ma 351, spring 2019, midterm 2 preview solutions
TRANSCRIPT
MA 351, Spring 2019, Midterm 2 preview solutions:
1. (Section 14.1) Find and sketch the domain of the function
f(x, y) =
√y − x2
1− x2.
Solution. We needy − x2 ≥ 0 and 1− x2 6= 0
Basically this means y ≥ x2 and x 6= ±1
2. (Section 14.1) Match the following two functions with their graphs and contour map.
(a) z = sin(xy)
(b) z =x− y
1 + x2 + y2
Solution. z = sin(xy) should have normal looking sine curves along both the x andy axes. This eliminates all of the graphs except the one if the upper left. (It may behard to see how it eliminates the one in the lower right: this graph is only a normalsine curve along one axis: the y = −x axis. But along the y = x axis it’s constant.This is probably the graph of sin(x− y).)
This surface has contour curves given by the lower left graph. To see this, look justat the red part in the surface, or better yet, the very top, horizontal part of the red.You should see that we have hyperbolas. This is what the contour in the lower leftshows.
z =x− y
1/x2 + y2should have asymptotic behavior. As both x and y increase, the
bottom of this fraction dominates, and the whole thing goes towards 0. There’s onlyone surface that goes towards 0 in all directions except near the origin: the one inthe upper right corner.
This surface has contour curves given by the upper right graph. To see this, notethat a “mountain” should have positive contour circles that get closer together as weget closer to the peak. Similarly, a “round valley” or closed hole should have negativecontour circles that get closer together.
−3−2
−10
12
3
−3
−2
−1
0
1
2
3
−4
−2
0
2
4
x
sin(x y)
y
−2
−1
0
1
−10−5
05
10−4
−3
−2
−1
0
1
2
3
4
x
exp(x) cos(y)
y
−5
0
5
−8−6−4−202468
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
x
(x−y)/(1+x2+y
2)
y
−5
0
5
−5
0
5−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
x
sin(x−y)
y
−0.49985
−0.49985
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−0.5
3012
−0.35341
−0.35341
−0.17671
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7671
−0.17671−0.1
7671
0
0
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0
0.17671
0.1
7671
0.17671 0.17671
0.35341
0.35341
0.5
3012
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3. (Section 14.2) UPDATE: changed problem slightly. Find the following limit orshow that it does not exist:
lim〈x,y〉→〈0,0〉
x+ y√x2 + y2
.
Solution. We start by trying to “plug in” 〈0, 0〉.
x+ y√x2 + y2
∣∣∣x=0,y=0
=0
0
= indeterminate
Now we try taking different paths that go towards the origin:
y = x lim〈x, y〉 → 〈0, 0〉
y = x
x+ y√x2 + y2
= limx→0
2x√x2 + x2
= limx→0
2x√2x2
= limx→0
1√2
=1√2
y = −x lim〈x, y〉 → 〈0, 0〉
y = −x
x+ y√x2 + y2
= limx→0
0√x2 + x2
= limx→0
0
= 0
Since these limits are different, the original limit does not exist.
3. (Section 14.2) UPDATE: original problem, corrected solution. Show that thefollowing limit exists,
lim〈x,y〉→〈0,0〉
xy√x2 + y2
= 0.
Hint: Break it into cases (which actually end up being the same since x and y aresymmetric.) Case I: Suppose −x ≤ y ≤ x and x ≥ 0. Show that
−x ≤ −x2√x2 + y2
≤ xy√x2 + y2
≤ x2√x2 + y2
≤ x
and apply the Squeeze Theorem.
