the velocity vector rβ²(t) :
rβ²(t) = γ(1 + cos t)β², (sin t)β², 1γ = γβ sin t, cos t, 1γ.At t = Ο
3 ,
r(Ο
3) = γ1 + cos
Ο
3, sin
Ο
3,Ο
3γ = γ3
2,
β3
2,Ο
3γ,
rβ²(Ο
3) = γβ sin
Ο
3, cos
Ο
3, 1γ = γβ
β3
2,
1
2, 1γ.
The vector part of tangent line T is rβ²(Ο3 ) and a point on line is
r(Ο3 ).
The vector equation is: T(t) = r(Ο3 ) + trβ²(Ο
3 ).The parametric equations are:
x = 32 β
β32 t
y =β32 + 1
2 tz = Ο
3 + t.
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Problem 9(a) - Spring 2008
Suppose that r(t) has derivative rβ²(t) = γβ sin 2t, cos 2t, 0γ on theinterval 0 β€ t β€ 1. Suppose we know that r(0) = γ12 , 0, 1γ.Determine r(t) for all t.
Solution:
Find r(t) by integration:
r(t) =
β«rβ²(t) dt =
β«γβ sin 2t, cos 2t, 0γ dt
= γ12
cos(2t) + x0,1
2sin(2t) + y0, z0γ.
Now solve for the point (x0, y0, z0) using r(0) = γ12 , 0, 1γ:(12 cos(0) + x0,
12 sin(0) + y0, z0) = (12 + x0, y0, z0) = (12 , 0, 1).
So x0 = 0, y0 = 0, z0 = 1.
Thus,r(t) = γ1
2cos 2t,
1
2sin 2t, 1γ.
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Problem 9(b) - Spring 2008
Suppose that r(t) has derivative rβ²(t) = γβ sin 2t, cos 2t, 0γ on theinterval 0 β€ t β€ 1. Suppose we know that r(0) = γ12 , 0, 1γ. Showthat r(t) is orthogonal to rβ²(t) for all t.
Solution:
By part (a),r(t) = γ1
2cos 2t,
1
2sin 2t, 1γ,
Taking dot products, we get:
r(t) Β· rβ²(t) = β1
2cos 2t sin 2t +
1
2sin 2t cos 2t + 0 = 0.
Since the dot product is zero, then for each t, r(t) isorthogonal to rβ²(t).
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Problem 9(c) - Spring 2008
Suppose that r(t) has derivative rβ²(t) = γβ sin 2t, cos 2t, 0γ on theinterval 0 β€ t β€ 1. Suppose we know that r(0) = γ12 , 0, 1γ. Findthe arclength L of the graph of the vector function r(t) on theinterval 0 β€ t β€ 1.
Solution:
Recall that the length of r(t) on the interval [0, 1] is gotten byintegrating the speed |rβ²(t)|.Calculating, we get:
L =
β« 1
0|rβ²(t)| dt =
β« 1
0|γβ sin 2t, cos 2t, 0γ | dt
=
β« 1
0
βsin2 2t + cos2 2t dt =
β« 1
0|1| dt = t
β£β£β£10
= 1.
ThusL = 1.
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Problem 10(a) - Spring 2008
If r(t) = (2t)i + (t2 β 6)jβ (13 t3)k represents the position vector
of a moving object (where t β₯ 0 is measured in seconds anddistance is measured in feet),Find the speed s(t) and the velocity v(t) of the object at time t.
Solution:
Recall that the velocity v(t) vector of r(t) at time t is rβ²(t)and the speed s(t) is its length |rβ²(t)|.Calculating with r(t) = γ2t, t2 β 6,β1
3 t3γ :
v(t) = rβ²(t) = γ2, 2t,βt2γ,
s(t) = |rβ²(t)| =β
22 + (2t)2 + (βt2)2 =β
4 + 4t2 + t4.
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Problem 10(b) - Spring 2008
If r(t) = (2t)i + (t2 β 6)jβ ( 13 t
3)k represents the position vector of amoving object (where t β₯ 0 is measured in seconds and distance ismeasured in feet.)If a second object travels along a path given defined by the graph of thevector function w(s) = γ2, 5, 1γ+ sγ2,β1,β5γ, show that the paths ofthe two objects intersect at a common point P.
Solution:
Note that w(s) = γ2 + 2s, 5β s, 1β 5sγ andr(t) = γ2t, t2 β 6,β 1
3 t3γ.
Setting the x and y -coordinates of w(s) and r(t) equal, we obtain:
x = 2t = 2 + 2s =β t = s + 1
y = t2 β 6 = 5β s =β (s + 1)2 β 6 = s2 + 2s β 5 = 5β s
=β s2 + 3s β 10 = 0 =β (s + 5)(s β 2) = 0.
So, (s = 2 and t = 3) or (s = β5 and t = β4).
Sincer(3) = γ6, 3,β9γ = w(2),
the paths intersect at P = (6, 3,β9).
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Problem 10(c) - Spring 2008
If r(t) = (2t)i + (t2 β 6)jβ (13 t3)k represents the position vector
of a moving object (where t β₯ 0 is measured in seconds anddistance is measured in feet),If s = t in part (b), (i.e. the position of the second object is w(t)when the first object is at position r(t)), do the two objects evercollide?
Solution:
Set t = s in part (b).
Then the x-coordinate of r(t) is 2t and the x-coordinate ofw(t) = γ2 + 2t, 5β t, 1β 5tγ is 2 + 2t, and 2t 6= 2 + 2t for allt.
Since r(t) and w(t) have different x-coordinates for all valuesof t, then they never collide.
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Problem 11(a) - Spring 2007
Find parametric equations for the line L which contains A(7, 6, 4)and B(4, 6, 5).
Solution:
A vector parallel to the line L is:
v =ββAB = γ4β 7, 6β 6, 5β 4, γ = γβ3, 0, 1γ.
A point on the line is A(7, 6, 4).
Therefore parametric equations for the line L are:
x = 7β 3ty = 6z = 4 + t.
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Problem 11(b) - Spring 2007
Find the parametric equations for the line L of intersection ofthe planes x β 2y + z = 5 and 2x + y β z = 0.
Solution:A vector v parallel to the line is the cross product of thenormal vectors of the planes:
v = γ1,β2, 1γ Γ γ2, 1,β1γ =
β£β£β£β£β£β£i j k1 β2 12 1 β1
β£β£β£β£β£β£=
β£β£β£β£ β2 11 β1
β£β£β£β£ iβ β£β£β£β£ 1 12 β1
β£β£β£β£ j +
β£β£β£β£ 1 β22 1
β£β£β£β£ k = γ1, 3, 5γ.
A point on L is any (x0, y0, z0) that satisfies both of the planeequations. Setting z = 0, we obtain the equations x β 2y = 5and 2x + y = 0 and find such a point (1,β2, 0).Therefore parametric equations for L are:
x = 1 + ty = β2 + 3tz = 5t.
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Problem 12(a) - Spring 2007
Find an equation of the plane which contains the pointsP(β1, 0, 2), Q(1,β2, 1) and R(2, 0,β1).
Solution:A normal vector to the plane can be found by taking the crossproduct of any two vectors that lie in the plane. Two vectors
that lie in the plane areββPQ = γ2,β2,β1γ and
ββPR = γ3, 0,β3γ.So the normal vector is
n = γ2,β2,β1γ Γ γ3, 0,β3γ =
β£β£β£β£β£β£i j k2 β2 β13 0 β3
β£β£β£β£β£β£=β£β£β£β£ β2 β10 β3
β£β£β£β£ iβ β£β£β£β£ 2 β13 β3
β£β£β£β£ j +
β£β£β£β£ 2 β23 0
β£β£β£β£ k = γ6, 3, 6γ.
A point on the plane is P(β1, 0, 2). Therefore,
6(x β (β1)) + 3(y β 0) + 6(z β 2) = 0,
or simplified, 6x + 3y + 6z β 6 = 0.
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Problem 12(b) - Spring 2007
Find the distance D from the point P1 = (1, 0,β1) to the plane2x + y β 2z = 1.
Solution:
The normal to the plane is n = γ2, 1,β2γ and the pointP0 = (0, 1, 0) lies on this plane. Consider the vector from P0 toP1 = (1, 0,β1) which is b = γ1,β1,β1γ. The distance D from(1, 0,β1) to the plane is equal to:
|compn b| =
β£β£β£β£b Β· n
|n|
β£β£β£β£ = |γ1,β1,β1γ Β· 1
3γ2, 1,β2γ| = 1.
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Problem 12(c) - Spring 2007
Find the point P in the plane 2x + y β 2z = 1 which is closest to thepoint (1, 0,β1). (Hint: You can use part (b) of this problem to help findP or first find the equation of the line passing through P and the point(1, 0,β1) and then solve for P.)
