Download (3 r dy (3x2f

Lo",,-pv1--b
])L~
Name:
Class:
Date:
Calculate the Derivative
Multiple Choice
Identify the choice that best completes the statement or answers the question.
(3
1. If y::;: x + 1
a.
r
dy
, then d.;
=
(3x f
2
b. 2( x3 + 1)
2
c.
2( 3x + 1)
d.
3x (x
e.
6x (x
1
3
+ 1)
2
3
+ 1)
= In(x + 4+
2. If f(x)
a.
--25
b.
"5
c.
-
e-3x
),
thenf' (0) is
J
1
4
2
d·S
e.
nonexistent
'(1)
3. If f(x) ::;:tan(2x), thenf
a.
b.
c.
d.
e.
=
V
2{3
4
4{3
8
4
_
ID: A
r;
/-
Name:
_
ID: A
2 •
dy
4. Ify ==x sm2x, then dx ==
2xcos2x
a.
b. 4xcos2x
2x(sin2x + cos 2x)
c.
d. 2x(sin2x -xcos2x)
e.
5. Ify
2x(sin2x + xcos2x)
2x+3
==~
a.
b. C.
d.
e.
b.
dy
d; ==
12x+ 13
(3x+2)2
12x-13 _
(3x +2)2
5
(3x+2)2
-
-5
(3x+2)2
23
6. If f(x)
a.
then
== x
~ 2x - 3, thenf' (x)
=
3x-3
V2x-3
x
~2x-3
c.
d.
e.
I
'l2x-3
-x+3
--..}2x-3
5x-6
2Y2x-3
2
5
-)
.--?
Name:
_
ID: A
7. What is the slope of the line tangent to the curve 3y2 - 2x2 ::;:6 - 2xy at the point (3, 2)?
a.
0
4
b.9
c.d.
7
9
....L
7
e.23
2
8.
If x 2
+y2 ::;:25, what is the value of
a.
-~
27
L
b.
c.
d
--f
at the point (4, 3) ?
dx
27
..:L
27
d.
-L
4
e.~
27
ro'f..
9. If F(x) ::;: ~dt
a.
-3
b.
-2
c.
d.
e.
2
3
l8
io. ~
, then F' (2) ::;:
[(sm(t' )dt1 =
a.
-cos( x6)
b.
sin( x3
)
c.
sin( x6
)
d.
2x Sin(x3
)
e.
2x Sin(x6
)
3
6
Name:
_
In: A
7
6\
\
\
s
4
\
3
,
\
2
1
~
1
0
-I
·2
\2
\
-3
S
4
3
<;
../' V
6
7
<,
-'" V
Ompholl
11.
The graph of the functionfshown
g(x) :;::;
Jor f(t)dt,
in the figure above has horizontal tangents at
x = 3 and x = 6. If
2x
o
a.
0
b.
c.
d.
e.
-1
what is the value of g'(3) ?
-2
-3
-6
12.
/'(.1)
g(X)
g'(x)
5
3
-2
3
-3
-1
2
1
-2
2
3
x
1(.1)
-1
6
I
3
The table above gives values off,/" g, and g' at selected values of x. If h(x) =f(g(x), then h'(l) =
a.
5
b.
6
c.
9
d.
10
e.
12
2004 Ap· CALCULUS AS FREE-RESPONSE
QUESTIONS
CALCULUSAB
SECTION
Part B
n,
Time-4S
minutes
Number or problems-3
No calculator is allowed for these problems.
4. Consider the curve given by Xl
+ 4y2 = 7 + 3xy.
dy
3y-2x
(a) Show that dx "" 8y _ 3x'
(b) Show that there is a point P with z-coordinate 3 at which the line tangent to the curve at P is horizontal.
Find the y-coordinate of P.
2
at the point P found in part (b). Does the curve have a local maximum, a local
(c) Find the value of ~
minimum, or neither at the point P '! Justify your answer.
