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Page history last edited by Darren Kuropatwa 7 months, 1 week ago

AP CALCULUS

 

Ara's FINAL TAKE HOME EXAMINATION.

 

1.) Let f be the function given by f(x) = (x^3/4) - (x^2/3) - (x/2) + (3cos(x)). Let R be the shaded region in the second quadrant by the graph of f, and let S be the shaded region bounded by the graph of f and line l, the line tangent to the graph of f at x = 0.

 

a. Find the area of R.

 

 

Using the graphing calculator, i plugged in the function **f( x ) = (x^3 / 4) - (x^2 / 3) - (x / 2) + 3cos(PI * x). Then, I looked for its root between the interval -2 < x < -1. Doing this, I acquired the root x = -1.373122. 

So, the integral of f( x ) from the interval -1.373122 to 0 is equal to 2.9030944, which is the R region. 

 

 

b. Find the volume of the solid generated when R is rotated about the horizontal line y = -2**.

 

 

When R is rotated about the horizontal line y = -2, the volume would be :

PI * THE INTEGRAL from -1.373122 to 0 OF ( f(x) + 2 )^2 - 4 dx

PI * THE INTEGRAL from -1.373122 to 0 OF ( [ (x^3 / 4) - (x^2 / 3) - (x / 2) + 3cos(PI * x) ] + 2 )^2 - 4 dx

plugging the integral in the calculator, we get :

PI * 28.4656 = 89.4273 is the volume generated when R is rotated about y = -2.

 

 

c. Write, but do not evaluate an integral expression that can be used to find the area of S.

 

 

f(0) = 3

f'(0) = -.4999998 SO, the equation of the tangent line is y = -.4999998x + 3.

Knowing this, the intersection point of f(x) and the tangent line is at x = 3.3898678

So, the expression would be:

the INTEGRAL of (-.4999998x + 3) - f(x) from the interval 0 to 3.3898678.

 

 

 

2.) Let f be the function defined for x > 0 with f(0) = 5 and f', the first derivative of f, given by f'(x) = [e^(-x/4)] sin(x^2).

 

a. Use the graph of f' to determine whether the graph of f is concave up, concave down, or neither on the interval 1.7 < x < 1.9. Explain your reasoning.

 

 

On the interval 1.7 < x < 1.9, f is concave down because the values of f'' on the interval 1.7 < __x __< 1.9 (using the calculator) are negative numbers.

 

 

b. On the interval 0 < x < 3, find the value of x at which f has an absolute maximum. Justify your answer.

 

 

the roots of f(x) are x = 0, x = 1.7725 and x = 2.5066.

since we are looking for the absolute maximum, we can only consider the roots 0 and 1.7725 because at x = 2.5066, f(x) has a minimum.

Therefore, the absolute maximum on the interval 0 to 3 is located at x = 1.7725

 

 

c. Write an equation for the line tangent to the graph of f at x = 2.

 

 

To get the slope we calculate :

f'(2) = [ e^(-2 / 4) ] * sin (4)

f'(2) = [ e^(-1 / 2) ] * (.0698)

f'(2) = (.6065) (.0698)

slope at x = 2 is .0423

Now, i need to find f(2). I am given f(0) = 5 and the fundamental theorem of calculus states that the integral from a to b is equal to its antiderative at point b minus the antiderivative at point a. SO,

THE INTEGRAL from 0 to 2 of f'(x) dx is equal to .0321

Now, we know f(0) is 5 so, F(b) - 5 is equal to .0321, therefore, f(2) is equal to 5.0321

Using the coordinates ( 2, 5.0321) and slope = .0423, we get the equation :

y - 5.0321 = .0423 ( x - 2 )

y = .0423x - .0846 + 5.0321

The equation of the line tangent to x = 2 at f is

y = .0423x + 4.9475

 

 

3.) The graph of x models the height of a skateboard ramp and meets the following requirements:

 

 

(i) at x = 0, the value of the function is 0, and the slope of the graph of the function is 0.

(ii) at x = 4, the value of the function is 1, and the slope of the graph of the function is 1.

(iii) between x = 0 and x = 4, the function is increasing.

 

a. Let f(x) = ax^2, where a is a nonzero constant. show that it is NOT POSSIBLE to find a value for a so that f meets requirement (ii) above.

