Answer Key for the Review Packet for Exam #3
Professor Danielle Benedetto – Math 11
Max-Min Problems
1. Show that of all rectangles with a given area, the one with the smallest perimeter is a square.
y
x
A
Let A equal the given area. We know xy = A is fixed, so that y = .
x
A
Then the perimeter P = 2x + 2y = 2x + 2 must be minimized. The (common-sensex
bounds)domain of P is {x : x > 0}.
√
2A
Next P ′ = 2 − 2 . Setting P ′ = 0 we solve for x = + A. (We take the positive square root
x
here since we are talking about positive lengths).
Sign-testing the critical number does indeed yield a minimum for the perimeter function.
P′ ⊖ ⊕
√
P ց Aր
MIN √
√
A
Since x = A then y = √ = A. As a result, x = y and the smallest perimeter occurs
A
when the rectangle is a square .
2. A rectangle lies in the first quadrant, with one vertex at the origin, two sides along the
coordinate axes, and the fourth vertex on the line x + 2y − 6 = 0. Find the maximum area
of the rectangle.
3HH x + 2y − 6 = 0
H (x, y)
Ht
HyH
HH
x
6
The vertex point (x, y) lies on the line x + 2y − 6 = 0. The line x + 2y − 6 = 0 can be rewritten
x
as y = 3 − .
2
x2
x
= 3x −
and must be maximized.
The area of the rectangle is given as A = xy = x 3 −
2
2
The (common-sense-bounds)domain of A is {x : 0 ≤ x ≤ 6}.
Next, A′ = 3 − x. Setting A′ = 0 we solve for x = 3 as the critical number.
Sign-testing the critical number does indeed yield a maximum for the area function.
A′ ⊕
A
⊖
ր3 ց
MAX
1
3
9
3
Finally, x = 3 =⇒ y = . As a result the maximum area is Area=xy = 3 · =
.
2
2
2
3. A farmer wants to use a fence to surround a rectangular field, using an existing stone wall as
one side of the plot. She also wants to divide the field into 5 equal pieces using fence parallel
to one side. The farmer must use exactly 1200 feet of fence. What is the maximum area
possible for this field?
x
y y y y y y
wall
We know that the length of fence used L = x + 6y = 1200 is fixed so that x = 1200 − 6y.
Then the area A = xy = (1200 − 6y)y = 1200y − 6y 2 must be maximized.
The (common-sense-bounds)domain of A is {y : 0 ≤ y ≤ 200}.
Next A′ = 1200 − 12y. Setting A′ = 0 we solve for y = 100.
Sign-testing the critical number does indeed yield a maximum for the area function.
A′ ⊕
⊖
ր100
ց
MAX
Since y = 100 then x = 1200 − 6(100) = 600. As a result, the maximum area possible is
60, 000 square feet.
A
4. You work for a soup manufacturing company. Your assignment is to design the newest can
in the shape of a cylinder. You are given a fixed amount of material, 600 cm2 , to make your
can. What are the dimensions of your can which will hold the maximum volume of soup?
2πr
r - h
r
r
top
bottom
h
side of can
that the amount of material used M = 2πr 2 + 2πrh = 600 is fixed so that
We know
h=
600 − 2πr2
.
2πr
Then the volume V =
maximized.
πr2 h
=
πr2
600 − 2πr2
2πr
=r
The (common-sense-bounds)domain of V is {r : 0 < r ≤
600 − 2πr2
2
q
= 300r − πr3 must be
300
π }.
10
Next V ′ = 300 − 3πr2 . Setting V ′ = 0 yields r = √ as the critical number.
π
Sign-testing the critical number does indeed yield a maximum for the volume function.
2
V′ ⊕
V
⊖
10
ր√πց
MAX
600 − 2π( √10π )2
10
600 − 200
20
√
Since r = √ then h =
=
=√ .
10
√
π
20 π
π
2π( π )
10
20
The dimensions of the can with the largest volume are r = √ and h = √ in cm.
π
π
A bit of extra work does indeed check that the amount of material used is 600 square cm.
M = 2πr2 + 2πrh = 2π( √10π )2 + 2π( √10π )( √20π ) = 200 + 400 = 600.
5. A rectangular box with square base cost $ 2 per square foot for the bottom and $ 1 per square
foot for the top and sides. Find the box of largest volume which can be built for $36.
y
x
x Then the Cost of materials, which is fixed is given as
Cost = cost of base + cost of top + cost of 4 sides
= x2 ($2) + x2 ($1) + 4xy($1)
= 3x2 + 4xy = 36
36 − 3x2
=⇒ y =
4x
3
36 − 3x2
2
2
= 9x − x3 must be maximized.
Then the volume of the box V = x y = x
4x
4
√
The (common-sense-bounds)domain of V is {x : 0 < x ≤ 12}.
9
Next V ′ = 9 − x2 . Setting V ′ = 0 we solve for x = 2 as the critical number.
4
Sign-testing the critical number does indeed yield a maximum for the volume function.
V′ ⊕
⊖
ր2 ց
MAX
36 − 12
Since x = 2 then y =
= 3. As a result, the box of largest volume will measure
8
2x2x3 , each in feet.
V
6. Among all the rectangles with given perimeter P, find the one with the maximum area.
y
x
P − 2x
.
Let P equal the given perimeter. We know 2x + 2y = P is fixed, so that y =
2
3
Then the area A = xy = x
P − 2x
2
=
P
x − x2 must be maximized.
2
The (common-sense-bounds)domain of A is {x : 0 ≤ x ≤
P
2 }.
P
P
− 2x. Setting A′ = 0 we solve for x = .
2
4
Sign-testing the critical number does indeed yield a maximum for the area function.
Next A′ =
P′ ⊕
P
ր4
⊖
ց
MAX
P − 2( P4 )
P
P
Since x =
then y =
= . As a result, x = y and the largest area occurs when
4
2
4
the rectangle is a square .
P
7. Consider a cone such that the height is 6 inches high and its base has diameter 6 in. Inside this
Lcone we inscribe a cylinder whose base lies on the base of the cone and whose top intersects
L cone in a circle. What is the maximum volume of the cylinder?
the
L
L
A
L
6 − hAA
r -L
L
rA
L $A
'
h
L
A
h
L
A
&
%
3
r
6−h
Using similar triangles, for the cross slice of the cone and cylinder, we see =
which
3
6
implies that 6r = 18 − 3h =⇒ h = 6 − 2r.
Then the volume of the cylinder, given by, V = πr2 h = πr2 (6 − 2r) = 6πr2 − 2πr3 must be
maximized.
The (common-sense-bounds)domain of V is {r : 0 ≤ r ≤ 3}.
Next V ′ = 12πr − 6πr2 . Setting V ′ = 0 we see 6πr(2 − r) = 0 and solve for r = 0 or r = 2 as
critical numbers. Of course r = 0 will not lead to a maximum since no cylinder exists there.
Sign-testing the critical number r = 2 does indeed yield a maximum for the volume function.
V′ ⊕
⊖
ր2 ց
MAX
Since r = 2, then h = 2, and the maximum volume V = π(2)2 2 = 8π. As a result the