Solution. Note that “plugging in” 〈0, 0〉 doesn’t work
xy√x2 + y2
∣∣∣x=0,y=0
=0
0
= indeterminate
There are four cases where (x, y) can be in the (x, y)-plane, shown the four regionsshaded in below:
x
y
case 1
case 2
case 3
case 4
Case 1: Suppose −x ≤ y ≤ x and x ≥ 0. Then
−x ≤ y ≤ x−x2 ≤ xy ≤ x2 (mult. prev. ineq. by x)
−x2√x2 + y2
≤ xy√x2 + y2
≤ x2√x2 + y2
(div. prev. line by√x2 + y2)
−x2√x2 + 0
≤ −x2√x2 + y2
andx2√x2 + y2
≤ x2√x2 + 0
(smaller denom. means larger fraction)
−x2√x2 + 0
≤ −x2√x2 + y2
≤ xy√x2 + y2
≤ x2√x2 + y2
≤ x2√x2 + 0
(combine prev. two lines)
−x2√x2≤ xy√
x2 + y2≤ x2√
x2(simplify prev. line)
−x ≤ xy√x2 + y2
≤ x (simplify prev. line)
limx→0−x = 0 lim
x→0x = 0
lim〈x,y〉→〈0,0〉
xy√x2 + y2
= 0 by Squeeze Thm. (in Case 1)
All four cases have basically the same algebraic argument, so in all four cases we findthe limit equals 0.
4. (Section 14.2) UPDATE: The solution has been changed (before it was justthe solution from #3(?))Find the following limit or show that it does not exist
lim〈x,y〉→〈0,0〉
y2
x4 + y4.
Solution. We start by trying to “plug in” 〈0, 0〉.
y2
x4 + y4=
0
0
= indeterminate
Now we try taking a path= that goes towards the origin:
y = x lim〈x, y〉 → 〈0, 0〉
y = x
y2
x4 + y4= lim
x→0
x2
x4 + x4
= limx→0
x2
2x4
= limx→0
1
2x2
= DNE (vertical asymptote)
Since the limit does not exist along (at least one) path, the limit for all 〈x, y〉 doesnot exist either.
5. (Section 14.3) Find all of the second partial derivatives of
z = tan−1(x+ y
1− xy
)Solution.
∂z
∂x=
1
1 +(x+y1−xy
)2 ∂
∂x
(x+ y
1− xy
)
=1
1 +(x+y1−xy
)2 1(1− xy)− (x+ y)(−y)
(1− xy)2
=1
1 +(x+y1−xy
)2 1 + y2
(1− xy)2
= . . .
=1
1 + x2
∂z
∂y=
1
1 + y2
∂2f
∂x2= − 2x
(1 + x2)2
∂2f
∂y2= − 2y
(1 + y2)2
∂2f
∂x∂y= 0
∂2f
∂y∂x= 0
6. (Section 14.4) Find the equation of the tangent plane at (3, 1, 0) for the surface
z = ln(x− 2y)
Solution. The tangent plane is just a generalization of the tangent line:
y = m(x− x0) + y0 tangent line
z = mx(x− x0) +my(y − y0) + z0 tangent plane
(x0, y0) = (3, 1)
fx =1
x− 2y
mx = fx(3, 1)
=1
3− 2
= 1
fy =−2
x− 2y
my = fy(3, 1)
=−2
1= −2
z = 1(x− 3)− 2(y − 1) + 0
7. (Section 14.4) Find the linearization of the function f(x, y) =√x+ e4y at the point
(3, 0).
Solution. This is basically just the equation of the tangent plane, solved for z as afunction of x and y
z = mx(x− x0) +my(y − y0) + z0
(x0, y0) = (3, 0)
fx =1
2√x+ e4y
mx = fx(3, 0)
=1
4
fy =2e4y√x+ e4y
my = fy(3, 0)
= 1
z0 = f(3, 0)
= 2
z =1
4(x− 3) + (y − 0) + 2
8. (Section 14.4) Use the linearization√y + cos2(x) ≈ 1 +
1
2y
to estimate the value of√
0.1 + cos2(0.1).
Solution.√0.1 + cos2(0.1) ≈ 1 +
1
2(0.1)
= 1.05
9. (Section 14.4) If z = 5x2 + y2 and if (x, y) changes from (1, 2) to (1.05, 2.1), comparethe values of ∆z and dz.
Solution.
∆z = f(1.05, 2.1)− f(1, 2)
= 5(1.05)2 + (2.1)2 −(5(1)2 + 22
)
= 9.9225− 9
= 0.9225
dz = 10x dx+ 2y dy
≈ 10x∆x+ 2y∆y
= 10(1)(0.05) + 2(2)(0.1)
= 0.9
10. (Section 14.5) Let w be a function of x, y and z, and let x, y and z be functions ofr, and θ as shown
w = xy + yz + zx,
x = r cos(θ),
y = r sin(θ),
z = rθ
Find∂w
∂r
∣∣∣r=2,θ=π/2
and∂w
∂θ
∣∣∣r=2,θ=π/2
Solution.
