2003 AP Calculus BC Scoring Guidelines - College Board

[Pages:7]AP? Calculus BC 2003 Scoring Guidelines

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AP? CALCULUS BC 2003 SCORING GUIDELINES

Question 1

Let R be the shaded region bounded by the graphs of y = x and y = e3x and the vertical line x = 1, as shown in the figure above. (a) Find the area of R. (b) Find the volume of the solid generated when R is revolved about the horizontal

line y = 1. (c) The region R is the base of a solid. For this solid, each cross section

perpendicular to the x-axis is a rectangle whose height is 5 times the length of its base in region R. Find the volume of this solid.

Point of intersection e3x = x at (T, S) = (0.238734, 0.488604)

? (a) Area = 1( x e3x )dx T = 0.442 or 0.443

1: Correct limits in an integral in (a), (b), or (c)

2 :

???????

1 1

: :

integrand answer

( ) ? (b) Volume = Q 1 (1 e3x )2 (1 x )2 dx T = 0.453 Q or 1.423 or 1.424

3

:

???????????????????????

2 1

: :

integrand < 1 > reversal < 1 > error with constant < 1 > omits 1 in one radius < 2 > other errors answer

(c) Length = x e3x

Height = 5( x e3x )

? Volume = 1 5( x e3x )2 dx = 1.554 T

3 :

???????????????????

2 : integrand < 1 > incorrect but has x e3x as a factor

1 : answer

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2

AP? CALCULUS BC 2003 SCORING GUIDELINES

Question 2

A particle starts at point A on the positive x-axis at time t = 0 and travels

along the curve from A to B to C to D, as shown above. The coordinates of

( ) the particle's position x (t ), y (t ) are differentiable functions of t, where

x a(t )

=

dx dt

=

9cos

Qt 6

??

sin

Q

t+ 2

1

??

and y a(t ) = dy dt

is not explicitly given.

At time t = 9, the particle reaches its final position at point D on the positive x-axis.

(a) At point C, is dy positive? At point C, is dx positive? Give a reason for each answer.

dt

dt

(b) The slope of the curve is undefined at point B. At what time t is the particle at point B ?

( ) (c)

The line tangent to the curve at the point

x (8), y (8)

has

equation

y

=

5 9

x

2.

Find

the

velocity vector and the speed of the particle at this point.

(d) How far apart are points A and D, the initial and final positions, respectively, of the particle?

(a)

At

point

C,

dy dt

is not positive because y(t) is

decreasing along the arc BD as t increases.

At

point

C,

dx dt

is not positive because x(t) is

decreasing along the arc BD as t increases.

2 :

???????????

1 : 1 :

dy dt dx dt

not positive with reason not positive with reason

( ) (b)

dx dt

= 0;

cos

Qt 6

= 0 or sin Q

t+ 2

1 ??

=

0

Qt 6

=

Q 2

or

Q

t+ 2

1

=

Q

;

t

=

3

for both.

Particle is at point B at t = 3.

2

:

?????????

1 1

: :

sets t=

dx dt 3

=

0

( ) ( ) (c)

x a(8) = 9 cos

4Q 3

sin

3Q 2

=

9 2

y a(8) x a(8)

=

dy dx

=

5 9

y a(8)

=

5 9

x a(8)

=

5 2

The velocity vector is < 4.5,2.5 >.

Speed = 4.52 + 2.52 = 5.147 or 5.148

? (d) x(9) x(0) = 9 x a(t)dt 0 = 39.255 The initial and final positions are 39.255 apart.

3

:

????????????

1 1 1

: : :

x a(8) y a(8) speed

2

:

???????

1 1

: :

integral answer

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3

AP? CALCULUS BC 2003 SCORING GUIDELINES

Question 3

The figure above shows the graphs of the line

x

=

5y 3

and the curve

C

given by

x = 1 + y2 . Let S be the shaded region bounded by the two graphs and the

x-axis. The line and the curve intersect at point P.

(a) Find the coordinates of point P and the value of dx for curve C at point P. dy

(b) Set up and evaluate an integral expression with respect to y that gives the

area of S.

(c) Curve C is a part of the curve x 2 y2 = 1. Show that x 2 y2 = 1 can be written as the polar

equation

r2

=

cos2 R

1

sin2 R .

(d) Use the polar equation given in part (c) to set up an integral expression with respect to the polar angle R that represents the area of S.

(a) At P, 5 y = 1 + y2, so y = 3 .

3

4

Since

x

=

5 3

y,

x

=

5 4

.

2

:

?????????

1 1

: :

coordinates

dx dy

at P

of

P

dx = dy

y 1 + y2

=

y. x

At P,

dx dy

=

34 54

=

3 5

.

( ) ? (b) Area =

34 0

1 + y2

5 3

y

dy

= 0.346 or 0.347

(c) x = r cos R ; y = r sin R

x 2 y2 = 1 ? r 2 cos2 R r2 sin2 R = 1

r2

=

cos2 R

1

sin2

R

3

:

????????????

1 1 1

: : :

limits integrand answer

2

:

?????????????

1 1

: substitutes x = r cos R and y = r sin R into x 2 y2 = 1 : isolates r 2

(d) Let C be the angle that segment OP makes with

the

x-axis.

Then

tan C

=

y x

=

34 54

=

3 5

.

? Area =

tan1( 35)

0

1 2

r

2

dR

? =

1 2

tan1 ( 3 5 )

0

cos2

R

1 sin2

R dR

2

:

???????

1 1

: :

limits integrand

and

constant

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4

AP? CALCULUS BC 2003 SCORING GUIDELINES

Question 4

Let f be a function defined on the closed interval 3 b x b 4 with f ( 0) = 3. The graph of f a, the derivative of f, consists of one line segment and a semicircle, as shown above. (a) On what intervals, if any, is f increasing? Justify your answer. (b) Find the x-coordinate of each point of inflection of the graph of f

on the open interval 3 < x < 4. Justify your answer. (c) Find an equation for the line tangent to the graph of f at the

point (0, 3).

(d) Find f (3) and f ( 4 ). Show the work that leads to your answers.

(a) The function f is increasing on [3, 2] since f a > 0 for 3 b x < 2 .

2

:

???????

1 1

: :

interval reason

(b) x = 0 and x = 2 f a changes from decreasing to increasing at x = 0 and from increasing to decreasing at x =2

(c) f a(0) = 2 Tangent line is y = 2x + 3.

2

:

???????

1 1

: :

x = 0 and x justification

=

2

only

1 : equation

? (d) f (0) f (3) = 0 f a(t)dt 3

=

1 2

(1)(1)

1 2

(2)(2)

=

3 2

f (3)

=

f (0)

+

3 2

=

9 2

? f (4) f (0) = 4 f a(t)dt 0

( ) =

8

1 2

(2)2

Q

= 8 + 2Q

f (4) = f (0) 8 + 2Q = 5 + 2Q

(( ) ) 4

:

??????????????????????????????????????????????????

1 1 1 1

:

?

1 2

2

(difference of areas

of triangles)

: answer for f (3) using FTC

:

?

8

1 2

(2)2

Q

(area of rectangle

area of semicircle)

: answer for f (4) using FTC

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5

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