maximum volume of the cylinder is 8π cubic inches.
V
8. Consider the right triangle with sides 6, 8 and 10. Inside this triangle, we inscribe a rectangle
such that one corner of the rectangle is the right angle of the triangle and the opposite corner
of the rectangle lies on the hypotenuse. What is the maximum area of the rectangle?
4
6H
6 − y HHH10
H
yHH
H
8
x
6−y
6
=
which implies that 48 − 8y =
x
8
6x =⇒ 8y = 48 − 6x =⇒ y = 6 − 68 x. Another way to think about this is that this picture
corresponds to the line with y-intercept 6 with slope − 86 If you draw the triangle with width
6 and height 8 instead, the math all still works out the same in the end.
6
The area A = xy = x 6 − 86 x = 6x − x2 must be maximized.
8
The (common-sense-bounds)domain of A is {x : 0 ≤ x ≤ 8}.
Using similar triangles we find the relationship
Next A′ = 6 − 32 x. Setting A′ = 0 we solve for x = 4 as the critical number.
Sign-testing the critical number does indeed yield a maximum for the area function.
A′ ⊕
⊖
ր4 ց
MAX
Finally, x = 4 =⇒ y = 6 − 86 (4) = 3 As a result the maximum area is Area=xy = 4(3) = 12
square units.
A
9. A toolshed with a square base and a flat roof is to have volume of 800 cubic feet. If the
floor costs $6 per square foot, the roof $2 per square foot, and the sides $5 per square foot,
determine the dimensions of the most economical shed.
y
x
x 800
We know the volume of the toolshed is given by V = x2 y = 800 is fixed, so that y = 2 .
x
Then the Cost of materials, which must be minimized, is given as
Cost = cost of floor + cost of top + cost of 4 sides
= x2 ($6) + x2 ($2) + 4xy($5)
= 8x2 + 20xy
800
2
= 8x + 20x
x2
16000
= 8x2 +
x
The (common-sense-bounds)domain of Cost is {x : x > 0}.
16000
. Setting Cost′ = 0 we solve x3 = 1, 000 =⇒ x = 10.
Next Cost′ = 16x −
x2
Sign-testing the critical number does indeed yield a maximum for the area function.
5
′
Cost⊖
⊕
Costց10ր
MIN
800
Since x = 10 then y =
= 8. As a result, the most economical shed has dimensions
(10)2
10x10x8 , each in feet.
10. A rectangular sheet of metal 8 inches wide and 100 inches long is folded along the center to
form a triangular trough. Two extra pieces of metal are attached to the ends of the trough.
The trough is filled with water.
100
@
@
4@@4
(a). How deep should the trough be to maximize the capacity of the trough?
(b). What is the maximum capacity?
x
@
@
x
2
h
4@@
4
@
We will maximize the volume of the trough, by maximizing the area of the metal pieces on
the ends of the trough.
q
2 q
2
By the Pythagorean Thrm., h = 16 − x2 = 16 − x4
Then the area, which must be minimized, is given as
1
A = base · height
2
1
= xh
2 r
1
x2
= x 16 −
2
4
The (common-sense-bounds)domain of A is {x : 0 ≤ x ≤ 8}.
Next,
A′
r
x
x2 1
+ 16 −
−
2
4 2
x2
4
r
x2
1
x2
=− q
+
16 −
2
2
4
8 16 − x4
1
1
= x q
2 2 16 −
6
Setting A′ = 0 implies
r
1
x2
x2
q
=
16 −
2
2
4
8 16 − x4
so that
q
2
x =4
16 −
x2
4
2
which implies
2
x2 = 4 16 − x4 = 64 − x2 .
As a result, x2 = 32 and finally x =
√
√
32 = 4 2.
Sign-testing the critical number does indeed yield a maximum for the area function.
A′ ⊕ ⊖
√
A ր 32ց
q
MAX √
√
2
Since x = 32 maximizes the area, then the corresponding height is h = 16 − ( 32)
=
4
√
√
√
8 = 2 2. The trough should be 2 2 inches deep. The maximum capacity is 800 cubic
√
√
inches because Max Volume=(Max Area of end panel)(100)= 12 (4 2)(2 2)(100) = 50(8)(2) =
800 .
11. A manufacturer wishes to produce rectangular containers with square bottoms and tops, each
container having a capacity of 250 cubic inches. The material costs $2 per square inch for
the sides. If the material used for the top and bottom costs twice as much per square inch as
the material for the sides, what dimensions will minimize the cost?
y
x
x 250
We know the volume of the toolshed is given by V = x2 y = 250 is fixed, so that y = 2 .
x
Then the Cost of materials, which must be minimized, is given as
Cost = cost of floor + cost of top + cost of 4 sides
= x2 ($4) + x2 ($4) + 4xy($2)
= 8x2 + 8xy
250
= 8x2 + 8x
x2
2000
= 8x2 +
x
The (common-sense-bounds)domain of Cost is {x : x > 0}.
2000
2000
Next Cost′ = 16x − 2 . Setting Cost′ = 0 we solve x3 =
= 125 =⇒ x = 5.
x
16
Sign-testing the critical number does indeed yield a minimum for the Cost function.
7
′
Cost⊖
⊕
Costց5 ր
MIN
250
Since x = 5 then y =
= 10. As a result, the dimension that will minimize the cost are
(5)2
5x5x10 in inches.
12. An outdoor track is to be created in the shape shown and is to have perimeter of 440 yards.
Find the dimensions for the track that maximize the area of the rectangular portion of the
field enclosed by the track.
r
x
r
We know the perimeter of the track is given by P = 2x + 2πr = 440 is fixed, so that
440 − 2πr
x=
= 220 − πr.
2
Then the Area of the rectangular portion of the field, which must be minimized, is given as
A = 2rx
= 2r(220 − πr)
= 440r − 2πr2
The (common-sense-bounds)domain of A is {r : 0 ≤ r ≤ 220
π }.
110
.
Next, A′ = 440 − 4πr. Setting A′ = 0, we solve for r =
π
Sign-testing the critical number does indeed yield a maximum for the area function.
A′ ⊕
A
⊖
110
ր πց
MAX110
, then x = 220 − π
Since r =
π
110
π
= 110.
The dimensions that maximize the area of the rectangular portion are x = 110 and r =
110
π
in yards.
13. Show that the entire region enclosed by the outdoor track in the previous example has maximum area if the track is circular.
From the previous problem, we still have x = 220 − πr.
Now the entire area enclosed by the track, which must be maximized, is given by
8
A = πr2 + 2rx
= πr2 + 2r(220 − πr)
= πr2 + 440r − 2πr2
= −πr2 + 440r
The (common-sense-bounds)domain of A is {r : 0 ≤ r ≤ 200
π }.
220
Next, A′ = 440 − 2πr. Setting A′ = 0, we solve for r =
.
π
Sign-testing the critical number does indeed yield a maximum for the area function.
A′ ⊕
A
⊖
220
ր πց
MAX
220
= 0 , in yards, resulting in a circular track.
