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8/4/2019 Last Lecture Practice Questions
http://slidepdf.com/reader/full/last-lecture-practice-questions 1/6
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= =
8/4/2019 Last Lecture Practice Questions
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(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)
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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 ≤ ≤
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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)
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(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:
8/4/2019 Last Lecture Practice Questions
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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: