Calculator Review #2

1. What is the average value of the function

on the interval [1, 3]?

A. 0.146B. 0.914C. 0.964D. 0.987E. 1.928

Use Average Value for Integrals

Ans: C

( ) sinf x x3 12

1

1( )

1sin( )

3 1Use the calculator to fi nd the value

(nI nt(sin(x (̂1/ 2)),x,1,3))/ (3-1) = 0.964

b

af x dx

b a

x dx

1

Calculator Review #2

2. Right triangle ABC has its right angle at A. Leg AB is decreasing at a constant rate of 3cm per minute. Leg AC is increasing at a constant rate of 4 cm per minute. How is the hypotenuse BC changing at the moment when AB = 12 and AC =5?

A. It is decreasing at the rate of 16 cm per minute

B. It is decreasing at the rate of 16/13 cm per minute

C. It is not changingD. It is increasing at the rate of 33/13 cm

per minuteE. It is increasing at the rate of 16 cm

per minute.

This is a derivative/related rate problem

C

a

b

A c B

2 2 2

Given: 3, 4

Find:

Equation:

Find the derivative: 2 2 2

When 12, 5, 13 (by pythagorean triple)

13 5(4) 12( 3)

20 36 1613 13

decreasing at the rate

dc dbdt dtdadt

a b c

da db dca b cdt dt dt

c b a

dadt

dadt

of 16/ 13 cm/ min

2

Calculator Review #2

3.

A. 2.163B. 2.660C. 2.780D. 5.163E. 5.660

The fundamental theorem of calculus

Ans: D

3

2I f ' ( ) (ln ) 3,what is g(3)?g x x x

3 2

2

( ) ( ) ' ( )

(ln ) 2.163

nI nt(x (ln x) 2̂, x, 2, 3)

But we know that g(2) = 3,

so g(3) = 5.163

b

af b f a f x dx

x x dx

Calculator Review #2

4. The derivative of a function f is given by f’(x) = sin (cos x) – 0.1x. How many critical points does f have on the open interval (0, 8)?

A. None B. OneC. TwoD. ThreeE. Four

Max/Min problem This function is the derivative

of f(x) Graph f’(x) then look for zeros,

that is, where the function crosses x axis.

1/2

-1/2 2 4 6 8

Ans: D

4

Calculator Review #2

5. What is the area of the region bounded by the graphs of

A. 1.523B. 2.358C. 2.493D. 4.783E. 7.409

Area - Integrals

Ans: A

5

2

and tan

1and the vertical lines x = - and x = 1?

2

xy e y x 2

2

2

1

12

Graph y = e and tan

1on [ ,1]

2

(e tan )

Use calculator:

1nI nt(e tan , , ,1) 1.523

2

x

x

x

y x

x dx

x x

Calculator Review #2

6. What is the x-value of the point at which the tangent line to the graph of

A. -0.732B. -0.589C. -0.162D. 0.236E. 0.361

Tangent Lines - Deriv

Ans: A

6

2

perpendicular

to the line 3 2 1?

xy e x

y x

2

2

2

2

2

3 2 1

2 13 3

2 3slope is , perpendicular:

3 2Find y' f or

' 2 1 <-- Slope

32 1

2Use calculator to fi nd x

F2: Algebra, Solve(

3solve( 2 1, )

2

5Find the zeros of y = 2

2F2: Algeb

x

x

x

x

x

y x

y x

y e x

y xe

xe

xe x

OR

xe

2

ra, Zeros(

5zeros(2 , )

2xxe x

Calculator Review #2

7. The region enclosed by the graphs of

and the vertical lines x = 0 and x = 2 is rotated about the y = -3. Which of the following gives the volume of the generated solid?

Volume – Integration

outer

inner

y=-3

7

1 and xy e y x

22 1 2

0

22 1 2

0

22 1 2 2

0

2 2

0

2 2

0

A. 3 ( 3)

B. 3 ( 3)

C. ( ) 3)

D. ln 2 ( 3)

E. ln 4 ( 3)

x

x

x

e

e

e x dx

e x dx

e x dx

y y dx

y y dx

2 2 2

02 1 2 2

0( 3) ( 3)x

V outer inner

V e x dx

Calculator Review #2

8. Which of the following is an equation of the tangent line to the graph of the function

at the point where f’(x) = 2?

A. y = 2x – 0.630B. y = 2x + 0.537C. y = 2x + 0.839D. y = 2x + 0.926E. y = 2x + 1.469

8

2( ) xf x e x

2( )

' ( ) 2

2 2

x

x

x

f x e x

f x e x

e x

GraphAnd y = 2Look for the points of intersection

When x = .315, find f(.315)

2( ) xf x e x

.315 2(.315) .315 1.469

1.469 2( .315)

2 .63 1.469

2 .839

:

f e

y x

y x

y x

Ans C

Calculator Review #2

9. The population of bacteria given by y(t) grows according to the equation

where k is a constant and t is measured in minutes. If y(10) = 10 and

y(30) = 25, what is the value of k?