, then x = 220 − π 220
π
π
Initial Valued Differential Equations
Since r =
14. Find a function f (x) that satisfies f ′′ (x) = 12x2 + 5, f ′ (1) = 5 and passes through the point
(1, 3).
Antidifferentiating yields f ′ (x) = 4x3 + 5x + C1 . The initial condition f ′ (1) = 5 yields
4 + 5 + C1 = 5 =⇒ C1 = −4 =⇒ f ′ (x) = 4x3 + 5x − 4.
5
Antidifferentiating one last time yields f (x) = x4 + x2 − 4x + C2 . The other initial condition
2
5
with f passing through the point (1, 3) implies f (1) = 3. As a result, 1 + − 4 + C2 = 3 =⇒
2
7
7
5
5
C2 = 6 − = . Finally, f (x) = x4 + x2 − 4x +
2
2
2
2
√
15. Find a function f (x) that satisfies f ′ (x) = 3 x and f (4) = 20.
3
3
Antidifferentiating yields f (x) = 2x 2 +C. The initial condition f (4) = 20 implies 2(4) 2 +C =
3
20 =⇒ C = 4. As a result, f (x) = 2x 2 + 4 .
16. Find a function f (x) that satisfies f ′′ (x) = x + sin x, f ′ (0) = 6 and f (0) = 4.
x2
− cos x + C1 . The initial condition f ′ (0) = 6 implies
2
x2
0 − cos 0 + C1 = 6 =⇒ C1 = 7. Then, f ′ (x) =
− cos x + 7
2
x3
−sin x+7x+C2 . The other initial condition
Antidifferentiating one last time yields f (x) =
6
x3
with f (0) = 4 implies 0−sin 0+0+C2 = 4 =⇒ C2 = 4. As a result, f (x) =
− sin x + 7x + 4 .
6
Antidifferentiating yields f ′ (x) =
17. Find a function f (x) that satisfies f ′′ (x) = 2 − 12x, f (0) = 9 and f (2) = 15.
Antidifferentiating yields f ′ (x) = 2x − 6x2 + C1 . Antidifferentiating one last time yields
f (x) = x2 − 2x3 + C1 x + C2 . The first initial condition f (0) = 9 implies C2 = 9 so f (x) =
x2 − 2x3 + C1 x + 9. The second initial condition f (2) = 15 implies 4 − 16 + 2C1 + 9 = 15 =⇒
2C1 = 18 =⇒ C1 = 9. As a result f (x) = x2 − 2x3 + 9x + 9 .
9
18. Find a function f (x) that satisfies f ′′ (x) = 20x3 + 12x2 + 4, f (0) = 8 and f (1) = 5.
Antidifferentiating yields f ′ (x) = 5x4 + 4x3 + 4x + C1 . Antidifferentiating one last time yields
f (x) = x5 + x4 + 2x2 + C1 x + C2 . The first initial condition f (0) = 8 implies C2 = 8 so
f (x) = x5 +x4 +2x2 +C1 x+8. The second initial condition f (1) = 5 implies 1+1+2+C1 +8 =
5 =⇒ C1 = −7. As a result f (x) = x5 + x4 + 2x2 − 7x + 8 .
Area and Riemann Sums
Z 1
x dx using Riemann Sums.
19. Evaluate
−1
2
1 − (−1)
= and xi = −1 + i
Here a = −1, b = 1, ∆x =
n
n
Z
1
−1
x dx = lim
n→∞
n
X
i=1
2i
2
= −1 + .
n
n
n
X
2i 2
f (xi )∆x = lim
f −1 +
n→∞
n n
i=1
= lim
n→∞
n X
i=1
2i
−1 +
n
2
n
n
n
i=1
i=1
2X
2 X 2i
−1 +
n
n
n
= lim
n→∞
!
!
n
4 X
2
i
· (−n) + 2
n
n
= lim
n→∞
i=1
= lim
2
4 n(n + 1)
· (−n) + 2
n
n
2
= lim
4 n
−2 +
2 n
= lim
1
−2 + 2 · 1 · 1 +
n
n→∞
n→∞
n→∞
= −2 + 2 = 0
20. Evaluate
Z
0
2
x2 − 5x dx using Riemann Sums.
2−0
2
Here a = 0, b = 2, ∆x =
= and xi = 0 + i
n
n
10
2i
2
= .
n
n
n+1
n
Z
0
2
2
x − 5x dx = lim
n→∞
n
X
i=1
n
X
2i 2
f (xi )∆x = lim
f
n→∞
n n
i=1
2
!
n
X
2i
2
2i
= lim
−5
n→∞
n
n
n
i=1
n
n
i=1
i=1
2 X 10i
2 X 4i2
−
n
n2
n
n
= lim
n→∞
!
n
n
8 X 2 20 X
i − 2
i
n3
n
= lim
n→∞
i=1
= lim
n→∞
!
i=1
8 n(n + 1)(2n + 1) 20 n(n + 1)
·
− 2
n3
6
n
2
n+1
20 n n + 1
2n + 1
8 n
−
= lim
n→∞ 6 n
n
n
2 n
n
= lim
n→∞
=
21. Use Riemann Sums to estimate
Z
1
1
1
1
4
(1) 1 +
2+
− 10(1) 1 +
3
n
n
n
22
8
8 30
− 10 = −
= −
3
3
3
3
x2 + 1 dx using 4 equal-length subintervals and right
0
endpoints.
Z 1
x2 + 1 dx ≈ f (x1 )∆x + f (x2 )∆x + f (x3 )∆x + f (x4 )∆x
0
1 1
1 1
3 1
1
=f
+f
+f
+ f (1) ·
4 4
2 4
4 4
4
1
5 1
25 1
+
+2·
4 4
16 4
4
17
16
1
=
4
17 20 25 32
+
+
+
16 16 16 16
1
4
94
16
=
=
1
+
4
=
47
94
=
64
32
22. Sand is added to a pile at aZ rate of 10 + t2 cubic feet per hour for 0 ≤ t ≤ 8. Compute the
8
10 + t2 dt using 4 subintervals and the left endpoint of each
Riemann Sum to estimate
0
subinterval. Finally, what two things does this Riemann Sum approximate?
11
Z
0
8
10 + t2 dt ≈ f (t0 )∆t + f (t1 )∆t + f (t2 )∆t + f (t3 )∆t
= f (0) · 2 + f (2) · 2 + f (4) · 2 + f (6) · 2
= 2(10 + 14 + 26 + 46) = 2(96) = 192
This Riemann sum approximates at least two things: the area under the curve y = 10+t2 from
t = 0 to t = 8, that is the definite integral. By the Net Change Thrm,
Z it also approximates
8
rate of change dt =
the net change of sand from time t = 0 to time t = 8. Recall,
0
Amount Sand(t = 8) − Amount Sand(t = 0).
Z 4
x − 1 dx using three different methods: (a) using Area interpretations of the
23. Compute
1
definite integral, (b) Fundamental Theorem of Calculus, and (c) Riemann Sums.
1 |{z} 4
3
3
1
9
1
(base) (height) = (3)(3) =
2
2
2
Z 4
4
1
x2
1
9
− x = (8 − 4) − ( − 1) = 4 + =
x − 1 dx =
(b)
2
2
2
2
1
(a) Area=
1
(c)
3
4−1
= and xi = 1 + i
Here a = 1, b = 4, ∆x =
n
n
12
3i
3
=1+ .
n
n
Z
4
1
x − 1 dx = lim
n→∞
n
X
i=1
n
X
3i 3
f (xi )∆x = lim
f 1+
n→∞
n n
i=1
n X
3
3i
= lim
−1
1+
n→∞
n
n
i=1
n
= lim
n→∞
3 X 3i
n
n
i=1
= lim
n→∞
!
!
n
9 X
i
n2
i=1
9 n(n + 1)
n2
2
9 n n + 1
= lim
n→∞ 2 n
n
= lim
n→∞
9
1
= lim (1) 1 +
n→∞ 2
n
=
24.
25.
26.
27.
28.
9
2
Differentiation Compute each of the following derivatives:
Z x
d
−t − 1 dt = −x − 1
dx 1
Z 7
Z x
d
d
1 − sin t dt = −
1 − sin t dt = −(1 − sin x) = sin x − 1
dx x
dx 7
Z sin x p
p
d
1 − t2 dt =
1 − sin2 x(cos x)
dx 0
Z 2
Z 3x
d
d
cos t dt = −
cos t dt = − cos(3x)(3) = −3 cos(3x)
dx 3x
dx 2
Z 7
7t2 + sin t dt.
Find g ′′ (x) if g(x) =
3x
d
dx
3 sin(3x)
g ′ (x) =
Z
7
3x
7t2 + sin t dt = −
d
dx
Z
3x
7
7t2 + sin t dt = −(7(3x)2 + sin(3x))(3) = −189x2 −
g ′′ (x) = −378x − 9 cos(3x)
13
Z x
Z x
√
√
sec2 x
d √
3
3
3
√
= tan x (x3 − 3) +
29.
tan x ·
t − t dt = tan x (x − 3) +
t − t dt
dx
2 tan x
0
0
4
x 4
2
2
2
√
t
t sec x
sec2 x
3 − 3) + x − x
√
√
tan
x
(x
−
=
4
2 2 tan x
4
2
2 tan x
0
x !2
Z x
3
Z x
2
d
t2
30.
(x + sin x) =
t + sin t dt
− cos t
= 3
t + sin t dt (x + sin x) = 3
dx
2
0
0
0
2
2
2
2
x
x
3
− cos x − (0 − cos 0) (x + sin x) = 3
− cos x + 1 (x + sin x)
2
2
Z x2
d
31. Find
3t − 1 dt using two different methods.
dx 1
Z x2
d
(a).
3t − 1 dt = (3x2 − 1)(2x) = 6x3 − 2x We used the FTC Part I.
dx 1

x2 
Z x2
d
d 3 2
d
3 4
3

2
=
3t − 1 dt =
(b).
t − t
x −x −
−1
= 6x3 − 2x
dx 1
dx 2
dx
2
2
1
We used FTC Part II.
Z 0
d
32. Find
t + 3 sin t dt using two different methods.
dx x
Z 0
Z x
d
d
(a).
t + 3 sin t dt = −
t + 3 sin t dt = −(x + 3 sin x) We used the FTC Part I.
dx x
dx 0

0 
2
Z 0
d
d  t2
d
x

=
t + 3 sin t dt =
(0 − 3 cos 0) −
(b).
− 3 cos t
− 3 cos x
=
dx x
dx 2
dx
2
x
−x − 3 sin x We used FTC Part II.
Z x
f (t)dt = 2x − 2
33. Find f (x) if
1
Differentiating both sides yields:
34. Find f (x) if
Z
0
d
dx
Z
x
f (t)dt =
1
d
(2x − 2). That is, f (x) = 2 .
dx
f (t)dt = x2 .
x
Z 0
d
d 2
Differentiating both sides yields:
f (t)dt =
x or
dx x
dx
Z x
d 2
d
f (t)dt =
x . That is, −f (x) = 2x ⇒ f (x) = −2x .
−
dx 0
dx
√
35. Differentiate y = (sin x)e
√
y ′ = (sin x)e
x+2
x+2
√
1
√
+ e x+2 (cos x)
2 x+2
14
x
36. Differentiate y = (x − e− cos x ) · e 2
x
x
1
′
−
cos
x
+ e 2 (1 − e− cos x (sin x))
y = (x − e
) · e2
2
x
√
37. Differentiate y = ee · cos(e x )
√
√
√
1
x
x
′
e
x
x
√
− sin(e )e
y =e
+ cos(e x )ee ex
2 x
38. Differentiate y =
y′ =
3xe−3x − 1 + e−3x
x(−e−3x (−3)) − (1 − e−3x )(1)
=
x2
x2
39. Differentiate y =
y′ =
1 − e−3x
x
1 + e−2x
1 − e7x
−2e−2x + 9e5x + 7e7x
(1 − e7x )e−2x (−2) − (1 + e−2x )(−e7x (7)
−2e−2x + 2e5x + 7e7x + 7e5x
=
=
(1 − e7x )2
(1 − e7x )2
(1 − e7x )2
40. Differentiate y = sin(ex ) cos(e−x )
y ′ = sin(ex )(− sin(e−x ))e−x (−1)+cos(e−x ) cos(ex )ex = sin(ex ) sin(e−x )e−x + cos(e−x ) cos(ex )ex
41. Differentiate y = (e2x − e−3x )7
y ′ = 7(e2x − e−3x )6 (e2x (2) − e−3x (−3)) = 7(e2x − e−3x )6 (2e2x + 3e−3x )
Integration Evaluate each of the following integrals:
Z
Z
Z
1
1
2
1
3
1
1
1
p
dz =
u− 3 du = − (3)u 3 + C = − (7 − 5z) 3 + C
42.
2 dz = −
3
2
5
5
5
(7 − 5z)
(7 − 5z) 3

