2005 AP Calculus BC Free-Response Questions Workshop ...

2005 AP? Calculus BC Free-Response Questions

Workshop, Tuesday, Aug. 2, 2005

1.

Let

f

and

g

be

the

functions

given

by

f (x)

=

1 4

+

sin(x)

and

g(x)

=

4-x.

Let

R

be the shaded region in the first quadrant enclosed by the y-axis and the graphs

of f and g, and let S be the shaded region in the first quadrant enclosed by the

graphs of f and g, as shown in the figure above.

(a) Find the area of R. (solution, (a)): This area is the integral

a

4-x -

1 + sin(x)

dx,

0

4

where a is the smallest non-negative solution of f (x) = g(x). This is evaluated numerically with the result a 0.178. Therefore,

Area of R

0.178

4-x -

1 + sin(x)

dx

0

4

4-x 1 1

0.178

= - - x + cos(x)

ln 4 4

0

0.065.

(b) Find the area of S. (solution, (b)): This area is the integral

b 1 + sin(x) - 4-x dx, a4

where b > a is the next positive solution of f (x) = g(x). This is clearly given by b = 1; therefore,

Area of S

1

1 + sin(x)

- 4-x dx

0.178 4

11

4-x 1

= x - cos(x) + -

4

ln 4 0.178

0.410.

(c) Find the volume of the solid generated when S is revolved about the horizontal line y = -1

(solution, (c)): This is easily evaluated by the method of washers:

1

Volume = (f (x) + 1)2 - (g(x) + 1)2 dx 4.559

a

2. The curve above is drawn in the xy-plane and is described by the equation in polar coordinates r = + sin(2), 0 , where r is measured in meters and is measured in radians. The derivative of r with respect to is given by dr = 1 + 2 cos(2). d

(a) Find the area bounded by the curve and the x-axis.

(solution, (a)): This is given by the integral

Area

1 =

r2 d

20

1 =

( + sin 2)2 d

20

1 =

(2 + 2 sin 2 + sin2 2) d

20

1 =

2 + 2 sin 2 + 1 - cos 4

d

20

2

1 3

1

sin 4

=

- cos 2 + sin 2 + -

23

2

2

8

0

1 3

=

- cos 2 +

= 3 - 4.382 meters2

23

20 6 4

(b) Find the angle that corresponds to the point on the curve with x-coordinate -2.

(solution, (b)): Since x = r cos , and since r = + sin 2, we are to solve the equation

( + sin 2) cos = -2 for , where, again, 0 . The relevant solution is found numerically to be 2.786 0.887 .

2 dr

(c) For < < , is negative. What does this fact say about r? What

3

3 d

does this fact say about the curve?

2

(solution, (c)): This fact says that on the interval < < , r is decreasing;

3

3

in turn, since r > 0 on this interval, this means that over this interval the

curve is approaching the origin.

(d) Find the value of in the interval 0 that corresponds to the point 2

on the curve in the first quadrant with greatest distance from the origin.

Justify your answer.

(solution, (d)): The problem is, then to maximize the function r = + sin 2

on the given interval. Since r is differentiable on its domain of definition, dr

the critical values of are precisely those values for which = 0. Since we d

dr are given that = 1 + 2 cos 2, we are to solve the equation

d

1 + 2 cos 2 = 0,

for

.

This

reduces

to

cos 2

=

-

1 2

;

the

only

relevant

solution

for

is

then

=

3

.

Finally, we have r(0) = 0,

r(

2

)

=

2

,

and

r(

3

)

=

3

+

sin

2 3

=

3

+

3 2

on the

>

2

,

and

so

interval 0

we conclude

2

.

that

=

2

affords

the

maximum

value

of

r

Distance x (cm)

0 1568

Temperature 100 93 79 62 55 T (x) ( C)

3. A metal wire of length 8 centimeters (cm) is heated at one end. The table above give selected values of the temperature T (x), in degrees Celsius ( C), of the wire

x cm from the heated end. The function T is decreasing and twice differentiable.

(a) Estimate T (7). Show the work that leads to your answer. Indicated units of measure.

(solution, (a)): Since T is known to be differentiable, we may apply the symmetric difference quotient to compute the derivative:

T (a + h) - T (a - h)

T (a) = lim

.

h0

2h

This gives a convenient approximation for T (7) :

T (7)

T (8) - T (6)

=

55 - 62

=

7 -

C

per

cm

of

length.

2

2

2

(b) Write an integral expression in terms of T (x) for the average temperature of the wire. Estimate the average temperature of the wire using a trapezoidal sum with the four subintervals indicated by the data in the table. Indicate units of measure.

(solution, (b)): First of all, the average temperature--expressed as an integral-- is given by

18

Avg Temp =

T (x) dx.

80

We may approximate this via four trapezoids constructed from the given data; the result is then

8

1

T (x) dx [(100 + 93) + 4(93 + 70) + (70 + 62) + 2(62 + 55)] 605.5,

0

2

and so the average temperature over the length of wire is given by

1 Avg temperature ? 605.5 = 75.6875.

8

8

(c) Find T (d) dx, and indicate units of measure. Explain the meaning of

0 8

T (d) dx in terms of the temperature of the wire.

0

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