Unformatted text preview: QUIZ 2 (CHAPTER 15, 16.1, 16.2)
SOLUTIONS
MATH 252 – FALL 2008 – KUNIYUKI
SCORED OUT OF 125 POINTS
MULTIPLIED BY 0.84 105% POSSIBLE Clearly mark vectors, as we have done in class. I will use boldface, but you don’t!
When describing vectors or vectorvalued functions, you may use either
or
“ i j k ” notation. Assume we are in our usual 2 and 3dimensional Cartesian
coordinate systems. Give exact answers, unless otherwise specified.
1) Find the length of the curve parameterized by: x = t sin t + cos t , y = t cos t sin t , z = 32
t, 0 t
2 4. Leave your answer as a simplified exact answer; do not approximate it using a
calculator. Distance is measured in meters. Show all work! (20 points)
L=
= 4
0 4
0 dx
dt (D t 2 dy
+
dt 2 2 dz
+
dt t sin t + cos t dt ) + (D
2 t t cos t sin t Use the Product Rule for
=
=
=
=
=
= 4
0
4
0
4
0
4
0
4
0
4
0 = 10 ( sin t + t cos t
(t cos t ) 2 + ( ) 2 dx
dy
and
.
dt
dt ) + ( cos t
2 sin t
t sin t ) 2 cos t dt ) + (3t ) dt
2 t sin t 2 3
+ Dt t 2
2 2 + 9t 2 dt t 2 cos 2 t + t 2 sin 2 t + 9t 2 dt ( ) t 2 cos 2 t + sin 2 t + 9t 2 dt
t 2 + 9t 2 dt
10t 2 dt
4
0 t dt Observe: t 2 = t = t , since t 0 on the interval 0, 4 . Now, apply the Fundamental Theorem of Calculus. t2
= 10
2 4 0 ( 4) = 10 (0) 2 2 2 2 () = 10 8 = 8 10 meters This is about 25.2982 meters. ( ) 2) Find parametric equations for the tangent line to C at the point 25, 0, 478 , ( t), z = 2t 5 It helps that x is a onetoone function of t. Solve t 3 2 = 25 . where C is parameterized by: x = t 3 2, y = sin t2 + 1. (12 points) () r t = t3 2, sin ( t ), 2t 5 t2 + 1 . Find the tvalue corresponding to the given point. 2 = 25 t3 t 3 = 27
t=3
Check: ( ) (3) r3= 3 ( ) , 2 (3) (3)
5 2, sin 3 2 +1 = 25, 0, 478
Find a direction vector for the desired tangent line. ()
r (t ) = ( t ) , 2t t + 1
, 10t 2t
3t , cos ( t )
= 3t , cos ( t ) , 10t 2t
r ( 3) = 3( 3) , cos ( 3 ) , 10 ( 3) 2 ( 3)
= 27, ( 1) , 804
= 27,
, 804 ( You can use any non0 scalar multiple of this.)
r t = t3 2, sin 5 2 2 4 2 4 2 4 Write parametric equations for the desired tangent line.
x = 25 + 27u
y= 0 u (u in R ) z = 478 + 804u
x = 25 + 27u
y= u (u in R ) z = 478 + 804u 3) A curve C in 3space is smoothly parameterized by the position vectorvalued
function (VVF) rule r(t ) . The position vector r(t ) and the tangent vector r (t ) () () are orthogonal for all real t. Simplify Dt r t • r t for all real t. Use a differentiation rule discussed in class. (5 points) () () Dt r t • r t () () = r t •r t =0
by orthogonality () () + r t •r t =0
by orthogonality ( by a Product Rule for VVFs: (iii) on p.756) =0 () ( )(
Find the position vectorvalued function (VVF rule) r(t ) if r ( 0 ) = 2i
v ( 0 ) = i + 3j . (15 points) ) 4) The acceleration of a moving particle is given by a t = 3sin t i + 5cos t j . () () v t = a t dt 6 j and ( one member ) = 3sin t , 5cos t dt = 3cos t , 5sin t + C () Solve for C by plugging in t = 0 and using the initial condition v 0 = i + 3j, or 1, 3 . () 1, 3 = () ()
3(1) , 5( 0 ) + C 1, 3 = 3, 0 + C v0= 3cos 0 , 5sin 0 + C C = 4, 3 Therefore,
vt = ()
v (t ) = Now, 3cos t , 5sin t + 4, 3
3cos t + 4, 5sin t + 3 () () ( one member ) r t = v t dt
= 3cos t + 4, 5sin t + 3 dt = 3sin t + 4t , 5 cos t + 3t + D () Solve for D by plugging in t = 0 and using the initial condition r 0 = 2i 6 j, or 2, 6 . () r0= () () 3sin 0 + 4 0 , 2, 6 = 0, () () 5 cos 0 + 3 0 + D 5 +D D = 2, 1 Therefore, () 3sin t + 4t , () 3sin t + 4t + 2, rt =
rt = ( 5 cos t + 3t + 2, 1
5 cos t + 3t 1 , or )( 3sin t + 4t + 2 i + ) 5 cos t + 3t 1 j () t3 t2
,
5) The curve C is determined by r t =
, where t > 0 . Show all work,
32
simplify radicals, and simplify completely, as we have done in class. Do not
eliminate the parameter. Messy and/or undisciplined work may not be graded!