u = 7 − 5z


du
= −5dz
Here
1

 − du = dz
5
√
π
Z π
3
π
3
√
3
sin
π
3
2
sec2 θ dθ = tan θ = tan − tan 0 =
−
0
=
43.
= 3
π
1
3
cos 3
0
0
2
44.
Z
3
−3
|x2 − 1| dx
We can solve this one two different ways. We will detail both: First
15
Z
3
−3
Z
|x2 − 1| dx
−1
=
−3
2
x − 1 dx +
Z
1
−1
2
1 − x dx +
Z
1
3
x2 − 1 dx
−1
3
x3
x3 1
x3
=
− x + x − +
− x
3
3 −1
3
−3
1
1
1
1
1
− −1 +
+ (9 − 3) −
−1
− + 1 − (−9 + 3) + 1 −
3
3
3
3
=
=
2
2 2
2
+6+ + +6+
3
3 3
3
=
8
+ 12
3
=
8 36
+
3
3
=
44
3
Second, we could use symmetry
Z 3
|x2 − 1| dx
−3
=2
Z
0
1
2
1 − x dx +
Z
3
1
2
x − 1 dx
"
3 #
x3 1 x3
=2 x− +
− x
3 0
3
1
1
1
=2 1−
−1
− 0 + (9 − 3) −
3
3
44
2
4 18
22
2
=2
=2
=
+6+
+
=2
3
3
3
3
3
3
Z 2
(|x| − 4) dx
45.
−1
16
=
Z
0
−1
(−x − 4) dx +
Z
2
0
(x − 4) dx
0
2
2
x2
x
= − − 4x +
− 4x
2
2
−1
0
1
19
7
7 12
= 0 − − + 4 + (2 − 8) − 0 = − − 6 = − −
= −
2
2
2
2
2
Z
(x + 1)(x + 2)
√
46.
dx
x
Z
(x2 + 3x + 2
√
dx
=
x
Z
3
1
1
= x 2 + 3x 2 + 2x− 2 dx
3
1
2 5
2
2
x 2 + 2(2)x 2 + C
= x +3
5
3
3
1
2 5
= x 2 + 2x 2 + 4x 2 + C
5
√
Z
u+ u+7
47.
du TYPO FIXED HERE!
u3
Z
5
= u−2 + u− 2 + 7u−3 du
2 3 7
= −u−1 − u− 2 − u−2 + C
3
2
1
7
2
= − −
− 2 +C
u 3u 23
2u
48.
49.
Z
Z
3
x|x| dx =
−3
Z
0
−3
2
−x dx +
2π
| sin x| dx
0
=
Z
Z
π
Z
0
3
x3 0
x3 3
x dx = − + = (0 − 9) + (9 − 0) = 0
3 −3
3 0
2
2π
sin x dx +
− sin x dx
0
π
π
2π
= − cos x + cos x
0
π
= − cos π − (− cos 0) + cos(2π) − cos π
= −(−1) + 1 + 1 − (−1) = 4
Z
1 −1 u=4
1 u=4 −2
1
3
1
cos x
1
u du = − u dx =
=
=− + =
50.
2
6 u=1
6
24 6
24
8
0 (1 + 6 sin x)
u=1

u = 1 + 6 sin x


x = 0 =⇒ u = 1
du
=
6
cos
xdx
and
Here
x = π6 =⇒ u = 1 + 6 sin( π6 ) = 1 + 6( 12 ) = 4

 1 du = cos xdx
6
Z
π
6
17
51.
Z
π
3
sin(2x) dx =
π
4



1
2
Z
u= 2π
3
u= π2
u= 2π
3
1
1
2π
π
1
1
1
−
=
+ cos = −
= − cos
sin u du = − cos u
2
2
3
2
2
2
4
π
u=
2
u = 2x
x = π4 =⇒ u = π2
du
=
2dx
and
Here
x = π3 =⇒ u = 2π

 1 du = dx
3
2
u= 1
Z π
Z u= 1
x
2
2
1
2 x
2
3
sin
=
52.
u du = 2u cos
dx = 6
6
6
4
0
u=0
u=0

x
 u = sin 6 x = 0 =⇒ u = 0
1
x
du = 6 cos 6 dx and
Here
x = π =⇒ u = 12

6du = cos x6 dx
Z
Z π
8
1 u=1 3
1 u4 u=1 1
1
3
2
tan (2x) sec (2x) dx =
u du =
53.
= −0=
2 u=0
2 4 u=0 8
8
0

u = tan(2x)

x = 0 =⇒ u = 0
2
du = 2 sec (2x)dx and
Here
x = π8 =⇒ u = tan π4 = 1
 1
2
2 du = sec (2x)dx
√
√
u=1
Z π2
Z 1
4
1
cos t
cos t
− 21
p
54.
du = 2(2)u 2 u
= 4−
√ dt = π2 √ p √ dt = 2
π2
1
1
u=
t
sin
t
t
sin
t
u=
36
36
2
2
√


q

 u = sin √t 1
 x = π2 =⇒ u = sin π2 = sin π = 1
36
6
2
√ dt
t
du
=
cos
q 36
Here
and
√ 2 t

 x = π2 =⇒ u = sin π2 = sin π = 1
 2du = cos√ t dt
4
4
2
π2
4
Z
√4
2
t
13
Z
1 u= 4 1
3 sin x cos x
dx =
55.
du =
2
2 u=1 u
0 1 + 3 sin x