Solution:
First find the parametric equations of the line that goes throughthe point (1, 0,β1) that is normal to the plane: x = 1 + 2t, y = t,z = β1β 2t; here n = γ2, 1,β2γ is a normal to the plane.The point P in the plane closest to (1, 0,β1) is the intersection ofthis line and the plane.Substitute the parametric equations of the line into the planeequation: 2(1 + 2t) + (t)β 2(β1β 2t) = 1.
Simplifying and solving for t,
9t + 4 = 1 =β t = β1
3.
Plugging this t-value into the parametric equations, we get thecoordinates of the point of intersection: x = 1 + 2(β 1
3 ) = 13 ,
y = β 13 , z = β1β 2(β 1
3 ) = β 13 .
So the point on the plane closest to (1, 0,β1) is P = ( 13 ,β
13 ,β
13 ).
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Problem 13(a) - Spring 2007
Consider the two space curvesr1(t) = γcos(tβ1), t2β1, 2t4γ, r2(s) = γ1+ln s, s2β2s +1, 2s2γ,where t and s are two independent real parameters. Find thecosine of the angle ΞΈ between the tangent vectors of the twocurves at the intersection point (1, 0, 2).
Solution:
The point (1, 0, 2) corresponds to the t-value t = 1 for r1 ands-value s = 1 for r2.
rβ²1(t) = γβ sin(t β 1), 2t, 8t3γ is the tangent vector to r1(t).
At t = 1, rβ²1(1) = γβ sin(1β 1), 2(1), 8(13)γ = γ0, 2, 8γ.rβ²2(s) = γ1s , 2s β 2, 4sγ is the tangent vector to r2(s).
At s = 1, rβ²2(1) = γ11 , 2(1)β 2, 4(1)γ = γ1, 0, 4γ.Therefore,
cos(ΞΈ) =γ0, 2, 8γ Β· γ1, 0, 4γ|γ0, 2, 8γ||γ1, 0, 4γ|
=32β
68β
17.
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Problem 13(b) - Spring 2007
Find the center and radius of the sphere
x2 + y2 + 2y + z2 + 4z = 20.
Solution:
Completing the square in the y and z variables, we get
x2 + (y2 + 2y + 1) + (z2 + 4z + 4) = 20 + 1 + 4.
Rewriting, we have
x2 + (y + 1)2 + (z + 2)2 = 25 = 52.
Hence, the center is C = (0,β1,β2) and the radius is r = 5.
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Problem 14(a) - Spring 2007
The velocity vector of a particle moving in space equalsv(t) = 2tiβ 2tj + tk at any time t β₯ 0.At the time t = 4, this particle is at the point (0, 5, 4). Find anequation of the tangent line T to the position curve r(t)at thetime t = 4.
Solution:
This line goes through the point (0, 5, 4) and has vector partparallel to the tangent vector v(4) = γ8,β8, 4γ.The vector equation is: T(t) = γ0, 5, 4γ+ tγ8,β8, 4γSo the line T has the parametric equations:
x = 8ty = 5β 8tz = 4 + 4t.
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Problem 14(b) - Spring 2007
The velocity vector of a particle moving in space equalsv(t) = 2tiβ 2tj + tk at any time t β₯ 0.Find the length L of the arc traveled from time t = 2 to timet = 4.
Solution:
Using the arclength formula,
L =
β« 4
2|v(t)| dt =
β« 4
2
β(2t)2 + (β2t)2 + t2 dt
=
β« 4
2
β9t2 dt =
β« 4
23t dt
=3
2t2β£β£β£β£42
=3
2(16β 4) = 18.
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Problem 14(c) - Spring 2007
Find a vector function r(t) which represents the curve ofintersection of the cylinder x2 + y2 = 1 and the planex + 2y + z = 4.
Solution:
Since the first equation is the equation of a circular cylinder,parametrize the x and y coordinates by setting x = cos(t) andy = sin(t).
Next use the second equation z = 4β x β 2y to solve for z interms of t:
z = 4β x β 2y = 4β cos(t)β 2 sin(t).
Therefore,
r(t) = γcos(t), sin(t), 4β cos(t)β 2 sin(t)γ.
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Problem 15(a) - Spring 2008
Consider the points A(2, 1, 0), B(1, 0, 2) and C (0, 2, 1). Find thearea A of the triangle ABC . (Hint: If you know how to find thearea of a parallelogram spanned by 2 vectors, then you should beable to solve this problem.)
Solution:
The area of the parallelogram is
|ββAB Γ
ββAC | =
β£β£β£β£β£β£i j kβ1 β1 2β2 1 1
β£β£β£β£β£β£=
β£β£β£β£β£β£β£β£ β1 21 1
β£β£β£β£ iβ β£β£β£β£ β1 2β2 1
β£β£β£β£ j +
β£β£β£β£ β1 β1β2 1
β£β£β£β£ kβ£β£β£β£= |γβ3,β3,β3γ| =
β27.
So the area of the triangle ABC is
A =
β27
2.
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Problem 15(b) - Spring 2008
Suppose a particle moving in space has velocity
v(t) = γsin t, cos 2t, etγ
and initial position r(0) = γ1, 2, 0γ. Find the position vectorfunction r(t).
Solution:We find r(t) by integrating rβ²(t) = v(t):
r(t) =
β« t
0v(t) dt + r(0) = γβ cos t,
1
2sin 2t, etγ
β£β£β£β£t0
+ γ1, 2, 0γ
= γβ cos t,1
2sin 2t, etγ β γβ cos 0,
1
2sin 0, e0γ+ γ1, 2, 0γ
= γβ cos t,1
2sin 2t, etγ β γβ1, 0, 1γ+ γ1, 2, 0γ
= γβ cos t,1
2sin 2t, etγ+ γ2, 2,β1γ.
So, r(t) = γ2β cos t, 2 +1
2sin 2t, β1 + etγ.
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Problem 16 - Fall 2007
Find the equation of the plane containing the linesx = 4β 4t, y = 3β t, z = 1 + 5t and
x = 4β t, y = 3 + 2t, z = 1
Solution:
To find the equation of a plane, we need to find its normal nand a point on it. Setting t = 0, we find the point (4, 3, 1) onthe first line.The part vector v1 of the first line is γβ4,β1, 5γ and thevector part v2 of the second line is γβ1, 2, 0γ.Since the vector
n = v1 Γ v2 =
β£β£β£β£β£β£i j kβ4 β1 5β1 2 0
β£β£β£β£β£β£ = γβ10,β5,β9γ,
is orthogonal to both v1 and v2, it is the normal to the plane.The equation of the plane is:
γβ10,β5,β9γ Β· γx β 4, y β 3, z β 1γ
= β10(x β 4)β 5(y β 3)β 9(z β 1) = 0.
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Problem 17 - Fall 2007
Find the distance D from the point P1 = (3,β2, 7) and the plane4x β 6y β z = 5.
Solution:
Recall the distance formula D = |ax1+by1+cz1+d |βa2+b2+c2
from a point
P = (x1, y1, z1) to a plane ax + by + cz + d = 0.In order to apply the formula, rewrite the equation of theplane in standard form: 4x β 6y β z β 5 = 0.So, the distance from (3,β2, 7) to the plane is:
D =|(4 Β· 3) + (β6 Β· β2) + (β1 Β· 7)β 5|β
42 + (β6)2 + (β1)2=
12β53.
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Problem 18 - Fall 2007
Determine whether the lines L1 and L2 given below are parallel,skew or intersecting. If they intersect, find the point ofintersection.
L1 :x
1=
y β 1
2=
z β 2
3
L2 :x β 3
β4=
y β 2
β3=
z β 1
2
Solution:Rewrite these lines as vector equations:
L1(t) = γt, 2t + 1, 3t + 2γL2(s) = γβ4s + 3,β3s + 2, 2s + 1γ
Equating x and y -coordinates:
x = t = β4s + 3y = 2t + 1 = β3s + 2.
Solving gives s = 1 and t = β1.L1(β1) = γβ1,β1,β1γ 6= γβ1,β1, 3γ = L2(1). So these lines donot intersect.Since the lines are clearly not parallel (the direction vectors γ1, 2, 3γand γβ4,β3, 2γ are not parallel), the lines are skew.
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Problem 19(a) - Fall 2007
Suppose a particle moving in space has the velocity
v(t) = γ3t2, 2 sin(2t), etγ.Find the acceleration of the particle. Write down a formula forthe speed of the particle (you do not need to simplify theexpression algebraically).
Solution:
Recall the acceleration vector a(t) = vβ²(t). Hence,
a(t) = γ6t, 4 cos(2t), etγ.
Recall that the speed(t) is the length of the velocity vector.Hence,
speed(t) =
β9t4 + 4 sin2(2t) + e2t .
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Problem 19(b) - Fall 2007
Suppose a particle moving in space has the velocity
v(t) = γ3t2, 2 sin(2t), etγ.If initially the particle has the position r(0) = γ0,β1, 2γ, what isthe position at time t?