 

 

f(4) = a (4)^2

1 = a(16) SO, a = 1/16

f'(x) = 2a(x)

f'(4) = 2 a (4)

1 = 2 (1/16) (4)

(1/2) is not equal to 1

 

 

b. Let g(x) = (cx^3) - (x^2/16), where c is a nonzero constant. Find the value of c so that g meets the requirement (ii) above. show the work that leads to your answer.

 

 

g(x) = (c x^3) - (x^2 / 16)

g(4) = c(64) - (16 / 16)

1 = c(64) - 1

(2 / 64) = c

c = 1 / 32

g'(x) = (3 c x^2) - (x / 8)

g'(4) = [3 (1 / 32) (16)] - (1 / 2)

1 = (3 / 2) - (1 / 2)

1 = 1

 

 

c. Using the function g and your value of c from part (b), show that g does not meet requirement (iii) above.

 

 

g(x) = [(1 / 32)(x^3)] - (x^2 / 16)

g'(x) = (3x^2 / 32) - (x / 8) plugging this function in the calculator, we can see that the graph of g'(x) on the interval 0 < x < 4 are negatively valued, therefore, g is decreasing, defying requirement (iii).

 

d. let h(x) = (x^n/k), where k is a nonzero constant and n is a positive integer. Find the values of k and n so that h meets requirement (ii) above. show that h also meets requirements (i) and (iii) above.

 

 

h(x) = (x^n) / k

h(4) = (4^n) / k

1 = (4^n) / k

k = 4^n

256 = 4^4

n = 4 , k = 256

h'(x) = (n / k)(x ^ n-1)

h'(4) = (4 / 256) (4 ^ 3)

1 = (1 / 64) (64)

1 = 1

h(0) = 0 / k

h(0) = 0

 

h'(0) = (4 / 256) (0)

h'(0) = 0

Plugging the function h'(x) in the calculator, not using 4 as x, we can see that the values of the graph are positively valued. Therefore, with k = 256 and n = 4, h meets requirement (iii).

 

 

4.) The rate, in calories per minute, at which a person using an exercise machine burns calories is modeled by the function f. The graph, f(t) = (-t^3/4) + (3t^2/2) + 1 for 0 < t__ <__ 4 and f is piecewise linear for 4 < t < 24.

 

a. find f'(22). Indicate units of measure.

 

 

f'(22) = (3 - 15) / (24 - 20)

f'(22) = -3 calories per minute ^2

 

b. for the time interval 0 < t < 24, at what time t is f increasing at its greatest rate? show the reasoning that supports your answer.

 

 

Logically speaking, there was no increase between the intervals, (4,12) and (16, 24). So, for the interval (0,4),

f'(t) = [(-3/4) (t^2)] + 3t

0 = [(-3/4) (t^2)] + 3t

t = 2 ON THE INTERVAL (0,4), THE GREATEST RATE IS AT T = 2.

f'(2) = [(-3/4) (2^2)] + 3(2)

f'(2) = 2 (greatest increase rate)

For the interval (12,16) :

since the graph is linear, my estimation of the greatest slope would be 1.5.

Therefore, the greatest rate of increase in the graph of f is at t = 2 which is 2.

 

 

c. Find the total number of calories burned over the time interval 6 < t < 18 minutes

 

 

Looking at the graph :

On the interval (6,12), 6 * 9 = 56 calories are burned

On the interval (12,16), (4 * 9) + [(4 * 6) / 2] = 48 calories are burned

On the interval (16,18), 2 * 15 = 30 calories are burned

So from t = 6 to t = 18, 56 + 48 + 30 = 134 calories are burned

 

 

d. The setting on the machine is now changed so that the person burns f(t) + c calories per minute. For this setting, find c so that an average of 15 calories per minute is burned during the time interval 6 < t < 18.

 

 

To obtain an average of 15 calories, the total calories burned between the intervals (6,18) should be 180, that is :

18 - 6 = 12 , 12 * 15 = 180

Changing the y intercept of f will also change the total calories burned. So to come up with a 180 total on the (6,18) interval, i used the trial and error method by adding a step or a no. of steps until i reach the total 180. And for that i have acquired:

f(t) = (-t^3 / 4) + (3t^2 / 2) + 5 making the graph 4 units higher than the original graph.