u = 1 + 3 sin2 x


du = 6 sin x cos xdx and
Here

 1 du = 3 sin x cos xdx
2
Z 5
√
56.
3x5 x sin(3x) dx = 0 why?
Z
π
3
u= 13
4
1 13 1
1 13
1
= ln
ln |u|
− ln 1 = ln
2
2
4
2
2
4
u=1
x = 0 =⇒ u = 1
x = π3 =⇒ u = 1 + 3 sin2 ( π3 ) = 1 + 3( 34 ) =
13
4
5
57.
58.
Z
(w3 cos w4 + 2009) dw =
Z
Z
y 3 − 9y sin y + 26y −1
dy =
y
w3 cos w4 dw +
1
sin u + 2009w + C = 14 sin w4 + 2009w + C
4

u = w4

du = 4w3 dw
Here
 1
3
4 du = w dw
Z
Z
2009 dw =
y 2 − 9 sin y + 26y −2 dy =
18
1
4
Z
cos u du +
Z
2009 dw =
y3
+ 9 cos y − 26y −1 + C
3
Z
√
Z
1
u2
Z
√
3
2
2
2
59.
x cos(x x) dx =
cos u du = sin u + C = sin(x 2 ) + C
3
3
3

3

u = x2

1
Here
du = 32 x 2 dx

√
 2 du = xdx
3
Z
Z
1
1
dt = − sin u du = cos u + C = cos 1t + C
sin
60.
t2
t

u = 1t

du = − t12 dt
Here

−du = t12 dt
61.
Here
62.
Z
r
1
1 − du =
u
3
3 4
3
w dw = w 3 + C =
4
4
1
3
4
1 3
+C
1−
u
w = 1 − u1
dw = u12 du
2
−2
Z
|1 − x| dx =
Z
1
−2
1 − x dx +
Z
2
1
1
1
1
(2 − 2) − ( − 1) = + 4 + = 5
2
2
2
Z 2π
63.
| cos x − sin x| dx
2
x2 1
x2
1
x − 1 dx = x − +
− x = (1 − ) − (−2 − 2) +
2 −2
2
2
1
0
=
Z
π
4
cos x − sin x dx +
0
Z
5π
4
π
4
sin x − cos x dx +
Z
2π
cos x − sin x dx
5π
4
π
2π
5π
4
4
= sin x + cos x + (− cos x − sin x) + sin x + cos x
0
π
4
π
4
5π
4
5π
4 )
5π
4
− (− cos π4 − sin π4 )
= sin + cos − (sin 0 + cos 0) + (− cos − sin
5π
5π
+ sin
2π √
+ cos 2π − (sin 4 + cos
4 )
√
√
√
√
√
√
2
2
2
= √2 + 2 −√(0 +√
1) + (−(−√2 ) − (− 22 )) − (− 22 − 22 ) + 0 + 1 − (− 22 −
= 2−1+ 2+ 2+1+ 2
√
= 4 2+2
π
4
Z
1
1
1
2t5 + 1
dt =
du = ln |u| + C =
64.
6
t + 3t
3
u
3

6
u = t + 3t

du = 6t5 + 3dt
Here
 1
5
3 du = 2t + 1dt
Z
65.
Z
14
x(x+1)
Here
dx =
Z
(u−1)u
14
du =
Z
1
3
19
2
2 )
ln |t6 + 3t| + C
u15 −u14 du =
u = x + 1 =⇒ x = u − 1
du = dx
√
(x + 1)16 (x + 1)15
1 16 1 15
u − u +C =
−
+C
16
15
16
15
66.
Z
7 cos(5x) − 5 sin(7x) dx =
7
5
sin(5x) + cos(7x) + C
5
7


u = 5x


du = 5dx and
Here

 1
5 du = dx
7
5
Z
cos u du −
5
7
Z
sin w dw =
5
7
sin u + cos w + C =
5
7
w = 7x
dw = 7dx
1
7 dw = dx
Z
p
3
3
1 3
1 √
1
1 2
u du = −
x 2 − 3x2 dx = −
u 2 + C = − u 2 + C = − (2 − 3x2 ) 2 + C
6
6 3
9
9

u = 2 − 3x2

du = −6xdx
Here
 1
− 6 du = xdx
Z
Z Z
√
3
3
1
1
2 5
2
2−u √
1
1
68.
x 2 − 3x dx = −
2
u2 − u2 +
u du = −
2u 2 − u 2 du = −
3
3
9
9
3
5
3
5
2
4
C = − (2 − 3x) 2 + (2 − 3x) 2 + C
27
45

2−u

u = 2 − 3x =⇒ x =

3
Here
du = −3dx

 1
− 3 du = dx
Z
Z
Z 5
5
12
5
1
1
1 7 19
7 12
u+1
69.
x(3x−1) 7 dx =
u 7 + u 7 +C =
u 7 + u 7 du =
u 7 du =
3
3
9
9 19
12
19
12
7 12
7
7
7 19
u7 +
u7 +C =
(3x − 1) 7 +
(3x − 1) 7 + C
171
108
171
108

u+1

u = 3x − 1 =⇒ x =

3
Here
du = 3dx

 1
3 du = dx
67.
Z
70.
Z
71.
Z
72.
1 √
7
(x 2 + x− 3 ) x dx =
2 −x3
x e


1
dx = −
3
Z
Z
3
−3x2
xe
1


1
dx = −
6
Z
x5 6 7
+ x6 + C
5
7
1
1
3
eu du = − eu + C = − e−x + C
3
3
u = −x3
du = −3x2 dx
Here
 1
− 3 du = x2 dx
Z
1
x4 + x 6 dx =
u=−27
u=−3
1 u u=−27
1
1
1
e du = − e − 27
= − (e−27 − e−3 ) =
3
6 u=−3
6
6e
6e
u
u = −3x2
x = 1 =⇒ u = −3
du = −6xdx and
Here
x = 3 =⇒ u = −27
 1
− 6 du = x dx
20
73.
1
Z
1
1
e− x
1 1
dx
=
eu du = eu + C = e− x + C
2
7x
7
7
7
u = − x1
Here
du = x12 dx
Z
Z
1
1
ex
dx =
du = −u−1 + C = − x
+C
74.
x
2
2
(e − 1)
u
e −1
u = ex − 1
Here
du = ex dx
Z
Z
1
1
3
3
3e7x
6√
6p
√
√ du = − (2)u 2 + C = −
dx = −
75.
u+C = −
1 − e7x + C
7
7
7
7
u
1 − e7x