Solution:
To find the position r(t), we first integrate the velocity v(t)and second use the initial position value r(0) = γ0,β1, 2γ tosolve for the constants of integration.
r(t) =
β«γ3t2, 2 sin 2t, etγ dt = γt3+x0,β cos(2t)+y0, e
t+z0γ.
Plugging in the position at t = 0, we get:
γ03+x0,β cos(0)+y0, e0+z0γ = γx0,β1+y0, 1+z0γ = γ0,β1, 2γ.
Thus, x0 = 0, y0 = 0 and z0 = 1.
Hence,r(t) = γt3,β cos 2t, et + 1γ.
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Problem 20(a) - Fall 2007
Three of the four vertices of a parallelogram β are P(0,β1, 1),Q(0, 1, 0) and R(3, 1, 1). Two of the sides are PQ and PR.Find the area of the parallelogram.
Solution:
Consider the vectorsββPQ = γ0, 2,β1γ and
ββPR = γ3, 2, 0γ. Then the
area of the parallelogram β spanned byββPQ and
ββPR is:
Area(β) = |ββPQ Γ
ββPR| =
β£β£β£β£β£β£β£β£β£β£β£β£
i j k0 2 β13 2 0
β£β£β£β£β£β£β£β£β£β£β£β£
= |γ2,β3,β6γ| =β
4 + 9 + 36 = 7
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Problem 20(b) - Fall 2007
Three of the four vertices of a parallelogram are P(0,β1, 1),Q(0, 1, 0) and R(3, 1, 1). Two of the sides are PQ and PR.Find the cosine of the angle between the vector PQ and PR.
Solution:
Note that:ββPQ = γ0, 2,β1γ
ββPR = γ3, 2, 0γ.
By our formula for dot products:
cos ΞΈ =
ββPQ Β·
ββPR
|ββPQ||
ββPR|
=γ0, 2,β1γ Β· γ3, 2, 0γβ
5β
13=
4β5β
13.
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Problem 20(c) - Fall 2007
Three of the four vertices of a parallelogram are P(0,β1, 1),Q(0, 1, 0) and R(3, 1, 1). Two of the sides are PQ and PR. Findthe coordinates of the fourth vertex.
Solution:
Denote the fourth vertex by S. Then
ββOS =
ββOQ +
ββPR = γ0, 1, 0γ+ γ3, 2, 0γ = γ3, 3, 0γ ,
where O is the origin. That is,
S = (3, 3, 0).
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Problem 21(a) - Fall 2007
Let C be the parametric curve
x = 2β t2, y = 2t β 1, z = ln t.
Determine the point(s) of intersection of C with the xz-plane.
Solution:
The points of intersection of C with the xz-plane correspondto the points where the y -coordinate of C is 0.
When y = 0, then 0 = 2t β 1 or t = 12 .
Hence,
γ2β (1
2)2, 2 Β· 1
2β 1, ln
1
2γ = γ13
4, 0,β ln 2γ
is the unique point of the intersection of C with xz-plane.
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Problem 21(b) - Fall 2007
Let C be the parametric curve
x = 2β t2, y = 2t β 1, z = ln t.Determine parametric equations of tangent line to C at (1, 1, 0).
Solution:
Using the y -coordinate of C, note that t = 1 when(1, 1, 0) β C.The velocity vector to
C(t) = γ2β t2, 2t β 1, ln tγis:
Cβ²(t) = γβ2t, 2,1
tγ.
Thus, Cβ²(1) = γβ2, 2, 1γis the vector part of the tangent line to C at (1, 1, 0).The parametric equations are:
x = 1β 2ty = 1 + 2tz = t.
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Problem 21(c) - Fall 2007
Let C be the parametric curve
x = 2β t2, y = 2t β 1, z = ln t.Set up, but not solve, a formula that will determine the length L ofC for 1 β€ t β€ 2.
Solution:
The vector equation of C is r(t) = γ2β t2, 2t β 1, ln tγ withvelocity vector
v(t) = rβ²(t) = γβ2t, 2,1
tγ.
Since the length of L is the integral of the speed |rβ²(t)|,
L =
β« 2
1|γβ2t, 2,
1
tγ| dt =
β« 2
1
β4t2 + 4 +
1
t2dt.
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Problem 22(a) - Fall 2006
Find parametric equations for the line r which contains A(2, 0, 1)and B(β1, 1,β1).
Solution:
Note thatββAB = γβ3, 1,β2γ and the vector equation is:
r(t) = ~A+ tββAB = γ2, 0, 1γ+ tγβ3, 1,β2γ = γ2β 3t, t, 1β 2tγ.
The parametric equations are:
x = 2β 3ty = tz = 1β 2t.
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Problem 22(b) - Fall 2006
Determine whether the lines L1 : x = 1 + 2t, y = 3t, z = 2β tand L2 : x = β1 + s, y = 4 + s, z = 1 + 3s are parallel, skew orintersecting.
Solution:Vector part of line L1 is v1 = γ2, 3,β1γ and for line L2 isv2 = γ1, 1, 3γ. Clearly, v1 is not a scalar multiple of v2 and sothese lines are not parallel.If these lines intersect, then for some values of t and s:
x = 1 + 2t = β1 + s =β 2t = β2 + s,
y = 3t = 4 + s =β 3t = 4 + s.
Solving yields: t = 6 and s = 14.
Plugging these values into z = 2β t = 1 + 3s yields theinequality β4 6= 43, which means the z-coordinates are neverequal and the lines do not intersect.Thus, the lines are skew.
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Problem 23(a) - Fall 2006
Find an equation of the plane which contains the pointsP(β1, 2, 1), Q(1,β2, 1) and R(1, 1,β1).
Solution:
Consider the vectorsββPQ = γ2,β4, 0γ and
ββPR = γ2,β1,β2γ
which are parallel to the plane.
The normal vector to the plane is:
n =ββPQ Γ
ββPR =
β£β£β£β£β£β£i j k2 β4 02 β1 β2
β£β£β£β£β£β£ = 8i + 4j + 6k.
Since P(β1, 2, 1) lies on the plane, the equation of theplane is:
γ8, 4, 6γΒ·γx+1, yβ2, zβ1γ = 8(x+1)+4(yβ2)+6(zβ1) = 0.
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Problem 23(b) - Fall 2006
Find the distance D from the point (1, 2,β1) to the plane2x + y β 2z = 1.
Solution:
The normal to the plane is n = γ2, 1,β2γ and the pointP0 = (0, 1, 0) lies on this plane. Consider the vector from P0 toP1 = (1, 2,β1) which is b = γ1, 1,β1γ. The distance D from(1, 2,β1) to the plane is equal to:
|compn b| =
β£β£β£β£b Β· n
|n|
β£β£β£β£ = |γ1, 1,β1γ Β· 1
3γ2, 1,β2γ| =
5
3.
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Problem 24(a) - Fall 2006
Let two space curvesr1(t) = γcos(t β 1), t2 β 1, t4γ, r2(s) = γ1 + ln s, s2 β 2s + 1, s2γ,be given where t and s are two independent real parameters. Findthe cosine of the angle between the tangent vectors of the twocurves at the intersection point (1, 0, 1).
Solution:
When r1(t) = γ1, 0, 1γ, then t = 1.When r2(s) = γ1, 0, 1γ, then s = 1.Calculating derivatives, we obtain:rβ²1(t) = γβ sin(t β 1), 2t, 4t3γrβ²1(1) = γ0, 2, 4γrβ²2(s) = γ1s , 2s β 2, 2sγrβ²2(1) = γ1, 0, 2γ.Hence,
cos ΞΈ =rβ²1(1) Β· rβ²2(1)
|rβ²1(1)||rβ²2(1)|=γ0, 2, 4γ Β· γ1, 0, 2γβ
20β
5
=1β100
(0 Β· 1 + 2 Β· 0 + 4 Β· 2) =8
10=
4
5.
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Problem 24(b) - Fall 2006
Suppose a particle moving in space has velocity
v(t) = γsin t, cos 2t, etγand initial position r(0) = γ1, 2, 0γ. Find the position vector function r(t).
Solution:
The position vector function r(t) is the integral of its derivativerβ²(t) = v(t):
r(t) =
β«v(t) dt
=
β«γsin t, cos 2t, etγ dt = γβ cos(t) + x0,
1
2sin(2t) + y0, e
t + z0γ.
Now use the initial position r(0) = γ1, 2, 0γ to solve for x0, y0, z0.
β cos(0) + x0 = β1 + x0 = 1 =β x0 = 2.
1
2sin(0) + y0 = 0 + y0 = 2 =β y0 = 2.
e0 + z0 = 1 + z0 = 0 =β z0 = β1.