Note: When differentiating, avoid using the Product Rule as an alternative to
the Quotient Rule, unless you simplify your result to the most compact form!
(18 points total)
a) Find the unit tangent VVF (rule) T(t ) for C. () rt = t3 t2
,
32 () r t = t 2 , t , or t t , 1 () rt = t2, t =t
=t
=t , or t t , 1 t, 1 ( because we can assume t > 0)
(t ) + (1)
t, 1
2 2 = t t2 + 1 () Tt =
=
= () Tt = ()
r (t ) rt t t, 1
t t2 + 1
t, 1
t2 + 1 t, 1
t2 + 1 () , or T t = t
t2 + 1 , 1
t2 + 1 Note: If we had eliminated the parameter, we would have obtained
3 y= 9x2
, which can be analyzed directly.
2 () b) Find the VVF (rule) T t for C. () t Remember, T t = t2 + 1 () Tt= , () t2 + 1 Dt t ( 1 or t2 + 1 ( t2 + 1 ) Dt t 2 + 1 t
t2 + 1 t ) ( ) , t +1
2 1
2 1/ 2 ( 2 ) ( 2t) 12
t +1
2 , 3/ 2 ( for the first component, we used the Quotient Rule for Diff.)
t2 + 1 1
= (
t +1
2 t2 + 1 , = (t + 1) (t + 1)
(t + 1) t2 + 1 ( t2 3/ 2 1 ) t2 + 1 3/ 2 t2 + 1 t2 (t + 1)
2 2 t t2 + 1 2 2 = ) (t + 1) (t ) , 2 2 t2 + 1
= t2 + 1 1/ 2 t2 t2 + 1
= ) ( 2t) ( 12
t +1
2 t (t + 1) 3/ 2 2 t , 3/ 2 t , (t + 1) 3/ 2 2 , ( t ) t2 + 1 3/ 2 , or ( 1 ) t2 + 1 3/ 2 1, t 3/ 2 () You didn’t have to do this, but here’s how we can find N t : () Tt = =
= (
( 1 (1) + ( t )
2 (t + 1) 3/ 2 1 1+ t2 (t + 1) 3/ 2 1 (t + 1) 1/ 2 2 (t + 1) 3/ 2 2 1
t +1
2 ()
T (t ) () Tt Nt = = 3/ 2 1 2 = 1, t ) t2 + 1 2 = ) 1 2 = 1, t 3/ 2 t2 + 1 ( 1 1, t ) t2 + 1 3/ 2 1
t +1
2
t +1
1
=
1, t
3/ 2
1
t2 + 1
2 ( = 1 (t + 1)
2 = 1/ 2 1
t2 + 1 ) 1, t , t
t2 + 1 Observe: For all t > 0 , N(t ) • T(t ) = 0 , which reflects the fact that () Nt () Tt. 6) Assume that r is a position VVF of t in 3space that is twice differentiable
everywhere (i.e., second derivatives exist for all real t). Write a curvature
formula we discussed for (t ) that involves a cross product. (4 points) ( ) a (t )
v (t ) () vt () t= 3 () rt rt
or () rt 3 () 7) A twisted cubic curve C is determined by r t = t , t 2 , t 3 , where t > 0 .
(19 points total)
a) Find a general curvature formula, (t ) , for every point on C. Use your formula from Problem 6), and simplify your answer completely. Show all
work! (15 points) ()
Then, r ( t ) or v ( t ) =
r ( t ) or a ( t ) = Let r t = t , t 2 , t 3 . ( ) a (t ) = 1, 2t , 3t 2 , and
0, 2, 6t . 1, 2t , 3t 2 vt 0, 2, 6t ijk
= 1 2t 3t 2
0 2 6t
= ( 2t 3t 2
i
2 6t = 12t 2 1 2t
1 3t 2
j+
k
02
0 6t ) (6t 0) j + ( 2 0) k 6t 2 i = 6t 2 i 6t j + 2k
= 6t 2 , 6t , 2 ( ) a (t ) vt or 2 3t 2 , 3t , 1 = 6t 2 , 6t , 2 = (6t ) + ( 6t ) + ( 2)
2 2 2 = 36t 4 + 36t 2 + 4 ( ) = 4 9t 4 + 9t 2 + 1
= 2 9t 4 + 9t 2 + 1 2 = 2 3t 2 , 3t , 1
or = 2 (3t ) + ( 3t ) + (1)
2 2 = 2 9t 4 + 9t 2 + 1 2 2 () 3 vt = (1) + ( 2t ) + (3t )
2 =
= 3 1, 2t , 3t 2 2 3
2 ( 1 + 4t + 9t ) 3 2 4 ( ) = 9t 4 + 4t 2 + 1 (t ) = 2 ( ) a (t )
v (t ) vt 3 3/ 2 2 9t 4 + 9t 2 + 1 = (9t 4 ) + 4t 2 + 1 3/ 2 b) Use your formula in part a) to find the curvature of the twisted cubic
curve C at the point 2, 4, 8 . Approximate your answer to four decimal ( ) places. (4 points) ( ) The point 2, 4, 8 corresponds to t = 2 . ( 2) = 2 9t 4 + 9t 2 + 1 (9t 4 ) + 4t 2 + 1 3/ 2 = t=2 () ()
(9(2) + 4(2) + 1)
4 2 2 9 2 + 9 2 +1
4 0.0132 2 3/ 2 () 8) Sketch the level curves of f x , y = y x 2 for k = 3, 0, 3 on the grid below. Label the curves with their corresponding kvalues. Be reasonably accurate.