∂w
∂r=∂w
∂x
∂x
∂r+∂w
∂y
∂y
∂r+∂w
∂z
∂z
∂r
∂w
∂θ=∂w
∂x
∂x
∂θ+∂w
∂y
∂y
∂θ+∂w
∂z
∂z
∂θ
∂w
∂x= y + z
∂w
∂y= x+ z
∂w
∂z= y + x
∂x
∂r= cos(θ)
∂y
∂r= sin(θ)
∂z
∂r= θ
∂x
∂θ= −r sin(θ)
∂y
∂θ= r cos(θ)
∂z
∂θ= r
∂w
∂r= (y + z) cos(θ) + (x+ z) sin(θ) + (y + x)θ
∂w
∂θ= −(y + z)r sin(θ) + (x+ z)r sin(θ) + (y + x)r
11. (Section 14.5) Use the chain rule method to find the implicit derivatives∂z
∂xand
∂z
∂ywhere
yz + x ln(y) = z2
Solution. Let F (x, y, z) = yz+x ln(y)−z2. Then the original equation is equivalentto F (x, y) = 0 and
∂z
∂y= −Fy/Fz
= −z + x/y
y − 2z
∂z
∂x= −Fx/Fz
= − ln(y)
y − 2z
12. (Section 14.6) Find the directional derivative of g(p, q) = p4−p2q3 at the point (2, 1)in the direction of v = i + 3j.
Solution.
∇g =⟨4p3 − 2pq3,−3p2q2
⟩u = v/ |v|
=1√10〈1, 3〉
Dug(2, 1) = ∇g(2, 1) · u
= 〈32− 2,−12〉 · 1√10〈1, 3〉
=1√10
(30− 36)
= −6/√
10
13. (Section 14.6) Find the maximal directional derivative of f(x, y, z) =√x2 + y2 + z2
at the point (3, 6,−2), and state the direction in which it occurs, using a unit vector.
Solution.
∇f =
⟨2x
x2 + y2 + z2,
2y
x2 + y2 + z2,
2z
x2 + y2 + z2
⟩∇f(3, 6,−2) =
⟨3√49,
6√49,−2√
49
⟩=
1
7〈3, 6,−2〉
Therefore, the maximum that Du can equal is∣∣∣∣17 〈3, 6,−2〉∣∣∣∣ =
1
7
√9 + 36 + 4 = 1
This occurs in the direction of ∇f(3, 6,−2) which is
1
7〈3, 6,−2〉 .
14. (Section 14.6) Find the maximal directional derivative of f(x, y, z) = arctan(xyz) atthe point (1, 2, 1), and state the direction in which it occurs, using a unit vector.
Solution.
∇f =
⟨yz
1 + (xyz)2,
xz
1 + (xyz)2,
xy
1 + (xyz)2
⟩∇f(1, 2, 1) =
⟨2
5,1
5,2
5
⟩=
1
5〈2, 1, 2〉
Therefore, the maximum that Du can equal is∣∣∣∣15 〈2, 1, 2〉∣∣∣∣ =
1
5|〈2, 1, 2〉|
=1
5
√4 + 1 + 4
= 3/5
This occurs in the direction of ∇f(1, 2, 1) =1
5〈2, 1, 2〉 which converts to a unit vector
as
u =15 〈2, 1, 2〉∣∣15 〈2, 1, 2〉
∣∣=
1/5
3/5〈2, 1, 2〉
=1
3〈2, 1, 2〉
15. (Section 14.7) Find the critical points and identify them as local max/min/neitherfor the function
f(x, y) = y2 − 2y cos(x), −1 ≤ x ≤ 7.
Solution.