Hence, r(t) = γβ cos(t) + 2,1
2sin(2t) + 2, et β 1.γ
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Problem 25(a) - Fall 2006
Let f (x , y) = ex2βy + x
β4β y2. Find partial derivatives fx , fy and
fxy .
Problem 25(b) - Fall 2006
Find an equation for the tangent plane of the graph of
f (x , y) = sin(2x + y) + 1
at the point (0, 0, 1).
Problem 26(a) - Fall 2006
Let g(x , y) = yex . Estimate g(0.1, 1.9) using the linearapproximation of g(x , y) at (x , y) = (0, 2).
Solutions to these problems:
These types of problems might not be on thisexam (check web site).
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Problem 26(b) - Fall 2006
Find the center and radius of the sphere x2 + y2 + z2 + 6z = 16.
Solution:
Complete the square in order to put the equation in the form:
(x β x0)2 + (y β y0) + (z β z0)2 = r2.We get:
x2 + y2 + (z2 + 6z) = x2 + y2 + (z2 + 6z + 9)β 9 = 16.
This gives the equation
(x β 0)2 + (y β 0)2 + (z + 3)2 = 25 = 52.
Hence, the center is C = (0, 0,β3) and the radius is r = 5.
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Problem 26(c) - Fall 2006
Let f (x , y) =β
16β x2 β y2. Draw a contour map of level curvesf (x , y) = k with k = 1, 2, 3. Label the level curves by thecorresponding values of k .
Solution:
A problem of this type might not be on this exam (check website).
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Problem 27Consider the line L through points A = (2, 1,β1) andB = (5, 3,β2). Find the intersection of the line L and the planegiven by 2x β 3y + 4z = 13.
Solution:
The vector part of L isββAB = γ3, 2,β1γ and the point A is on
the line.The vector equation of L is:
L = ~A+tββAB = γ2, 1,β1γ+tγ3, 2,β1γ = γ2+3t, 1+2t,β1βtγ.
Plugging x = 2 + 3t, y = 1 + 2t and z = β1β t into theequation of the plane gives:
2(2 + 3t)β 3(1 + 2t) + 4(β1β t) = β4t β 3 = 13
=β β4t = 16 =β t = β4.
So, the point of intersection is:
L(β4) = γ2β 12, 1β 8,β1β (β4)γ = γβ10,β7, 3γ.
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Problem 28(a)
Two masses travel through space along space curve described bythe two vector functions
r1(t) = γt, 1β t, 3 + t2γ, r2(s) = γ3β s, s β 2, s2γ
where t and s are two independent real parameters.Show that the two space curves intersect by finding the point ofintersection and the parameter values where this occurs.
Solution:
Equate the x and z-coordinates:
x = t = 3β s
z = 3 + t2 = 3 + (3β s)2 = 3 + 9β 6s + s2 = s2
Thus, the parameter values are:
12β 6s = 0 =β (s = 2 and t = 1).So, r1(1) = γ1, 0, 4γ = r2(2) is the desired intersection point.
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Problem 28(b)
Two masses travel through space along space curve described bythe two vector functions
r1(t) = γt, 1β t, 3 + t2γ, r2(s) = γ3β s, s β 2, s2γwhere t and s are two independent real parameters.Find parametric equation for the tangent line to the space curver1(t) at the intersection point. (Use the value t = 1 in part (a)).
Solution:The velocity vector of r1(t) at the intersection point is rβ²1(1).Since rβ²1(t) = γ1,β1, 2tγ,
rβ²1(1) = γ1,β1, 2γ.The vector equation of the tangent line is:T(t) = r1(1)+tγ1,β1, 2γ = γ1, 0, 4γ+tγ1,β1, 2γ = γ1+t,βt, 4+2tγ.The parametric equations are:
x = 1 + ty = βtz = 4 + 2t
.
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Problem 29
Consider the parallelogram with vertices A,B,C ,D such that Band C are adjacent to A. If A = (2, 5, 1), B = (3, 1, 4),D = (5, 2,β3), find the point C .
Solution:
After drawing a picture, the point C is easily seen to be:
ββOA +
ββBD = γ2, 5, 1γ+ γ2, 1,β7γ = γ4, 6,β6γ,
where O is the origin.
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Problem 30(a)
Consider the points A = (2, 1, 0), B = (1, 0, 2) and C = (0, 2, 1).
Find the orthogonal projection proj ~AB(ββAC ) of the vector
ββAC
onto the vectorββAB.
Solution:
We just plug in the vectors a =ββAB = γβ1,β1, 2γ and
b =ββAC = γβ2, 1, 1γ into the formula:
projab =a Β· ba Β· a
a.
Plugging in, we get:
projββAB
(ββAC ) =
γβ1,β1, 2γ Β· γβ2, 1, 1γγβ1,β1, 2γ Β· γβ1,β1, 2γ
γβ1,β1, 2γ =1
2γβ1,β1, 2γ.
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Problem 30(b)
Consider the points A = (2, 1, 0), B = (1, 0, 2) and C = (0, 2, 1).Find the area of triangle ABC .
Solution:
Consider the points A = (2, 1, 0), B = (1, 0, 2) andC = (0, 2, 1).
Then the area of the triangle β with these vertices can be
found by taking the area of the parallelogram spanned byββAB
andββAC and dividing by 2.
Thus:
Area(β) =|ββAB Γ
ββAC |
2=
1
2
β£β£β£β£β£β£β£β£β£β£β£β£
i j kβ1 β1 2β2 1 1
β£β£β£β£β£β£β£β£β£β£β£β£
=1
2|γβ3,β3,β3γ| =
1
2
β9 + 9 + 9 =
1
2
β27.
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Problem 30(c)
Consider the points A = (2, 1, 0), B = (1, 0, 2) and C = (0, 2, 1).Find the distance d from the point C to the line L that containspoints A and B.
Solution:
From the figure drawn on the blackboard, we see that thedistance d from C to L is the absolute value of the scalar
projection ofββAC in the direction
v =ββAC β projββ
AB
ββAC .
The vector v lies in the plane containing A,B,C and is
perpendicular toββAB.
Hence,
d = |v| =
β|ββAC |2 β |projββ
AB
ββAC |2.
Next, you the student, do the algebraic calculation of d.
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Problem 31
Find parametric equations for the line L of intersection of the planesx β 2y + z = 1 and 2x + y + z = 1.
Solution:
The vector part v of the line L of intersection is orthogonal to thenormal vectors γ1,β2, 1γ and γ2, 1, 1γ. Hence v can be taken to be:
v = γ1,β2, 1γ Γ γ2, 1, 1γ =
β£β£β£β£β£β£i j k1 β2 12 1 1
β£β£β£β£β£β£ = β3i + j + 5k.
Choose P β L so the z-coordinate of P is zero. Setting z = 0, weget: x β 2y = 1
2x + y = 1.
Solving, we find that x = 35 and y = β 1
5 . Hence, P = γ 35 ,β15 , 0γ lies
on the line L.The parametric equations are:
x = 35 β 3t
y = β 15 + t
z = 5t.
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Problem 32Let L1 denote the line through the points (1, 0, 1) and (β1, 4, 1)and let L2 denote the line through the points (2, 3,β1) and(4, 4,β3). Do the lines L1 and L2 intersect? If not, are they skewor parallel?
Solution:
The vector equations of the lines are:L1(t) = γ1, 0, 1γ+ tγβ2, 4, 0γ = γ1β 2t, 4t, 1γ
L2(s) = γ2, 3,β1γ+ sγ2, 1,β2γ = γ2 + 2s, 3 + s,β1β 2sγ
Equating z-coordinates, we find 1 = β1β 2s =β s = β1.
Equating y -coordinates with s = β1, we find4t = 3β 1 =β t = 1
2 .
Equating x-coordinates with s = β1 and t = 12 , we find:
L1(1
2) = γ0, 2, 1γ = L2(β1).
Hence, the lines intersect.
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Problem 33(a)
Find the volume V of the parallelepiped such that the followingfour points A = (1, 4, 2), B = (3, 1,β2), C = (4, 3,β3),D = (1, 0,β1) are vertices and the vertices B,C ,D are alladjacent to the vertex A.
Solution:
The volume V is equal to the absolute value of the determinant of
the matrix with rowsββAB = γ2,β3,β4γ,
ββAC = γ3,β1,β5γ,
ββAD = γ0,β4,β3γ.
V =2 β3 β43 β1 β50 β4 β3
= |2 Β· (β17) +β(β3) Β· (β9) + (β4) Β· (β12)| = | β 13| = 13.
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Problem 33(b)
Find an equation of the plane throughA = (1, 4, 2),B = (3, 1,β2),C = (4, 3,β3).
Solution:
Consider the vectorsββAB = γ2,β3,β4γ and
ββAC = γ3,β1,β5γ
which lie parallel to the plane.
