Download Solution of Quiz 3

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6/28/2017
Name (Print Clearly):
Student Number:
MTH132 Section 1 & 12, Quiz 3
Feb 20, 2009
Instructor: Dr. W. Wu
Instructions: Answer the following questions in the space provided. There is more than adequate
space provided to answer each question. The total time allowed for this quiz is 15 minutes.
1 [3 pts each]. Find
dy / dx for following functions (Do Not Simplify)
(a) y  3 4x
y  (4 x)1 / 3 （apply power chain rule）
So y ' 
1
d
4
(4 x)  2 / 3  (4 x)  (4 x)  2 / 3
3
dx
3
(b) y 
4
tan 3 x
3
(chain rule)
4
d
y '  sec 2 (3x)  (3x)  4 sec 2 (3 x)
3
dx
(c) y  cos x  cos x
3
2
y  cos( x3 )  (cos x)2 (use chain rule, the inside function of the first term is x 3 , and the inside function of
the second term is cos x ).
So y '   sin( x ) 
3
d 3
d
( x )  2 cos x  (cos x)   sin( x 3 )  (3x 2 )  2 cos x  ( sin x)
dx
dx
(d) x y  xy  1 (use implicit differentiation)
2
2
consider y as a function of x (implicitly)
apply
d
to this implicit equation:
dx
2 xy  x 2 y ' y 2  x 2 y  y ' 
y 2  2 xy
so ( x  2 xy) y'  ( y  2 xy)  y'   2
x  2 xy
2
2
d
(1)  0
dx
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2. [4 pts]. Show that (2,1) lies on the curve xy2  2 y 3  4 . Then find an equation for the tangent line to
the curve at this point.
Substitute x  2, y  1 into the left hand side  2(1) 2  2(1)3  2  2  4 . So it is on the curve.
Consider y as a function of x (implicitly), and use implicit differentiation
y 2  x  (2 yy ' )  6 y 2 y ' 
d
(4)  0
dx
 y2
 12
1
So y ' 
. At (2,1) , the slope is y' 

 1/ 10
2
2
2 xy  6 y
2  (2)  1  6  1  10
Use point-slope equation: the tangent line at (2,1) is y  1 
1
( x  2)
10
3. Suppose the parametric equations of a curve are x  2 cos t , y  3 sin t , 0  t  2 .
(a) [2 pts]. Find dy / dx by using parametric formula.
dy dy / dt
3 cos t


dx dx / dt  2 sin t
(b) [2 pts]. Let p be the point when t   / 6 . Find the slope of the tangent line to this curve at p .
3 cos( / 6)
3 3/2
3


 2 sin(  / 6)
 2 1/ 2
2
The slope is the derivative at t   / 6 , so it is
(c) [bonus 2 pts]. Find a Cartesian equation for this curve, then find
x  2 cos t , y  3 sin t  cos t 
dy
at p by using implicit differentiation.
dx
x
y
2
2
, sin t 
, by trigonometric identity cos t  sin t  1
2
3
2
2
A Cartesian equation for this ellipse is
x
y

 1 . (Ex 3.5 #69)
4
3
Using
1
2y
 2x 
 y'  0
4
3
implicit
differentiation:
y' 
 x/2
3x

2y /3
4y
x  2 cos( / 6)  3 , y  3 sin(  / 6)  3 / 2 . So the slope is y '  
.
When
t  /6 ,
3 3
3 3

 3 / 2 .
4 3 / 2
2 3