∇f(x, y) = 〈2y sin(x), 2y − 2 cos(x)〉∇f(x, y) = 〈0, 0〉
〈2y sin(x), 2y − 2 cos(x)〉 = 〈0, 0〉2y sin(x) = 0⇒ y = 0 or x = 0, π, 2π
2y − 2 cos(x) = 0⇒if y = 0 (from above) cos(x) = 0⇒ x = π/2, 3π/2
if x = 0, π, 2π (from above) y = cos(x)⇒y = 1,−1, 1 respectively
Combining the above cases:
(0, 1), (π,−1), (2π, 1), (π/2, 0), (3π/2, 0)
Now we identify them as local max/min/neither.
fxx = 2y cos(x)
fyy = 2
fxy = 2 sin(x)
D(x, y) = fxx(x, y)fyy(x, y)− (fxy(x, y))2{D(0, 1) = (2)(2)− 02 > 0
fxx(0, 1) = (2) > 0
}⇒ (0, 1) l.min{
D(π,−1) = (2)(2)− 02 > 0
fxx(π,−1) = (2) > 0
}⇒ (π,−1) l.min{
D(2π, 1) = (2)(2)− 02 > 0
fxx(2π, 1) = (2) > 0
}⇒ (2π, 1) l.min{
D(π/2, 0) = (0)(2)− 12 < 0
fxx(π/2, 0) doesn’t matter
}⇒ (π/2, 0) saddle point{
D(3π/2, 0) = 0(2)(−(−1)2 < 0
fxx(0, 1) = doesn’t matter
}⇒ (0, 1) saddle point
16. (Section 14.7) Find the absolute max/min of the function
f(x, y) = 4x− 6y − x2 − y2, D = {(x, y) | 0 ≤ x ≤ 4, 0 ≤ y ≤ 5}.
Solution.
∇f = 〈4− 2x,−6− 2y〉∇f = 〈0, 0〉
〈4− 2x,−6− 2y〉 = 〈0, 0〉〈x, y〉 = 〈2,−3〉
Thus, 〈2,−3〉 is a critical point. But, this critical point is not in the domain D, sowe do not bother evaluating f at this point.
Now we break up the boundary of D into 4 pieces:
bdy D = D1 ∪D2 ∪D3 ∪D4
D1 = {(0, y) | 0 ≤ y ≤ 5}D2 = {(x, 5) | 0 ≤ x ≤ 4}D3 = {(4, y) | 0 ≤ y ≤ 5}D4 = {(x, 0) | 0 ≤ x ≤ 4}
g1(y) = f(x, y) on D1
= −6y − y2, 0 ≤ y ≤ 5
g1(0) = 0, g1(5) = −55;
i.e. f(0, 0) = 0, f(0, 5) = −55
g2(y) = f(x, y) on D2
= −x2 + 4x− 55, 0 ≤ x ≤ 4
g2(2) = −51, g2(0) = −55, g2(4) = −55;
i.e. f(2, 5) = −51, f(0, 5) = f(4, 5) = −55
g3(y) = f(x, y) on D3
= −6y − y2, 0 ≤ y ≤ 5
g3(0) = 0, g3(5) = −55;
i.e. f(4, 0) = 0, f(4, 5) = −55
g4(x) = f(x, y) on D4
= 4x− x2, 0 ≤ x ≤ 4
g4(2) = 4, g4(0) = 0, g4(4) = 0;
i.e. f(2, 0) = 4, f(0, 0) = f(4, 0) = 0
abs min f = −55 at (0, 5), (4, 5)
abs max f = 4 at (2, 0)
17. (Section 14.8) Find the absolute max and min of the function, subject to the givenconstraint
f(x, y, z) = x2y2z2, x2 + y2 + z2 = 1
Solution.
g(x, y, z) = x2 + y2 + z2
∇f = λ∇g⟨2xy2z2, x22yz2, x2y22z
⟩= λ 〈2x, 2y, 2z〉
So we have four equations 2xy2z2 = λ2x
x22yz2 = λ2y
x2y22z = λ2z
x2 + y2 + z2 = 1
If any of x, y or z equal 0, then we can evaluate f :
f(0, y, z) = 0, f(x, 0, z) = 0, f(x, y, 0) = 0
Otherwise, they are all nonzero and we havey2z2 = λ
x2z2 = λ
x2y2 = λ
x2 + y2 + z2 = 1
and it’s easy to show that x = y = z = ±1/
√3. Note that the ± makes no difference
when we evaluate f :
f(±1/√
3,±1/√
3,±1/√
3)
= 1/27
Thus, we have:
abs min 0 at x or y or z equal 0
abs max 1/27 at x = y = z = ±1/√
27
18. (Section 15.1)Approximate the integral using the midpoint rule with n = m = 2∫∫R
sin(xy) dA, R = {(x, y) | 0 ≤ x ≤ π, 0 ≤ y ≤ π}.