The normal vector is:
n =ββAB Γ
ββAC =
β£β£β£β£β£β£i j k2 β3 β43 β1 β5
β£β£β£β£β£β£ = 11iβ 2j + 7k.
Since A = (1, 4, 2), is on the plane, then the equation of theplane is given by:
11(x β 1)β 2(y β 4) + 7(z β 2) = 0.
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Problem 33(c)
Find the angle between the plane throughA = (1, 4, 2), B = (3, 1,β2), C = (4, 3β 3) and the xy -plane.
Solution:The normal vectors of these planes are n1 = γ0, 0, 1γ,n2 = γ11,β2, 7γ.If ΞΈ is the angle between the planes, then:
cos ΞΈ =n1 Β· n2
|n1||n2|=
7β112 + (β2)2 + 72
=7β174
.
ΞΈ = cosβ1(
1β174
).
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Problem 34(a)
The velocity vector of a particle moving in space equalsv(t) = 2ti + 2t1/2j + k at any time t β₯ 0. At the time t = 0 thisparticle is at the point (β1, 5, 4). Find the position vector r(t) ofthe particle at the time t = 4.
Solution:
To find the position r(t), integrate the velocity vector fieldrβ²(t) = v(t).
r(t) =
β«v(t) dt =
β«γ2t, 2t
12 , 1γ dt
= γt2 + x0,4
3t32 + y0, t + z0γ.
Now use the initial position r(0) = γβ1, 5, 4γ to findx0 = β1; y0 = 5; z0 = 4.Thus, r(t) = γt2 β 1,
4
3t32 + 5, t + 4γ
r(4) = γ15,32
3+ 5, 8γ.
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Problem 34(b)
The velocity vector of a particle moving in space equalsv(t) = 2ti + 2t1/2j + k at any time t β₯ 0.Find an equation of the tangent line T to the curve at the timet = 4.
Solution:
Vector equation of the tangent line T to r(t) at t = 4 is:
T(s) = r(4) + srβ²(4) = r(4) + sv(4).
By part (a), r(4) = γ15, 323 + 5, 8γ.Since
v(4) = 8i + 4j + k = γ8, 4, 1γ,
thenT(s) = γ15,
32
3+ 5, 8γ+ sγ8, 4, 1γ.
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Problem 34(c)
The velocity vector of a particle moving in space equalsv(t) = 2ti + 2t1/2j + k at any time t β₯ 0.Does the particle ever pass through the point P = (80, 41, 13) ?
Solution:
From part (a), we have
r(t) = γt2 β 1,4
3t32 + 5, t + 4γ.
If r(t) = γ80, 41, 13γ, then t + 4 = 13 =β t = 9.
Hence the pointr(9) = γ80, 41, 13γ
is on the curve r(t).
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Problem 34(d)
The velocity vector of a particle moving in space equalsv(t) = 2ti + 2t1/2j + k at any time t β₯ 0.Find the length of the arc traveled from time t = 1 to time t = 2.
Solution:
Length =
β« 2
1|v(t)| dt =
β« 2
1
β4t2 + 4t + 1 dt.
Since we are not using calculators on our exam, then this is thefinal answer.
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Problem 35(a)
Consider the surface x2 + 3y2 β 2z2 = 1.What are the traces in x = k , y = k , z = k? Sketch a few.
Solution:
For x = k 6= 1, we get the hyperbolas 3y2 β 2z2 = k .
For x = 1, we get the 2 lines y = Β±32z .
For z = 0, we get the ellipse x2 + 3y2 = 1.
For z = 1, we get the ellipse x2 + 3y2 = 3.
I am leaving it to you to do the sketches!
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Problem 35(b)
Consider the surface x2 + 3y2 β 2z2 = 1.Sketch the surface in the space.
Solution:
Sorry, you need to do the sketch.
Problem 36
Find an equation for the tangent plane to the graph off (x , y) = y ln x at (1, 4, 0).
Solution:
A problem of this type might not be on this exam (check website).
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Problem 37
Find the distance D between the given parallel planes
z = 2x + y β 1, β4x β 2y + 2z = 3.
Solution:
The normal to the first plane is n = γ2, 1,β1γ and the pointP0 = (0, 0,β1) lies on this plane. The point P1 = γ0, 0, 32γ lies onthe second plane. Consider the vector from P0 to P1 which isb = γ0, 0, 52γ. The distance D from P1 to the first plane is equalto:|compn b| =
β£β£β£β£b Β· n
|n|
β£β£β£β£ = |γ0, 0, 5
2γ Β· 1β
6γ2, 1,β1γ| =
5
2β
6.
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Problem 38
Identify the surface given by the equation4x2 + 4y2 β 8y β z2 = 0. Draw the traces and sketch the curve.
Solution:
Sorry, no sketch given.
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Problem 39(a)
A projectile is fired from a point 5 m above the ground at an angleof 30 degrees and an initial speed of 100 m/s.Write an equation for the acceleration vector.
Solution:
Since the force due to gravity acts downward, we have
F = ma = βmg j,where g = |a| β 9.8 m/s2. Thus a = βg j.
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Problem 39(b) and 34(c)
A projectile is fired from a point 5 m above the ground at an angleof 30 degrees and an initial speed of 100 m/s.(b) Write a vector for initial velocity v(0).(c) Write a vector for the initial position r(0)
Solution:
Initial velocity is:
v(0) = 100(cos 30β¦i + sin 30β¦j) = 50β
3i + 50j,
in units of m/s.
The initial position is:
r(0) = 5j,
in units of meters m.
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Problem 39(d)
A projectile is fired from a point 5 m above the ground at an angleof 30 degrees and an initial speed of 100 m/s.At what time does the projectile hit the ground?
Solution:
We first find the velocity r(t) and position r(t) functions.rβ²(t) = v(t) = βgtj + v(0)
r(t) = β1
2gt2j + tv(0) + D.
Since D = r(0) = 5j, then r(t) = β12gt
2j + tv(0) + 5j.Hence,
r(t) = 50β
3ti + [50t β 1
2gt2 + 5]j.
The projectile hits the ground when 50t β 12gt
2 + 5 = 0.Applying the quadratic formula, we find
t =100 +
β1002 + 40g
2g.
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Problem 39(e)
A projectile is fired from a point 5 m above the ground at an angleof 30 degrees and an initial speed of 100 m/s.How far did it travel, horizontally, before it hit the ground?
Solution:
Recall r(t) = 50β
3ti + [50t β 12gt
2 + 5]j and the projectile
hits the ground when t =100+β
1002+40g2g .
The horizontal distance d traveled is the value of the
x-coordinate of r(t) at t =100+β
1002+40g2g :
d = 50β
3
(100 +
β1002 + 40g
2g
).
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Problem 40
Explain why the limit of f (x , y) = (3x2y2)/(2x4 + y4) does notexist as (x , y) approaches (0, 0).
Solution:
A problem of this type might not be on this exam (check website).
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Problem 41
Find an equation of the plane that passes through the pointP(1, 1, 0) and contains the line given by parametric equationsx = 2 + 3t, y = 1β t, z = 2 + 2t.
Solution:The direction vector a = γ3,β1, 2γ of the line is parallel to theplane.
For t = 0, the point Q = γ2, 1, 2γ on the line and the plane.
So b =ββPQ = γ1, 0, 2γ is also parallel to the plane.
To find a normal vector to the plane, take cross products:
n = aΓ b =i j k3 β1 21 0 2
= γβ2,β4, 1γ.
Since (1, 1, 0) is on the plane, the equation of the plane is:
γβ2,β4, 1γ Β· γx β 1, y β 1, zγ = β2(x β 1)β 4(y β 1) + z = 0.
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Problem 42(a)
Find all of the first order and second order partial derivatives of thefunction. f (x , y) = x3 β xy2 + y
Solution:
There is no problem of this type on this exam.
Problem 42(b)
Find all of the first order and second order partial derivatives of the
function. f (x , y) = ln(x +βx2 + y2)
Solution:
There is no problem of this type on this exam.
Problem 43
Find the linear approximation of the function f (x , y) = xyex at(x , y) = (1, 1), and use it to estimate f (1.1, 0.9).
Solution:
There is no problem of this type on this exam.
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Problem 44
Find a vector function r(t) which represents the curve ofintersection of the paraboloid z = 2x2 + y2 and the paraboliccylinder y = x2.
Solution:
Set t = x .
Since y = x2 = t2, we get from the equation of theparaboloid a vector function r(t) which represents the curveof intersection:
r(t) = γt, t2, 2t2 + (t2)2γ = γt, t2, 2t2 + t4γ.
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Problem 1(a) - Spring 2009
Given a = γ3, 6,β2γ, b = γ1, 2, 3γ.Write down the vector projection of b along a. (Hint: Useprojections.)
Solution:
We have |a| =β
9 + 36 + 4 =β
49 = 7.
Then
n =a
|a|=
1
7a = unit vector parallel to a.