(8 points)
Let f ( x , y ) = k . For any real k: k=y x2 x2 + k = y
y = x2 + k
The graph of this is a parabola that opens upward and that has the point (0, k ) as
its vertex and its yintercept. ()
has vertex ( 0, 0 ) .
+ 3 has vertex ( 0, 3) . k = 3 : The parabola y = x 2
k = 0 : The parabola y = x 2
k = 3 : The parabola y = x 2 5 3 has vertex 0, 3 . y 4
3 3 2 0 1
5 4 3 2 1
1
2 1 2 3 4 3 3
4
5
Here is the corresponding surface, the graph of z = y x2 : 5 x 9) Matching. (9 points total)
Fill in each blank with the best choice (AK) in the list below to indicate the
level surface of f for the given value of k.
A.
B.
C.
D.
E.
F.
G.
H.
I.
J.
K. A Sphere or Ellipsoid
A Hyperboloid of One Sheet
A Hyperboloid of Two Sheets
A Cone
A Circular or Elliptic Paraboloid
A Hyperbolic Paraboloid
A Right Circular or Elliptic Cylinder
A Plane
A Line (a “degenerate” surface)
A Point (a “degenerate” surface)
NONE (no surface) ( ) a) The level surface of f x , y , z = x 2 + z 2 , k = 7 is __G__.
Analyze: 7 = x 2 + z 2 .
Its graph is a right circular cylinder in xyzspace. Its axis is the yaxis.
Imagine taking a circle in the xzplane and sweeping it parallel to the yaxis. ( ) b) The level surface of f x , y , z = 3x + 4 y 5z , k = 0 is __H__.
Analyze: 0 = 3x + 4 y 5z .
Its graph is a plane in xyzspace that passes through the origin. ( ) c) The level surface of f x , y , z = x 2 + 4 y 2 + 9 z 2 , k = 25 is __A__.
Analyze: 25 = x 2 + 4 y 2 + 9 z 2 .
Its graph is an ellipsoid centered at the origin in xyzspace.
Below are graphs of a), b), and c), respectively. (Courtesy Mathematica.) 10) 3x 2 4 y 2
Show that lim
does not exist. (8 points)
( x , y ) (0 ,0) 2 x 2 + 3 y 2
Let ( x , y ) (0,0) along the yaxis (x = 0): ()
() 2 30
4 y2
3x 2 4 y 2
lim
= lim
( x , y ) (0 ,0) 2 x 2 + 3 y 2 ( x , y ) (0 ,0) 2 0 2 + 3 y 2
= lim
y =
Let ( x , y ) 0 4 y2
3y2 4
3 (0,0) along the xaxis (y = 0): ()
() 3x 2 4 0
3x 2 4 y 2
lim
= lim
( x , y ) (0 ,0) 2 x 2 + 3 y 2 ( x , y ) (0 ,0) 2 x 2 + 3 0 2
2 3x 2
= lim 2
x 0 2x
3
=
2 () We have found two paths approaching 0, 0 that yield different limit values for
3x 2 4 y 2
, so the indicated limit does not exist by the TwoPath Rule.
2x2 + 3y2 Here is a graph of z = 3x 2 4 y 2
:
2x2 + 3y2 11) x3 y3
Use polar coordinates to find lim
. (7 points)
( x , y ) (0 ,0) x 2 + y 2 ( )( ) 3 3 r cos
r sin
x3 y3
lim
= lim
2
2
2
r0
( x , y ) (0 ,0) x + y
r ( r cos )( r sin )
= lim
3 r 3 3 r2
sin 3 0 r 6 cos3
= lim
r0
r2
= lim r 4 cos3 sin 3
r 0 3 ( ) =0 Justifying the last step:
For all real , 1 cos As a result, 1 cos3 1 and 1 sin
1 and 1 sin 3 Therefore, 1 cos3 sin 3 1.
1. 1. () () Observe that r 4 > 0 along any path approaching 0, 0 that avoids 0, 0 ,
itself. Multiply all three parts by r 4 and apply the Sandwich / Squeeze
Theorem:
As r 0, r4 r 4 cos3 sin 3 0 r4 .
0 So, 0 Informally, the limit of something approaching 0 times something that is
bounded is 0.
Here is a graph of z = x3 y3
:
x2 + y2 ...
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This note was uploaded on 09/08/2011 for the course MATH 252 taught by Professor Staff during the Spring '11 term at Mesa CC.
 Spring '11
 staff
 Math, Calculus, Vectors

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