Solution.
(xi, yi) = (π/4, π/4), (π/4, 3π/4), (3π/4, π/4), (3π/4, 3π/4)∫∫R
sin(xy) dA ≈ sin(π/4 · π/4)× (π/2)2 + sin(π/4 · 3π/4)× (π/2)2
+ sin(3π/4 · π/4)× (π/2)2 + sin(3π/4 · 3π/4)× (π/2)2
19. (Section 15.1) Use basic geometry to find the volume contained in the triangularregion defined by z = 4− 2y, over the region R = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 2}.
Solution. This is a triangular solid
y
z
x
z = 4
z = 4− 2y
y = 2
x = 1
In other words we have
b
hl
V = area × length =
(1
2bh
)l
so we have1
2(2)(4)(1) = 4
20. (Section 15.1)Calculate the integral:∫∫Rx sin(x+ y) dA, R = [0, π/6]× [0, π/3].
Solution.∫ x=π/6
x=0
∫ y=π/3
y=0x sin(x+ y) dy dx =
∫ π/6
0−x cos(x+ y)
∣∣∣y=π/3y=0
dx
=
∫ π/6
0
x
2cos(x)− cos(x+ π/3) dx
=1
2cos(x) +
x
2sin(x)− sin(x+ π/3)
∣∣∣π/60
(note integration by parts)
= −3
2+
3√
3
4+
π
24
21. (Section 15.3)Find the volume of the solid under the paraboloid z = 3x2 + y2 andabove the region bounded by y = x and x = y2 − y.
Solution. The region R is of Type II:
x
y
2
2
D = {(x, y) | y2 − y ≤ x ≤ y, 0 ≤ y ≤ 2}
V =
∫ 2
0
∫ y
y2−y(3x2 + y2) dx dy
=
∫ 2
0
[x3 + xy2
∣∣∣x=yx=y2−y
]dy
=
∫ 2
0
[y3 + y3 −
((y2 − y)3 + (y2 − y)y2
)]dy
=
∫ 2
0
[2y3 −
(y6 − 3y5 + 3y4 − y3 + y4 − y3
)]dy
=
∫ 2
0(4y3 − 4y4 + 3y5 − y6) dy
= y4 − 4
5y5 +
1
2y6 − 1
7y7∣∣∣20
=144
35
22. (Section 15.3)Evaluate
∫∫R
ex2+y2dA where R = {(x, y) | 16 ≤ x2 + y2 ≤ 25, x ≥
0, y ≥ 0}.
Solution.
R = {(r, θ) | 4 ≤ r ≤ 5, 0 ≤ θ ≤ π/2}∫∫R
ex2+y2dA =
∫ π2
0
∫ 5
4er
2r dr dθ
=
∫ π2
0
[1
2er
2∣∣∣r=5
r=4
]dθ
=
∫ π2
0
[e25 − e16
2
]dθ
=(e25 − e16)π
4
23. (Section 15.6) Evaluate
∫∫∫D
πyz cos(π
2x5)dV where
D = {(x, y, z) | 1 ≤ x ≤ 2, 0 ≤ y ≤ x, x ≤ z ≤ 2x}.
Solution.∫∫∫D
πyz cos(π
2x5)dV =
∫ 2
1
∫ x
0
∫ 2x
xπyz cos
(π2x5)dz dy dx
=
∫ 2
1
∫ x
0
[π
2yz2 cos
(π2x5) ∣∣∣z=2x
z=x
]dy dx
=
∫ 2
1
∫ x
0
π
2y4x2 cos
(π2x5)− π
2yx2 cos
(π2x5)dy dx
=
∫ 2
1
∫ x
0
3π
2yx2 cos
(π2x5)dy dx
=
∫ 2
1
[3π
4y2x2 cos
(π2x5) ∣∣∣y=x
y=0
]dx
=
∫ 2
1
3π
4x4 cos
(π2x5)dx =
3
10sin(π
2x5) ∣∣∣2
1
=3
10
[sin(16π)− sin
π
2
]= − 3
10