So,projab = (b Β· n)n =
b Β· a|a|2
a =
1
49γ1, 2, 3γ Β· γ3, 6,β2γγ3, 6,β2γ =
9
49γ3, 6,β2γ.
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Problem 45(b) - Spring 2009
Given a = γ3, 6,β2γ, b = γ1, 2, 3γ.Write b as a sum of a vector parallel to a and a vector orthogonalto a. (Hint: Use projections.)
Solution:
We have
b = γ1, 2, 3γ = γ1, 2, 3γ β 9
49γ3, 6,β2γ+
9
49γ3, 6,β2γ
=1
49γ22, 44, 165γ+
9
49γ3, 6,β2γ.
Here 9
49γ3, 6,β2γ parallel to a = γ3, 6,β2γ
and1
49γ22, 44, 165γ orthogonal to a = γ3, 6,β2γ.
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Problem 45(b) Continuation - Spring 2009
Given a = γ3, 6,β2γ, b = γ1, 2, 3γ.Write b as a sum of a vector parallel to a and a vector orthogonalto a. (Hint: Use projections.)
Solution:
Why so? All we did was to write
b = bβ (b Β· n)n + (b Β· n)n
where n = a7 , n2 = 1.
Of course this is the same as
b = (bβ projab) + projab.
That is, we write b as projab plus βthe restβ. But βthe restβis orthogonal to n (and to a), since
(bβ (b Β· n)n) Β· n = b Β· nβ (b Β· n)(n Β· n) = 0, as n Β· n = 1.
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Problem 45(c) - Spring 2009
Given a = γ3, 6,β2γ, b = γ1, 2, 3γ.Let ΞΈ be the angle between a and b. Find cos ΞΈ.
Solution:
cos(ΞΈ) =a Β· b|a||b|
=γ3, 6,β2γ Β· γ1, 2, 3γ|γ3, 6,β2γ||γ1, 2, 3γ|
=9β
49β
14=
9
7β
14.
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Problem 46(a) - Spring 2009
Given A = (β1, 7, 5), B = (3, 2, 2) and C = (1, 2, 3).Let L be the line which passes through the points A = (β1, 7, 5)and B = (3, 2, 2). Find the parametric equations for L.
Solution:
To get parametric equations for L you need a point throughwhich the line passes and a vector parallel to the line. For
example, take the point to be A and the vector to beββAB.
The vector equation of L is
r(t) =ββOA+t
ββAB = γβ1, 7, 5γ+t γ4,β5,β3γ = γβ1 + 4t, 7β 5t, 5β 3tγ ,
where O is the origin.
The parametric equations are:β£β£β£β£β£β£x = β1 + 4ty = 7β 5t, t β R.z = 5β 3t
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Problem 46(b) - Spring 2009
Given A = (β1, 7, 5), B = (3, 2, 2) and C = (1, 2, 3).A, B and C are three of the four vertices of a parallelogram, whileCA and CB are two of the four edges. Find the fourth vertex.
Solution:
Denote the fourth vertex by D. Then
ββOD =
ββOA +
ββCB = γβ1, 7, 5γ+ γ2, 0,β1γ = γ1, 7, 4γ ,
where O is the origin. That is,
D = (1, 7, 4).
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Problem 47(a) - Spring 2009
Consider the points P(1, 3, 5), Q(β2, 1, 2), R(1, 1, 1) in R3.Find an equation for the plane containing P, Q and R.
Solution:
Since a plane is determined by its normal vector n and a point onit, say the point P, it suffices to find n. Note that:
n =ββPQ Γ
ββPR =
β£β£β£β£β£β£i j kβ3 β2 β30 β2 β4
β£β£β£β£β£β£ = γ2,β12, 6γ = 2γ1,β6, 3γ.
So the equation of the plane is:
(x β 1)β 6(y β 3) + 3(z β 5) = 0.
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Problem 47(b) - Spring 2009
Consider the points P(1, 3, 5), Q(β2, 1, 2), R(1, 1, 1) in R3.Find the area of the triangle with vertices P, Q, R.
Solution:
The area of the triangle β with vertices P, Q, R can be found by
taking the area of the parallelogram spanned byββPQ and
ββPR and
dividing it by 2. Thus, using a), we have:
Area(β) =|ββPQ Γ
ββPR|
2=
1
2|2γ1,β6, 3γ|
=β
1 + 36 + 9 =β
46.
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Problem 48 - Spring 2009
Find parametric equations for the line of intersection of the planesx + y + 3z = 1 and x β y + 2z = 0.
Solution:
A vector v parallel to the line is the cross product of thenormal vectors of the planes:
v = γ1, 1, 3γ Γ γ1,β1, 2γ =
β£β£β£β£β£β£i j k1 1 31 β1 2
β£β£β£β£β£β£ = γ5, 1,β2γ.
A point on L is any (x0, y0, z0) that satisfies the equations ofboth planes.
Setting z = 0, we obtain the equations x + y = 1 andx β y = 0 and find such a point (12 ,
12 , 0). Therefore
parametric equations for L are:β£β£β£β£β£β£x = 1
2 + 5ty = 1
2 + tz = β2t.
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Problem 49(a) - Spring 2009
Consider the parametrized curve
r(t) =β¨t, t2, t3
β©, t β R.
Set up an integral for the length of the arc between t = 0 andt = 1. Do not attempt to evaluate the integral.
Solution:
The velocity field is:
v(t) = rβ²(t) =β¨1, 2t, 3t2
β©.
Then the speed isβ£β£rβ²(t)β£β£ =
β1 + 4t2 + 9t4.
Therefore, the length of the arc is:
L =
β« 1
0
β1 + 4t2 + 9t4 dt.
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Problem 49(b) - Spring 2009
Consider the parametrized curve
r(t) =β¨t, t2, t3
β©, t β R.
Write down the parametric equations of tangent line to r(t) at(2, 4, 8).
Solution:
The parametrized curve passes through the point (2, 4, 8) ifand only if
t = 2, t2 = 4, t3 = 8ββ t = 2.The velocity vector field to the curve is given by
rβ²(t) =β¨
1, 2t, 3t2β©hence rβ²(2) = γ1, 4, 12γ .
The equation of the tangent line in question is:β£β£β£β£β£β£x = 2 + Οy = 4 + 4Ο, Ο β Rz = 8 + 12Ο
Caution: The parameter along the line, Ο , has nothing to do with
the parameter along the curve, t.
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Problem 50(a) - Spring 2009
Consider the sphere S in R3 given by the equation
x2 + y2 + z2 β 4x β 6z β 3 = 0.
Find its center C and its radius R.
Solution:
Completing the square we get
(x β 2)2 β 4 + y2 + (z β 3)2 β 9β 3 = 0
ββ
(x β 2)2 + y2 + (z β 3)2 = 16.
This gives:C = (2, 0, 3) R = 4
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Problem 50(b) - Spring 2009
What does the equation x2 + z2 = 4 describe in R3? Make asketch.
Solution:
This a (straight, circular) cylinder determined by the circle inthe xz-plane of radius 2 and center (0, 0) and parallel to they -axis.
![Page 101: 233 ReviewExam1 Print Sp 2013](https://reader030.vdocuments.site/reader030/viewer/2022020106/55cf912a550346f57b8b4a5d/html5/thumbnails/101.jpg)
Problem 51(a) - Spring 2009Jane throws a basketball at an angle of 45o to the horizontal at an initial speedof 12 m/s, where m denotes meters. It leaves her hand 2 m above the ground.Assume the acceleration of the ball due to gravity is downward with magnitude10 m/s2 and neglect air friction.
(a) Find the velocity function v(t) and the position function r(t) of the ball.
Use coordinates in the xy -plane to describe what is happening; assume Jane is
standing with her feet at the point (0, 0) and y represents the height.
Solution:
Acceleration due to gravity is a = γ0,βgγ = γ0,β10γ. Initialvelocity is v(0) = 12γcos
Ο
4, sin
Ο
4γ = γ6
β2, 6β
2γ.So the velocity function is
v(t) = v(0) +
β« t
0
adΟ = v(0) + at = γ6β
2, 6β
2β 10tγ.
One can recover the position by integrating the velocity:
r(t) =
β« t
0
v(Ο)dΟ + r(0).
Notice the initial position is r(0) = γ0, 2γ. This integral yields:
r(t) = r(0) + v(0)t + at2
2=β¨
6β
2t, 2 + 6β
2t β 5t2β©.
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Problem 51(b) - Spring 2009
Jane throws a basketball from the ground at an angle of 45o to thehorizontal at an initial speed of 12 m/s, where m denotes meters.It leaves her hand 2 m above the ground. Assume the accelerationof the ball due to gravity is downward with magnitude 10 m/s2
and neglect air friction.(b) Find the speed of the ball at its highest point.