u = 1 − e7x

du = −7e7x dx
Here
 1
− 7 du = e7x dx
Z
76.
Z
3x
e3x ee


dx =
1
3
Z
1 3x
1
eu du = eu + C = ee + C
3
3
u = e3x
du = 3e3x dx
Here
 1
3x
3 du = e dx
Area between Curves
77. Compute the area bounded by y = x3 and y = 4x.
You can solve this problem two different ways. I will detail both. First we will use symmetry
to integrate one side and then double the value to capture the entire area. Note that y = x3
and y = 4x intersect when x3 = 4x =⇒ x(x2 − 4) = 0 or when x = 0 or x = ±2.
21
Area = 2
Z
2
Z
2
top − bottom dx
0
=2
3
4x − x dx
0
"
#
4 2
x
= 2 2x2 − 4 0
= 2 [(8 − 4) − 0] = 8
If you don’t use symmetry we see that
Z 2
Z 0
top − bottom dx
top − bottom dx +
Area =
0
−2
=
0
0
x4
x4 2
2 2
− 2x + 2x −
4
4 0
−2
−2
=
Z
Z
3
x − 4x dx +
0
2
4x − x3 dx
= 0 − (4 − 8) + (8 − 4) + (8 − 4) − 0
=4+4= 8
78. Compute the area bounded by y = 2|x| and y = 8 − x2 .
You can solve this problem two different ways. I will detail both. First we will use symmetry
to integrate one side and then double the value to capture the entire area. Note that y = 2|x|
and y = 8−x2 intersect (for x > 0) when 2x = 8−x2 =⇒ x2 +2x−8 = 0 =⇒ (x+4)(x−2) = 0
or when x = −4 or x = 2. We will ignore x = −4 here since we are considering the side with
x > 0.
22
Area = 2
Z
2
Z
2
Z
2
top − bottom dx
0
=2
0
=2
0
(8 − x ) − 2x dx
2
−x2 − 2x + 8 dx
"
2 #
x3
2
= 2 − − x + 8x
3
0
8
= 2 (− − 4 + 16) − 0
3
8
= 2 12 −
3
36 8
=2
−
3
3
28
=2
3
=
56
3
If you don’t use symmetry we see that
23
Area =
Z
0
Z
0
−2
top − bottom dx +
Z
2
0
top − bottom dx
Z
2
(8 − x2 ) − 2x dx
(8 − x ) − 2(−x) dx +
0
−2
Z 2
Z 0
2
−x2 − 2x + 8 dx
−x + 2x + 8 dx +
=
=
2
0
−2
3
0
2
3
x
x
2
2
= − + x + 8x + − − x + 8x 3
3
−2
0
8
8
= 0 − − + 4 − 16 + − − 4 + 16 − 0
3
3
8
8
= − + 12 − + 12
3
3
=−
16
+ 24
3
=−
16 72
+
3
3
=
56
3
79. Compute the area bounded by y = 4 − x2 , y = x + 2, x = −3, and x = 0.
Note that these two curves intersect when 4−x2 = x+2 =⇒ x2 +x−2 = 0 =⇒ (x+2)(x−1) =
0 =⇒ x = −2 or x = 1.
24
Area =
−2
Z
−2
Z
−2
−2 3
0
x3 x2
x
x2
+
− 2x + − −
+ 2x 3
2
3
2
−3
−2
−3
=
−3
=
−3
=
Z
Z
top − bottom dx +
2
0
−2
(x + 2) − (4 − x ) dx +
x2 + x − 2 dx +
Z
0
−2
top − bottom dx
Z
0
−2
(4 − x2 ) − (x + 2) dx
−x2 − x + 2 dx
8
9
8
−2−4
= − + 2 + 4 − −9 + + 6 + 0 −
3
2
3
8
9 8
=− +6+3− − +6
3
2 3
=−
16
9
+ 15 −
3
2
=−
32 90 27
+
−
6
6
6
=
31
6
Position, Velocity, Acceleration
80. A ball is thrown upward with a speed of 128 ft/sec from the edge of a cliff 144 ft above the
ground. Find its height above the ground t seconds later. When does it reach its maximum
height? When does it hit the ground?
d
ft
Note v0 = 128 sec
, s0 = 144ft.
6?
144
a(t) = −32
v(t) = −32t + v0 =⇒ v(t) = −32t + 128
s(t) = −16t2 + 128t + s0 =⇒ s(t) = −16t2 + 128t + 144
Max height occurs when v(t) = 0. That is, −32t + 128 = 0 or when t = 4 seconds.
The ball hits the ground when s(t) = −16t2 + 128t + 144 = 0 or when −16(t2 − 8t − 9) = 0
so that −16(t − 9)(t + 1) = 0. Then t = 9 or t = −1. We will ignore the negative time here.
The ball hits the ground after 9 seconds.
81. The skid marks made by an automobile indicate that its brakes were fully applied for a
distance of 90 ft before it came to a stop. Suppose that it is known that the car in question
has a constant deceleration of 20 ft/sec2 under the conditions of the skid. Suppose also that
25
the car was travelling at 60 ft/sec when the brakes were first applied. How long did it take
for the car to come to a complete stop?
ft
, s0 = 0ft.
Note v0 = 60 sec
s0 = 0
v0 = 60
-
sstop = 90
tstop =?
a(t) = −20
v(t) = −20t + v0 =⇒ v(t) = −20t + 60
s(t) = −10t2 + 60t + s0 =⇒ s(t) = −10t2 + 128t + 0
You can solve this two ways. First, the car stops when v(t) = 0. Set v(t) = −20t + 60 = 0
and solve for the stopping time. That is, t = 3 seconds.
Second, the car stops when s(t) = 90. Set s(t) = −10t2 + 60t = 90 and solve for the stopping
time. We see that −10(t2 − 6t + 9) = 0 implies −10(t − 3)(t − 3) = 0 or the car stops when
t = 3 seconds.
82. Suppose that a bolt was fired vertically upward from a powerful crossbow at ground level,
and that it struck the ground 48 seconds later. If air resistance may be neglected, find the
initial velocity of the bolt and the maximum height it reached.
d
ft
Note v0 =? sec
, s0 = 0 ft, timpact = 48sec.
6?
a(t) = −32
v(t) = −32t + v0
s(t) = −16t2 + v0 t + s0 =⇒ s(t) = −16t2 + v0 t + 0 =⇒ s(t) = −16t2 + v0 t
First we use the impact information, s(48) = 0.
s(48) = 0
−16(48)2 + v0 (48) = 0 =⇒ 48v0 = 16(48)2 =⇒ v0 =
16(48)2
ft
= 16(48) = 768
,
48
sec
768
= 24
32
2
seconds. Finally the max height is s(24) = −16(24) + 768(24) = −16(576) + 18432 =
−9216 + 18432 = 9216 feet.
As a result v(t) = −32t + 768 and the max height occurs when v(t) = 0 =⇒ t =
