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