Solution:
At the highest point, the vertical component of the velocity is zero,so we only need to calculate the horizontal component which is6β
2. Thus the speed at the highest point is 6β
2.
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Problem 51(c) - Spring 2009
Jane throws a basketball from the ground at an angle of 45o to thehorizontal at an initial speed of 12 m/s, where m denotes meters.It leaves her hand 2 m above the ground. Assume the accelerationof the ball due to gravity is downward with magnitude 10 m/s2
and neglect air friction.(c) At what time T does the ball reach its highest point.
Solution:
When the ball reaches its highest point, the vertical component ofits velocity is zero. That is,
6β
2β 10t = 0,
so T = 3β2
5 .
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Problem 57(a) - Spring 2010
Consider the parallelogram with vertices A, B, C , D such that Band C are adjacent to A where A = (1, 2,β1), B = (3, 5, 1) andD = (2,β1, 2). Find the area of the parallelogram.
Solution:
Recall that:
area =
β£β£β£β£ββAB Γ ββBDβ£β£β£β£ .Since
ββAB = γ2, 3, 2γ and
ββBD = γβ1,β6, 1γ,
area =
β£β£β£β£β£β£i j k2 3 2β1 β6 1
β£β£β£β£β£β£ = |γ15,β4,β9γ| =β
322.
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Problem 57(b) - Spring 2010
Consider the parallelogram with vertices A, B, C, D such that Band C are adjacent to A where A = (1, 2,β1), B = (3, 5, 1) andD = (2,β1, 2). Find the coordinates of the point C.
Solution:
ββOC =
ββOA +
ββBD = γ1, 2,β1γ+ γβ1,β6, 1γ = γ0,β4, 0γ ,
where O is the origin. That is,
C = (0,β4, 0).
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Problem 58(a) - Spring 2010
Consider the points A = (0, 3,β3), B = (β1, 3, 2),
C = (β1, 2,β3). Find the orthogonal projection projββAB
(ββAC ) of
the vectorββAC onto the vector
ββAB.
Solution:
We just plug in the vectors a =ββAB = γβ1, 0, 5γ and
b =ββAC = γβ1,β1, 0γ into the formula:
projab =a Β· ba Β· a
a.
Plugging in, we get:
projββAB
(ββAC ) =
γβ1, 0, 5γ Β· γβ1,β1, 0γγβ1, 0, 5γ Β· γβ1, 0, 5γ
γβ1, 0, 5γ =1
26γβ1, 0, 5γ.
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Problem 58(b) - Spring 2010
Consider the points A = (0, 3,β3), B = (β1, 3, 2),C = (β1, 2,β3). Find the distance d from the point C to the lineL that contains points A and B.
Solution (Method 1):
From the figure drawn on the blackboard, we see that thedistance d from C to L is the absolute value of the scalar
projection ofββAC in the direction
v =ββAC β projββ
AB
ββAC .
The vector v lies in the plane containing A,B,C and is
perpendicular toββAB.
Hence,d = |v|.
Next, you the student, do the algebraic calculation of d.
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Problem 58(b) - Spring 2010
Consider the points A = (0, 3,β3), B = (β1, 3, 2),C = (β1, 2,β3). Find the distance d from the point C to the lineL that contains points A and B.
Solution (Method 1):
From the figure drawn on the blackboard and by thePythagorean Theorem, we see that the distance d from C toL is equal to:
d =
β|ββAC |2 β |projββ
AB
ββAC |2.
Next, you the student, do the algebraic calculation of d.
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Problem 58(b) - Spring 2010
Consider the points A = (0, 3,β3), B = (β1, 3, 2),C = (β1, 2,β3). Find the distance d from the point C to the lineL that contains points A and B.
Solution (Method 2):
Let P be the point on the line L closest to the point C .
From the figure drawn on the blackboard, we see that thedistance d from C to L is equal to the length of the side PCof the right triangle with legs AP, PC and hypotenuse AC .
Then, using the rule |aΓ b| = |a||b| sin(ΞΈ),
d = |ββPC | = |
ββAC | sin(ΞΈ) =
|ββAC Γ
ββAB|
|ββAB|
.
Next, you the student, do the algebraic calculation of d.
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Problem 59(a) - Spring 2010
Let P1 be the plane x + 3y + z = 0 and P2 be the plane2x + y β z = 1. Find the cosine of the angle between the planes.
Solution:
Note that the normals of the planes are:
n1 = γ1, 3, 1γ n2 = γ2, 1,β1γ.
By the formula for dot products of 2 vectors:
cos ΞΈ =n1 Β· n2
|n1||n2|=γ1, 3, 1γ Β· γ2, 1,β1γβ
11β
6=
4β11β
6.
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Problem 59(b) - Spring 2010
Let P1 be the plane x + 3y + z = 0 and P2 be the plane 2x + y β z = 1.Find the parametric equations of the line of intersection between the 2planes P1 and P2.
Solution:
The vector part v of the line L of intersection is orthogonal to thenormal vectors γ1, 3, 1γ and γ2, 1,β1γ. Hence v can be taken to be:
v = γ1, 3, 1γ Γ γ2, 1,β1γ =
β£β£β£β£β£β£i j k1 3 12 1 β1
β£β£β£β£β£β£ = β4i + 3jβ 5k.
Choose P β L so the z-coordinate of P is zero. Setting z = 0, weobtain: x + 3y = 0
2x + y = 1.
Solving, we find that x = 35 and y = β 1
5 . Hence, P = γ 35 ,β15 , 0γ lies
on the line L.
The parametric equations are:
x = 35 β 4t
y = β 15 + 3t
z = 0β 5t = β5t.
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Problem 59(c) - Spring 2010
Let P1 be the plane x + 3y + z = 0 and P2 be the plane2x + y β z = 1. Find the distance from the plane P2 to the origin.
Solution:
Recall the distance formula D =|ax1 + by1 + cz1 + d |β
a2 + b2 + c2from a
point P = (x1, y1, z1) to a plane ax + by + cz + d = 0.
In order to apply the formula, rewrite the equation of theplane in standard form: 2x + y β z β 1 = 0.
So, the distance from the origin to the plane is:
D =|(2 Β· 0) + (1 Β· 0) + (β1 Β· 0)β 1|β
22 + 12 + (β1)2=| β 1|β
6=
1β6
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Problem 60(a) - Spring 2010
Let r(t) = cos(2t)i + sin(2t)j + tk. What is the length L of thecurve starting at t = 0 and ending at t = 5.
Solution:
Recall that the length of r(t) on the interval [0, 5] is gotten byintegrating the speed |rβ²(t)|.Calculating, we get:
L =
β« 5
0|rβ²(t)| dt =
β« 5
0|γβ2 sin 2t, 2 cos 2t, 1γ | dt
=
β« 5
0
β4 sin2 2t + 4 cos2 2t + 1 dt =
β« 5
0
β5 dt
=β
5tβ£β£β£50
= 5β
5.
ThusL = 5
β5
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Problem 60(b) - Spring 2010
Let r(t) = cos(2t)i + sin(2t)j + tk. Find an equation for thetangent line to the graph at the point given by t = 0.
Solution:The parametrized curve passes through the pointr(0) = (1, 0, 0)
The velocity vector field to the curve is given by
rβ²(t) = γ β2 sin(2t), 2 cos(2t), 1γ hence rβ²(0) = γ0, 2, 1γ .The equation of the tangent line in question is:
x = 1y = 2Ο, Ο β R.z = Ο
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Problem 61 - Spring 2010
Show that the limit lim(x ,y)β(0,0)x2βy2
x2+y2 does not exist.
Solution:
Let f (x , y) = x2βy2
x2+y2 .
Along the line γt, tγ, t 6= 0, f (x , y) has the value 02t2
= 0.
Along the line γ0, tγ, t 6= 0, f (x , y) has the value t2
t2= 1.
Since f (x , y) has 2 different limiting values at (0, 0), it doesnot have a limit at (0, 0).
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Problem 62(a) - Spring 2010
Consider the sphere S in R3 given by the equation
x2 + y2 + z2 β 2x β 8y β 2 = 0.
Find the coordinates of its center and its radius.
Solution:
Completing the square we get
x2β2x+y2β8y+z2 = (x2β2x+1)β1+(y2+8y+16)β16+(z2)
= (x β 1)2 β 1 + (y + 2)2 β 16 + z2 = 2ββ
(x β 2)2 + (y + 2)2 + z2 = 19.
This gives:
Center = (1, 4, 0) Radius =β
19
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Problem 62(b) - Spring 2010
What does the equation x2 + y2 = 64 describe in R3. Make asketch.
Solution:
This equation describes a circular cylinder of radius 8 centeredalong the z-axis. Here is a sketch of a cylinder of radius 1 centered
along the z-axis.