AP Calculus AB-S /
no-s
2000
Consider the curve given by :cy2 - x3y = 6.
d
(e) Show that dxY =
3x2
2
Y - ~ .
2:cy - x
(b) Find all points on the curve whose 3rcoordinate is 1, and write an equation for the tangent line at
each of these points.
(c) Find the z-coordinete of each point on the curve where the tangent line is vertical.
9
-
---------------_._._----
/
r:o'-
Ape CALCULUS AB
2004 SCORING GUIDELINES
Question 4
Consider the curve given by x2
dy
(a) Show that dx
+ 4y2
= 7 + hy.
3y-2x
= 8y _ 3x'
(b) Show that there is a point P with x-coordinate
horizontal. Find the y-coordinate of P.
3 at
which the line tangent to the curve at
P
is
2
(c) Find the value of d ;
at the point P found in
dx
or neither at the point P?
local minimum,
(a)
part (b).
JustifY
2x + 8yy' = 3y + 3xy'
(8y - 3x)y'
3y - 2x
, 3y - 2x
y = 8y-3x
3y-2x
8y-3x
2 : { 1 : implicit diff~rentiation
I :solves for y
1::=0
;:::0; 3y - 2x ;:::0
When x = 3, 3y
3:
=6
y=2
32 + 4.22
Therefore,
= 25 and
P
a
your answer.
=
(b)
Does the curve have a local maximum.
1
l : shows slope is 0 at (3, 2)
1 : shows (3,2)
lies on curve
7 + ).3·2 ;:::25
= (3, 2)
is on the curve and the slope
is 0 at this point.
(c)
d2y
(8y - 3x)(3y' - 2) - (3y - 2x)(8y' - 3)
dx2 =
(8y - 3x)2
At P
Since
= (3, 2).
y' = 0
maximum
2
d y;::: (16 - 9)(-2)
dx2
(16 _ 9)2
and
y. < 0
=
d2y
2:-2
dx
4:
_1.
I : value of
7
at P, the curve has a local
d2
; at (3, 2)
dx
I : conclusion
with justification
at P.
Copyright \0 2004 by College Entrance Examination Board. All rights reserved.
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10
______
o··
_
7.
2000
AP Calculus AB-S / BC-S
Consider the curve given by xy2 - x3y = 6.
dy
(a)
3x2y
==
Show that dx
_ y2
2xy - x
3 .
(b) Find all points on the curve whose z-coordinate is 1, and write an equation for the tangent line at
each of these points.
(c)
Find the z-coordinate of each point on the curve where the tangent line is vertical.
(a)
y2
+ 2xy dy
_ 3x2y _ x3
dx
1: implicit differentiation
=0
dy
dx
2
: (2xy _ x3
dy
dx
=
= 3x2y
)
_ y2
1
1: verities expression for :
3x2y -,;
2xy-x3
(b) When x = 1, y2 - Y
y2 _ Y _ 6
dy
At (1,3), -
dx
=0
4
+ 2) =
(y - 3) (y
y
1:
=6
1: solves for y
1
0
y2_y=6
2:
tangent lines
= 3, y =-2
Note: 0/4 if not solving an equation of the
9- 9
form y2 - Y = k
== --
6-1
=0
Tangent line equation is y =3
dy
-6 - 4 -10
At (1,-2), -=--=-=2
dx
-4 -1
-5
Tangent line equation is y + 2
2(x - 1)
=
(c)
.
Tangent line is vertical when 2xy - x3 = 0
X
(2y -
X2 )
==
0 gives x
=
0 or y
=~
1: sets denominator
1: substitutes
X2
dy
of dx equal to 0
y :;:::~x2
or
3
x
== ±.fiY
into the equation for the curve
There is no point on the curve with
1: solves for z-coordinate
z-coordinate O.
When y
1 2
= 2"x
,
1"
-x
4
__1x5
4
1.
2
--xV
=6
=6
x = ~-24
Copyright © 2000 by College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP is a registered trademark of the College Entrance Examination Board.
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