AP
®
CALCULUS AB
2015 SCORING GUIDELINES
Question 6
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Consider the curve given by the equation
y
3
−xy 2.
=
It can be shown that
dy y
= .
dx
3y
2
− x
(a) Write an equation for the line tangent to the curve at the point
(
−1, 1
)
.
(b) Find the coordinates of all points on the curve at which the line tangent to the curve at that point is vertical.
(c) Evaluate
dy
2
dx
2
at the point on the curve where
x = −1
and
y = 1.
(a)
( ) ( )
( ) ( )
dy
11
= =
dx
2
xy, = −1, 1
31
−−
1
4
An equation for the tangent line is
1
y =
(
x +1
)
+1.
4
2:
{
1 : slope
1 : equation for tangent line
(b)
3y
2
−=x0 ⇒xy=3
2
So,
yx
3
−yy=2⇒y
32
−
(
32
)
(
y
)
=⇒=y−1
(
−1
)
3
−x
(
−=12
)
⇒x=3
The tangent line to the curve is vertical at the point
(
3, −1
)
.
1 : sets 30y
2
−=x
3 :
1 : equation in one variable
1 : coordinates
(c)
( )
( )
( )
dy
3
2
dy
2
y
−−
x y61y
−
dy
dx dx
=
dx
22
3
y
2
− x
( )
( )
(
)
( )
2
31
2
11
− · · ··
dy
2
· −1 −1 61 −1
44
=
dx
2
2
(
xy,
)
=
(
−1, 1
)
31· −
(
−1
)
1
1 −
2
1
= =
16 32
2 : implicit differentiation
dy
4 :
1 : substitution for
dx
1 : answer
©2015 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
©2015 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
©2015 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
©2015 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
©2015 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
©2015 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
AP
®
CALCULUS AB
2015 SCORING COMMENTARY
Question 6
Overview
In this problem students were given the equation of a curve,
y
3
−=xy 2,
with
dy y
=
.
dx
3
y
2
− x
In part (a)
students had to find an equation for the line tangent to the curve at the point
(
−1, 1
)
.
Students were expected to
use the given
dy
dx
to find the slope of the curve at the point
(
−1, 1
)
.
In part (b) students were asked to find the
co
ordinates of all points on the curve at which there is a vertical tangent line. These are the points on the curve
where
3y
2
−=x0,
but
y ≠ 0.
Students were expected to solve
y
3
−=xy 2
with the condition that
3y
2
−x
0=
and report only those pairs
(
xy,
)
where
y ≠ 0.
In part (c) students were asked to evaluate
dy
2
dx
2
at
the point
(
−1, 1
)
on the curve. Students had to use implicit differentiation with
dy
dx
to find an expression for
dy
2
,
dx
2
which required use of the chain rule and either the product rule or the quotient rule. The expression can be
written in terms of x and y or can involve
dy
dx
. In either case, students needed to evaluate the expression for
dy
2
dx
2
at
(
−1, 1
)
.
Sample: 6A
Score: 9
The response earned all 9 points.
Sample: 6B
Score: 6
The response earned 6 points: 2 points in part (a), 1 point in part (b), and 3 points in part (c). In part (a) the student’s
wor
k is correct. In part (b) the student considers the equation
3y
2
−=x0,
so the first point was earned. The student
does not present an equation in one variable. In part (c) the student correctly differentiates and substitutes for
dy
,
dx
so
the first 3 points were earned. The student makes an error in computation, so the answer point was not earned.
Sample: 6C
Score: 3
The response earned 3 points: 1 point in part (a), no points in part (b), and 2 points in part (c). In part (a) the student
make
s an arithmetic error in computing the slope, so the first point was not earned. The student uses the slope to
present a line that passes through
(
−1, 1
)
,
so the second point was earned. In part (b) the student does not consider
the equation
3y
2
−=x0.
In part (c) the student correctly differentiates
dy
,
dx
so the first 2 points were earned.
© 2015 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