![Page 118: 233 ReviewExam1 Print Sp 2013](https://reader030.vdocuments.site/reader030/viewer/2022020106/55cf912a550346f57b8b4a5d/html5/thumbnails/118.jpg)
Problem 63(a) - Spring 2012
Given A = (β1, 7, 5), B = (3, 0, 2) and C = (1, 2, 3).Let L be the line which passes through the points A = (β1, 7, 5)and B = (3, 0, 2). Find the parametric equations for L.
Solution:
To get parametric equations for L you need a point throughwhich the line passes and a vector parallel to the line. For
example, take the point to be A and the vector to beββAB.
The vector equation of L is
r(t) =ββOA+t
ββAB = γβ1, 7, 5γ+t γ4,β7,β3γ = γβ1 + 4t, 7β 7t, 5β 3tγ ,
where O is the origin.
The parametric equations are:β£β£β£β£β£β£x = β1 + 4ty = 7β 7t, t β R.z = 5β 3t
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Problem 63(b) - Spring 2012
Given A = (β1, 7, 5), B = (3, 0, 2) and C = (1, 2, 3).A, B and C are three of the four vertices of a parallelogram, whileAB and BC are two of the four edges. Find the fourth vertex.
Solution:
Denote the fourth vertex by D.
Then
ββOD =
ββOA +
ββBC = γβ1, 7, 5γ+ γβ2, 2, 1γ = γβ3, 9, 6γ ,
where O is the origin.
That is,D = (β3, 9, 6).
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Problem 64(a) - Spring 2012
Consider the points P(1, 1, 1), Q(β2, 1, 2), R(1, 3, 5) in R3.Find an equation for the plane containing P, Q and R.
Solution:
Since a plane is determined by its normal vector n and a pointon it, say the point P, it suffices to find n.
Note that:
n =ββPQ Γ
ββPR =
β£β£β£β£β£β£i j kβ3 0 10 2 4
β£β£β£β£β£β£ = γβ2, 12,β6γ.
So the equation of the plane is:
β2(x β 1) + 12(y β 1)β 6(z β 1) = 0.
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Problem 64(b) - Spring 2012
Consider the points P(1, 1, 1), Q(β2, 1, 2), R(1, 3, 5) in R3.Find the area of the triangle with vertices P, Q, R.
Solution:
The area of the triangle β with vertices P, Q, R can be
found by taking the area of the parallelogram spanned byββPQ
andββPR and dividing it by 2.
Thus, using part a), we have:
Area(β) =|ββPQ Γ
ββPR|
2=
1
2|γβ2, 12,β6γ|
=1
2
β4 + 144 + 36 =
1
2
β184.
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Problem 65 - Spring 2010
Let P1 be the plane x β 2y + 2z = 10 and P2 be the plane2x + y + 2z = 0. Find the cosine of the angle between the planes.
Solution:
Note that the normals of the planes are:
n1 = γ1,β2, 2γ n2 = γ2, 1, 2γ.
By the formula for dot products of 2 vectors:
cos ΞΈ =n1 Β· n2
|n1||n2|=γ1,β2, 2γ Β· γ2, 1, 2γβ
9β
9=
4
9.
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Problem 66(a) - Spring 2012
Find the distance d from the point Q = (1, 6,β1) to the plane2x + y β 2z = 19.
Solution:
Method 1
The distance from the point (x1, y1, z1) to the planeAx + By + Cz + D = 0 is:
d =|Ax1 + By1 + Cz1 + D|β
A2 + B2 + C 2.
In our case this gives
d =|2 + 6 + 2β 19|β
9=
9β9
= 3
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Problem 66(a) - Spring 2012
Find the distance d from the point Q = (1, 6,β1) to the plane2x + y β 2z = 19.
Solution:
Method 2
Note that P = (0, 19, 0) is on the plane.
The distance between Q and the plane equals the length of the
component of b =ββPQ which is orthogonal to the plane, i.e., which
is parallel to the normal vector to the plane.
If the normal vector of the plane is n, then this is distance is the
absolute value of the scalar projection: d =β£β£β£b Β· n
|n|
β£β£β£ .Here b =
ββPQ = γ1,β13,β1γ and n = γ2, 1,β2γ with
|n| =β
22 + 1 + (β2)2 = 3.
Then
d =| γ1,β13,β1γ Β· γ2, 1,β2γ |β
9=|2β 13 + 2|
3= 3.
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Problem 66(a) - Spring 2012
Find the distance d from the point (1, 6,β1) to the plane2x + y β 2z = 19.
Solution:
Method 3 Write the equation of the line through P(1, 6,β1) whichis orthogonal to the plane. Find the point Q where it intersects theplane. Find the distance d between P and Q from the distanceformula.The line in question passes through (1, 6,β1) and is parallel ton = γ2, 1,β2γ, so it has parametric equationsβ£β£β£β£β£β£
x = 1 + 2ty = 6 + t, t β P.z = β1β 2t
It intersects the plane 2x + y β 2z = 19 if and only if
2(1 + 2t) + (6 + t)β 2(β1β 2t) = 19β 9t = 9β t = 1.
Substituting t = 1 above gives the point Q(3, 7,β3).
So the distance between P and Q isd =
β(3β 1)2 + (7β 6)2 + (β3 + 1)2 =
β4 + 1 + 4 = 3.
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Problem 66(b) - Spring 2012
Write the parametric equations of the line L containing the pointT (1, 2, 3) and perpendicular to the plane 2x + y β 2z = 19.
Solution:
The line L passes through (1, 2, 3) and is parallel ton = γ2, 1,β2γ.So, L has parametric equations:
x = 1 + 2ty = 2 + t, t β R.z = 3β 2t
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Problem 66(c) - Spring 2012
Find the point of intersection of the line L in part b) with theplane 2x + y β 2z = 19.
Solution:
The line L has parametric equations:
x = 1 + 2ty = 2 + t, t β R.z = 3β 2t
So, L intersects the plane 2x + y β 2z = 19 if and only if
2(1 + 2t) + (2 + t)β 2(3β 2t) = 19ββ 9t = 21ββ t =7
3.
Substituting t = 73 in the parametric equations of L gives
the point Q = (173 ,133 ,β
53).
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Problem 67 - Spring 2012
Find the equation of the sphere with center the point (1, 2, 3) andwhich contains the point (3, 1, 5).
Solution:
The distance from P = (1, 2, 3) to Q = (3, 1, 5) is the length
of the vectorββPQ which is 3.
So the radius of the sphere is r = 3.
Hence the equation of the sphere is:
(x β 1)2 + (y β 2)2 + (z β 3)2 = 32 = 9
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Problem 68 - Spring 2012
Make a sketch of the surface in R3 described by the equationy2 + z2 = 36. In your sketch of the surface, include the labeledcoordinate axes and the trace curves on the surface for the planesx = 0 and x = 4.
Solution:
This equation describes a circular cylinder of radius 6 centeredalong the x-axis. Below is a sketch of a cylinder of radius 1centered along the z-axis. Th red curve should be removed.
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Problem 69 - Spring 2012
Find the equation of the plane which contains the points A(1, 2, 3) andB(1, 0, 4) and which is also perpendicular to the plane 4x β 2y + z = 8.
Solution:
Since a plane is determined by its normal vector n and a point on it,say the point A, it suffices to find n.
Note that the normal γ4,β2, 1γ to the plane 4x β 2y + z = 8 and
the vectorββAB = γ0,β2, 1γ are both perpendicular to the the normal
vector n that we want.
It follows that:
n =ββAB Γ γ4,β2, 1γ =
β£β£β£β£β£β£i j k0 β2 14 β2 1
β£β£β£β£β£β£ = γ0, 4, 8γ.
So the equation of the plane is:
0(x β 1) + 4(y β 2) + 8(z β 3) = 4(y β 2) + 8(z β 3) = 0.
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Problem 70(a) - Spring 2012
Suppose that a = γ2, 1, 2γ and b = γ8, 2, 0γ. Find the vector
projection, call it c, of b in the direction ofββa .
Solution:
We just plug in the vectors a = γ2, 1, 2γ and b = γ8, 2, 0γ intothe formula:
projab =a Β· ba Β· a
a.
Plugging in, we get:
projab =γ2, 1, 2γ Β· γ8, 2, 0γγ2, 1, 2γ Β· γ2, 1, 2γ
γ2, 1, 2γ = γ4, 2, 4γ.
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Problem 70(b) - Spring 2012
Suppose that a = γ2, 1, 2γ and b = γ8, 2, 0γ andc = projab = aΒ·b
aΒ·a a. Calculate the vector bβ c and then show thatit is orthogonal to a.
Solution:
By the previous problem, c = γ4, 2, 4γ.Calculating, we get
bβ c = γ8, 2, 0γ β γ4, 2, 4γ = γ4, 0,β4γ.
Sinceγ4, 0,β4γ Β· γ2, 1, 2γ = 0,
the vector bβ c is orthogonal to a.