83. Jack throws a baseball straight downward from the top of a tall building. The initial speed
of the ball is 25 feet per second. It hits the ground with a speed of 153 feet per second. How
tall is the building?
ft
ft
Note v0 = −25 sec
, s0 =?ft, vimpact = −153 sec
?
d
?
26
a(t) = −32
v(t) = −32t + v0 =⇒ v(t) = −32t − 25
s(t) = −16t2 − 25t + s0
The ball hits the ground when v(t) = −32t − 25 = −153 or when 32t = 153 − 25 = 128 which
is when timpact = 4 seconds.
Finally, we solve s(4) = 0 for s0 . The ball hits the ground when −16(4)2 − 25(4) + s0 = 0 or
when −256 − 100 + s0 = 0 which is when s0 = 356 feet. As a result, the building is 356 feet
tall.
84. A ball is dropped from the top of the building 576 feet high. With what velocity should a
second ball be thrown straight downward 3 seconds later so that the two balls hit the ground
simultaneously?
Note v0 = 0
ft
sec , s0
= 576ft.
d
576 ?
d
Ball 1 has the following motion equations:
a(t) = −32
v(t) = −32t + v0 =⇒ v(t) = −32t
s(t) = −16t2 + s0 =⇒ s(t) = −16t2 + 576
The first ball hits the ground when s(t) = −16t2 + 576 = 0 or when 16t2 = 576 which is when
576
or when timpact = 6 seconds.
t2 =
16
For the second ball to be thrown 3 seconds later and hit the ground at the same time as the
first ball, the second ball must travel just 3 seconds before hitting the ground.
Ball 2 has the following motion equations:
a(t) = −32
v(t) = −32t + v0
s(t) = −16t2 + v0 t + s0 =⇒ s(t) = −16t2 + v0 t + 576
We solve s(3) = 0 to find the second ball’s initial velocity. Set −16(3)2 + 3v0 + 576 = 0 =⇒
3v0 = −576 + 144 = −432 =⇒ v0 = −144. Finally, the second ball must have an initial
velocity of −144 feet per second, or we can say the second ball must be thrown straight down
with a speed of 144 feet per second.
85. A particle starts from rest at the point x = 10 and moves along the x-axis with acceleration
function a(t) = 12t. Find its resulting position function.
Note s0 = 10, v0 = 0.
a(t) = 12t
v(t) = 6t2 + v0 =⇒ v(t) = 6t2
s(t) = 2t3 + s0 =⇒ s(t) = 2t3 + 10
As a result, the resulting position function is given as s(t) = 2t3 + 10.
86. The skid marks made by an automobile indicate that its brakes were fully applied for a
distance of 160 ft before it came to a stop. Suppose that it is known that the car in question
27
has a constant deceleration of 20 ft/sec2 under the conditions of the skid. How fast was the
car travelling when its brakes were first applied?
ft
, s0 = 0ft, sstop = 160ft.
Note v0 =? sec
s0 = 0
v0 =?
-
sstop = 160
vstop = 0
a(t) = −20
v(t) = −20t + v0
s(t) = −10t2 + v0 t + s0 =⇒ s(t) = −10t2 + v0 t + 0
You can solve this in two parts. First, the car stops when v(t) = 0. Set v(t) = −20t + v0 = 0
v0
and solve for the stopping time in terms of the initial velocity v0 . That is, tstop =
seconds.
20
v v v 2
0
0
0
Second, the car stops when s(tstop ) = 160 or s
+ v0
= 160. Set −10
= 160
20
20
20
and solve for the unknown initial velocity. We see that
v v 2
0
0
+ v0
= 160
−10
20
20
v2
−10v02
+ 0
400
20
= 160
−10v02 20v02
+
400
400
= 160
10v02
400
= 160
10v02 = 160(400)
v02 = 16(400) = 6400
v0 = 80
Finally, the initial velocity v0 = 80 feet per second.
Displacement–Total Distance–Net Change
87. Suppose that the velocity of a moving particle is v(t) = t2 − 11t + 24 feet per second. Find
both the displacement and total distance it travels between time t = 0 and t = 10 seconds.
10
Z 10
t3
11t2
1000 1100
2
t − 11t + 24 dt =
Displacement=
−
+ 24t =
−
+ 240 − 0 =
3
2
3
2
0
0
70
1000
1000 930
1000
− 550 + 240 =
− 310 =
−
=
3
3
3
3
3
The displacement in question is
70
.
3
Note that if you sketch the parabola y = t2 − 11t + 24, it passes below the x-axis between
x = 3 and x = 8.
28
Total Distance=
Z
0
Z
3
10
|t2 − 11t + 24| dt
Z
8
Z
10
t2 − 11t + 24 dt
−(t − 11t + 24) dt +
t − 11t + 24 dt +
8
3
0
3 3
10
8
t3 11t2
t
11t2
t3 11t2
=
−
+ 24t + − +
− 24t + −
+ 24t
2
2
3
2
8
3
0
3
3
512 704
99
99
+ 72 − 0 + −
+
− 192 − −9 +
− 72
= 9−
2
3
2
2 1000 1100
512 704
+
−
+ 240 −
−
+ 192
3
2
3
2
99 1000
512
99 512
−
+ 352 − 192 + 81 −
+
− 550 + 240 −
+ 350 − 192
= 81 −
2
3
2
3
3
198 24
= 170 −
−
2
3
= 170 − 99 − 8 = 63
=
2
2
The total distance in question is 63 .
88. Supppose that water is pumped into an initially empty tank. The rate of water flow into the
tank at time t (in seconds) is 50 − t liters per second. How much water flows into the tank
during the first 30 seconds?
Using the Net Change Theorem,Z the amount of water in the tank between time t = 0 and
30
rate of water flow dt.
time t = 30 is W (30) − W (0) =
0
Using the FTC, we simply integrate
Z 30
900
t2 30
= 1500 − 450 = 1050.
50 − t dt = 50t − = 1500 −
2 0
2
0
The amount of water that flows into the tank during the first 30 seconds is 1050 liters.
29