On the interval (6,12), 6 * 13 = 78 calories are burned

On the interval (12,16), (4 * 13) + [(4 * 6) / 2] = 64 calories are burned

On the interval (16,18), 2 * 9 = 38 calories are burned

78 + 64 + 38 = 180 / 12 = 15 calories are burned per minute on the interval (6,18)

So c = 5

 

 

5.) Consider the differential equation **dy/dx = ( y - 1 )^2 cos ( PI * x ).

 

a. sketch of the slopefield

 

 

 

b. there is a horizontal line with equation y = c** that satisfies this differential equation. find the value of c.

 

 

horizontal line means the derivative is equal to zero, that is :

dy/dx = 0 SO,

0 = (y - 1)^2 [ cos (PI * x)

Since multiplication is involved here, whatever equation multiplied to 0 will always be zero, SO

(y - 1)^2 = 0

y = 1

Therefore y = 1 satisfies the differential equation, making c = 1

 

 

c. Find the particular solution y = f(x) to the differential equation with the initial condition f(1) = 0

 

 

dy/dx = (y - 1)^2 cos(PI*x)

INTEGRAL OF dy/(y - 1)^2 = INTEGRAL OF cos(PI*x)

ln (y - 1)^2 = (1 / PI) (sin (PI*x))

(y - 1)^2 = e ^ [(1 / PI) sin (PI*x)]

y - 1 = √e^ [(1 / PI) sin (PI*x)]

y = ( √e^ [(1 / PI) sin (PI*x)] ) + 1

y (1) = √e^ [(1 / PI) sin (PI)] + 1

y (1) = ( √e^ [.0174] ) + 1

y (1) = 1.0088 + 1

y (1) = 2.0088

With the condition y (1) = 0, then

y = ( √e^ [(1 / PI) sin (PI*x)] ) + 1 - 2.0088

**y = ( √e^ [(1 / PI) sin (PI*x)] ) - 1.0088 

 

 

6.) A car travels on a straight track. during the time interval 0 < t < 60 seconds, the car's velocity v, measured in feet per second, and acceleration a, measured in feet per second per second, are continuous functions. The table given shows selected values of these functions.

 

a. Using appropriate units, explain the meaning of THE INTEGRAL FROM 30 TO 60 OF v(t) dt in terms of the car's motion. Approximate (THE INTEGRAL FROM 30 TO 60 OF v(t) dt)** using trapezoidal approximation with three subintervals determined by the table.

 

 

THE INTEGRAL FROM 30 TO 60 OF v(t) dt is the distance traveled by the car from times t = 30 to t = 60.

LEFTHAND SUM:

(10 * 14) + (10 * 10) + (10 * 0) = 240

RIGHTHAND SUM:

(10 * 10) + (10 * 0) + (10 * 10) = 200

( 240 + 200 ) / 2 = 220

 

b. Using appropriate units, explain the meaning of THE INTEGRAL FROM 0 TO 30 OF a(t) dt in terms of the car's motion. Find the exact value of THE INTEGRAL FROM 0 TO 30 OF a(t) dt.

 

 

THE INTEGRAL FROM 0 TO 30 OF a(t) dt is the velocity of the car between times t = 0 and t = 30 measured in feet per second.

Since we are given exact values for the car's velocity, it will be easy to calculate the exact velocity from t = 0 to t = 30. Plotting the values in the graph, i have acquired the measures :

from 0 to 15, the measure is 45

from 15 to 25, the measure is 35

from 25 to 30, the measure is 7.5

So the velocity from t = 0 to t = 30 is 87.5 feet per second.

 

 

c. for 0 < t < 60, must there be a time t when v(t) = -5? Justify your answer.

 

 

Yes. Because in the table given, a change of v of negative to positive happened between times 35 and 50. Therefore, it is just right to say that there has been a velocity of -5.

 

 

d. For 0 < t < 60, must there be a time t when a(t) = 0? Justify your answer.

 

 

Yes. As we can see, from the interval 0 to 25, the velocity function had a minimum point. from its initial velocity -20, to -30 and back to -20 at time t = 15, clearly there has been a minimum point. and what does this indicate?

 

This means that there is a root in the function a(t) at t = 15.

therefore, yes, there is a time t where a(t) is equal to zero.

 

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