11
Class:
Name:
Date:
-----
Implicit Differentiation
Multiple Choice
Identify the choice that best completes the statement or answers the question.
1. The slope of the line tangent to the curve y2
3
2
3
4
a.
b.
c.
o
3
d.
4
3
e.
2
r.---;
dy
2.
If-
a.
b.
= '" 1-y-
dx
d2 Y
then
2
'dx
=
-2y
-y
-y
C.
~1_y2
d.
e.
~
.:>.
y
1
2
2
If y = xy + x + 1 , then when x
a.
b.
c.
d.
e.
-1
2
1
2
-1
-2
nonexistent
= -1,
dy
dx is
+ ( xy + 1) 3
= 0 at (2,
-1) is
ID: A
---------
-------------------
-----
l'1X-~ ~~f~~?L
--(
1~
2008 AB 6 Form B
Consider the closed curve in the xy-plane given by
X2
(a) Show that
dy
dx
+ 2x + y4 + 4y
= 5.
= 2(y3+1)'
-(x+l)
(b) Write an equation for the line tangent to the curve at the point (-2, 1).
(c) Find the coordinates of the two points on the curve where the line tangent to the curve is
vertical.
(d) Is it possible for this curve to have a horizontal tangent at points where it intersects the x-axis?
Explain your reasoning.
2005 AB 5 Form B
Consider the curve given by y2
(a) Show that
dy
dx
= 2 + xy.
= -y-.
2y-x
(b) Find all points (x, y) on the curve where the line tangent to the curve has slope~.2
(c) Show that there are no points (x, y) on the curve where the line tangent to the curve is
horizontal.
(d) Let x and y be functions of time t that are related by the equation y2
the value of y is 3 and
dy
dt
= 6.
Find the value of
dx
dt
at time t
------
-
---
= 2 + xy.
At time t
= 5,
= I(x)
, and
= 5.
-------
2001AB6
The function/is
differentiable for all real numbers. The point
the slope at each point (x, y) on the graph is given by dy
dx
2
(a)
Find:' d
(b)
Find y
1(3)
and evaluate it at the point
= I(x)
=1..
4
(
3,
(3,±)
= y2(6
is on the graph of y
- 2x).
1)
'4 .
by solving the differential equation :
= y2(6_
2x) with the initial condition
3>
Ap® CALCULUS AB
2008 SCORING GUIDELINES (Form B)
Question 6
Consider the closed curve in the xy-plane given by
x2
+ 2x + y4 + 4y = 5.
d
-(x+l)
y -- 2(i + 1)
(a) Show that dx
(b) Write an equation for the line tangent to the curve at the point (-2,1).
(c) Find the coordinates of the two points on the curve where the line tangent to the curve is vertical.
(d) Is it possible for this curve to have a horizontal tangent at points where it intersects the x-axis?
Explain your reasoning.
(a)
2x
+ 2 + 4i
(4i+4):
(b)
:
+ 4:
=-2x-2
dy
-2(x
+ 1)
-(x
dx =
4(i
+1) =
2(y3+1)
+ 1)
= -(-2+1) =!
dyl
dx
2 : { 1 : implicit differentiation
1 : verification
=0
(-2,1)
2(1 + 1)
Tangent line: y
I: slope
{
2: 1: tangent line equation
4
= 1+
±(x + 2)
(c) Vertical tangent lines occur at points on the curve where
i + 1 = 0 (or y = -1 ) and x *- -1.
On the curve, y
so x = -4 or x
= -1
=
I:
3:
implies that x2 + 2x + I - 4
= 5,
y =-1
I: substitutes y = -1 into the
equation of the curve
{
1 : answer
2.
Vertical tangent lines occur at the points (-4, -1) and
(2, -1).
(d) Horizontal tangents occur at points on the curve where
x = -1 and y *- -1.
The curve crosses the x-axis where y
4
(_1)2 + 2 ( -1) + 0 + 4 . 0
I: works with x = -lor
2' {
. I: answer with reason
= o.
*- 5
No, the curve cannot have a horizontal tangent where it
crosses the x-axis.
© 2008 The College Board. Allrights reserved.
Visit the College Board on the Web: www.collegeboard.com.
y
=0
4Ap® CALCULUS AB
2005 SCORING GUIDELINES (Form B)
Question 5
i = 2 + ~.
Consider the curve given by
(a) Show that dy = _y_.
dx
2y - x
(b) Find all points (x, y) on the curve where the line tangent to the curve has slope ~.
(c) Show that there are no points (x, y) on the curve where the line tangent to the curve is horizontal.
(d) Let x and y be functions of time t that are related by the equation y2
value of y is 3 and ~
(a)
=
6. Find the value of;;
at time t
,
=
At time t
=
5, the
5.
2 : {I :
2yy' = y + xy'
(2y - x)y'
=
= 2 +~.
implicit diff~rentiation
1 : solves for y
y
y
y=--2y-x
(b)
y
_ 1
2y - x -"2
l.-y
2:
2y =2y-x
{
. 2y - x
=.1
2
1 : answer
x=O
y=±J2
(0, J2), (0, -J2)
(c)
-y_=O
l:y=O
2· {
. 1: explanation
2y-x
y=O
The curve has no horizontal tangent since
02
(d)
* 2 + x-O for any x.
=
When y
3, 32
= 2 + 3x
so x
=
1.
I: solves for x
3:
dy_dy
dx_
y
dx
dt - dx . dt - 2 y - x . dt
At t=5
;;Ls
,
3
dx
9
6 =-_._=_.6 _ 7... dt
11
{
1: chain rule
1 : answer
dx
dt
3
=
2
23
Copyright © 2005 by College Board. All rights reserved.
Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents
6
(for AP students and parents).
5
Ap® CALCULUS AB
2001 SCORING GUIDELINES
Question
f is
The function
y
=
differentiable for all real numbers. The point
f(x) , and the slope at each point (x,y)
d;
2
(a)
Find
and evaluate it at the point
dx
(b) Find y = f(x)
condition
2
(a)
-.d y2 =
=
is on the graph of
=
y2 (6 - 2x).
(3,.!.).
4
~~ = y2 (6 - 2x) with the initial
f(3) = ~.
- 2x) - 2y
2
2:
2
"4
dx
3:
1
8
(1)2
d2y
-2
< - 2 > product rule or
2y3(6 - 2x? - 2y2
I(3 ~\ = 0 -
2
dy
dx2
(3, ~)
on the graph is given by dy
dx
by solving the differential equation
dy
2y-(6
dx
dx
6
chain rule error
1: value at
(3,~)
'41
1
(b) 2dy
y
-
=
(6 - 2x)dx
1 : separates variables
1
- = 6x -x
y
- 4
=
18 - 9
C
= -13
=
x2 - 6x
2
+C
6:
+C
=
9
+C
of dy term
1 rantiderivative
of dx term
1 : constant of integration
1 : uses initial condition
f(3) = ~
1 : solves for y
1
y
1 : antiderivative
Note: max 3/6
+ 13
[1-1-1-0-0-0] if no
constant of integration
Note: 0/6 if no separation of variables
Copyright © 2001 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.
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