A. -2.079

B. 0.046

C. 0.107

D. 0.125

E. 0.230

Differential Equation

Growth and Decay

9

dyky

dx

10 30

30

10

20

20

(10) 10

(30) 25

10 Ce 25

Ce 2510

2.5

ln ln2.5

20 ln2.5

ln2.50.046

20:

kt

k k

k

k

k

k

dykx y Ce

dxy

y

Ce

Cee

e

k

k

Ans B

Calculator Review #2

10. The function is continuous on the closed interval [0, 3] and has values that are given in the table below. Using the subintervals

[0, 1] [1, 2] [2, 3], what is the approximation to

6

4

2

0 1 2 3

10

3

0( )f x dx

x 0 1 2 3

f(x) 2 5 4 3 3

0f (x)dx

1(f (0)+2f (1)+2f (2)+f (3))

21

(2+2*5+2*4+3)21 23

(2+10+8+3)=2 2

Calculator Review #2

11. Because of rainfall from 9am on a given day, water enters an open reservoir at the rate of

where R(t) is measured in gallons per hour and t is measured in hours after 9am (so 0 <= t <= 12). To prevent overflow, water is pumped out of the reservoir at the constant rate of k gallons per hour. The reservoir holds 500,000 gallons of water both at 9am and 9pm.

A. What is k? Round your answer to the nearest whole number.

11

2

73000( )

8 25R t

t t

12 12

0 0

12 12

2 00

12

20

12

20

Amount added = Amount leaving

( )

73000

8 2573000

128 25

1 7300012 8 25Use calculator: k 4338

R t dt kdt

ktt t

kt t

kt t

Calculator Review #2

11. Because of rainfall from 9am on a given day, water enters an open reservoir at the rate of

where R(t) is measured in gallons per hour and t is measured in hours after 9am (so 0 <= t <= 12). To prevent overflow, water is pumped out of the reservoir at the constant rate of k gallons per hour. The reservoir holds 500,000 gallons of water both at 9am and 9pm.

B. To the nearest gallon, how much water is the reservoir at 7pm (t = 10)?

Want to find the adding water and leaving water. The combined rate is W’(t) = R(t) – k

Know: W(0)=500000

k = 4338.07

12

2

73000( )

8 25R t

t t

10

0

10

20

10

20

(10) (0) ( )

73000(10) 500000 4338.07

8 2573000

(10) 4338.07 5000008 25

(10) 506124.13 506124

W W R t kdt

W dtt t

W dtt t

W gallons

Calculator Review #2

11. Because of rainfall from 9am on a given day, water enters an open reservoir at the rate of

where R(t) is measured in gallons per hour and t is measured in hours after 9am (so 0 <= t <= 12). To prevent overflow, water is pumped out of the reservoir at the constant rate of k gallons per hour. The reservoir holds 500,000 gallons of water both at 9am and 9pm.

C. At what time t, for 0<= t <= 12, does the reservoir hold the greatest amount of water?

W’(t) = R(t) – k Graph this function and find the

zeros.t = 1.202 and t = 6.798These are between 0 and 12Determine what happens at these

points.

At x = 6.798 the function increases then decreases, so this is a max

13

2

73000( )

8 25R t

t t

2

73000'( ) 4338.07

8 25W t

t t

Calculator Review #2

12. The table below gives the velocity v(t) at selected times t of a car traveling along a straight road.

A. Use the values of the table to approximate the acceleration of the car at time t = 6. Show the work that leads to your answer and indicate units of measure.

Find the average acceleration by using

14

T 0 2 5 7 8 10

V(t) 50 55 60 70 65 75

2

(7) (5)7 5

70 60 105 /

7 5 2

v va

m hr

Calculator Review #2

12. The table below gives the velocity v(t) at selected times t of a car traveling along a straight road.

B. Use a right Riemann sum with the subintervals given in the table to approximate

Indicate units of measure. What physical quantity does this integral represent?

80

70

60

50

40

30

20

10

2 3 2 1 2

1 2 3 4 5 6 7 8 9 10

15

T 0 2 5 7 8 10

V(t) 50 55 60 70 65 75

10

0( )v t dt

10

0( ) 2 * 55 3* 60 2* 70 1* 65 2* 75

110 180 140 65 150

645 miles

v t

Calculator Review #2

12. The table below gives the velocity v(t) at selected times t of a car traveling along a straight road.

C. The function v(t) is twice differentiable on the interval [0,10]. Show that there must be a moment of time when the acceleration of he car is equal to zero.

Choose a small interval on [5,8]

V(7) = 70

V(8) = 65

V(5) = 60

V(7) is greater than this small interval of [5,8]

Intermediate value theorem: find some c within [5,7] with

v ( c ) = 65

The mean value theorem guarantees a point somewhere on [c,8] where v’ is zero.

16

T 0 2 5 7 8 10

V(t) 50 55 60 70 65 75