last lecture practice questions

8/4/2019 Last Lecture Practice Questions

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Practice Questions and Solutions(Not covered by the written assignments )

Parametric curves, Lengths, Surface Area, Curvature, Frenet’s Frame

1. Find a parametric representation which does not involve radicals, for the curve that is the

intersection of the following surfaces:

(a) The cone 2 2 z x y= + and the plane 1 z y= +

(b) The cylinder and the plane2 21 x y+ = 2 y z+ =

(c) (c) The cylinders: 2 2, 1 z x y x= = − .

Solution:

(a) We have 2 2 2 2 2 211 1 2 (

2 x y y x y y y y x+ = + ⇒ + = + + ⇒ = −1) . If set x t = we obtain

2 21 1( ) , ( 1), ( 1)

2 2r t t t t =< − + >

.

(b) If set cos x t = , then andsin y t = 2 sin z t = − .

(c) First we notice that . If set2 21 z y+ = cos z t = we get sin y t = and

21 sin x t = −

REMARK: The above parameterizations are not unique!

2. (a) Show that the curve ( ) ( sin ) ( cos )t t t r e i e t j e t k = + +

is on a cone.

(b) Find an equation for the tangent line at 0t = .

Solution:

(a) , hence the cure is on the cone2 2 2 2 2[ ( )] [ ( )] ( sin ) ( cos ) [ ( )]t t t y t z t e t e t e x t + = + = = 2 22 2 x y z= + .

(b) 0

0

, (cos sin ), (cos sin ) 1,1,1t t t

t t

dr u e e t t e t t

dt =

=

= = < + − > =<

> ; At 0t = we get the point (1,0,1)

on the curve. An equation for the tangent line at (1,0,1) is

0r r tu= +

, , 1,0,1 1,1,1 x y z t ⇔< >=< > + < > 1 , , 1 x t y t z t ⇔ = + = = + .

3. (a) At what point do the curves2

1 (1 ) (3 )r ti t j t k = + − + +

,2

2 (3 ) ( 2) ( )r s i s j s= − + − + k

intersect?

(c)

Find their angle of intersection at the intersection point (i.e. the angle between the tangent vectorsat the point of intersection).

Solution:

(a) We must find t and s which satisfy the following equations: 23, 1 2, 3t s t s t s2

− = − = − + = . We

obtain and the point of intersection (1,0,4).1 and 2t s= =



(b) Let’s find the angle θ of intersection at this point. The tangent vectors at (1,0,4) are

and1 (1) 1, 2, 2r ′ =< − > 2 (1) 1,1, 4r ′ =< − > .So

1 1cos ( 1 1 8)

6 18 3θ = − − + = 55θ ⇒ ≈

.

4. Find the length of the following curves:

(a)5

3

1

6 10

x y

x= + ,1 2 x≤ ≤

(b)1

( 3)3

x y y= − y≤ ≤,1 9

(c) 2r e

θ = , 0 2θ π ≤ ≤

(d) , 02

( ) ,sin cos ,cos sinr t t t t t t t t =< − + >

t π ≤ ≤

Solution:

(a)

(b)

(c)

(d)



5. Find the area of the surface obtained by rotating the curve about the x-axis.

(a)3 1

6 2

x y

x= + , 1/ 2 1 x≤ ≤

(b) , 0 / cos2 y x= 6 x π ≤ ≤

(c) 33 x t t = − , ,2

3 y t = 0 1t ≤ ≤

Solution:

(a)

(b)

(c)

6. Find the area of the surface obtained by rotating the curve about the y-axis.

(a) cosh( / ) x a y a= a x a− ≤ ≤,

(b) t x e t = − , ,

/ 24

t y e= 0 1t ≤ ≤



Solution:

(a)

(b)

7. Reparametrize the curve with respect to the arclength measured

from the point where in the direction of increasing t .

2 2( ) cos 2 ,2, sin 2

t t r t e t e t =< >

0t =Solution:

8. (a) Find the curvature of at the point .( ) cos , sin ,t t

r t e t e t t =< >

(1,0,0)

(b) Find the curvature of the ellipse 3cos , 4sin x t y t = = at the points (3 and (0,4).,0)

Solution:

(a)



(b) ,( ) 3cos , 4sin ,0r t t t =< >

( ) 3sin ,4cos ,0r t t t ′ =< − >

, ( ) 3cos , 4sin , 0r t t t ′′ =< − − >

9.

At what point does the curve ln y x=

have maximum curvature?Solution:

10. Find the vectors and for the curve, ,T N

B

( ) , sin , cost t t r t e e t e t =< >

at the point (1, .0,1)

Solution:



11. Find equations for the normal and osculating plane of the curve2 3

( ) , ,r t t t t =< >

at (1,1,1)

Solution:

12. At what point on the curve3

( ) , ( ) 3 , ( )4 x t t y t t z t t = = = is the normal plane parallel to the

plane 6 6 8 1? x y z+ − =

Solution:

13. The helix intersects the curve1( ) cos ,sin ,r t t t t =< >2 3

2 ( ) 1 , ,r t t t t =< + > at the point . Find

the angle of the intersection of these curves.

(1,0,0)

Solution:

last lecture practice questions

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