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How to Solve Related Rate “Word Problems”
Step 1: Draw a sketch , if appropriate.
Step 2: Make a list of variables .
have to come back and add more.
Don’t expect to get it right immediately, you may
Step 3: Make a list of symbols .
Step 4: Make a list of relations .
develops.
Again, you may have to add more as the problem
Step 5: Identify the variable whose rate of change is to be found, and the variable
whose rate of change is given.
Step 6:
Diﬀerentiate the relations.
Step 7:
Solve the mathematical problem by inserting the given value of the variables.
This must NOT be done BEFORE diﬀerentiating!
Step 8: Translate the problem back into English, and be sure that it is a reasonable
solution.
A Collection of Useful Formulas
Areas of plane regions:
Area A of a square of side s:
A = s2
Area A of a rectangle of width w and height h:
Area A of a triangle of base b and height h:
A = wh
A=
1
bh
2
Area A of a trapezoid with bottom width wB , top width wT and height h:
A=
wB + w T
h
2
Area A of a circle of radius r :
A = πr2
1
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Word Problems
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Volumes of Solids:
3
Volume V of a cube with side s: V = s
Volume V of a rectangular parallelopiped (box) of length , width w and height h:
V = wh
V = π r 2h
π
V = r 2h
3
Volume V of a cylinder with circular base of radius r and height h:
Volume V of a cone with circular base of radius r and height h:
Volume V of a sphere of radius r :
V =
4
πr2
3
Surface Areas of Solids:
2
Surface Area S of a cube with side s: S = 6s
Surface Area S of a rectangular parallelopiped (box) of length , width w and height
h:
S = 2(wh + w + h)
Surface Area S of a cylinder with circular base of radius r and height h:
Surface Area S of a cone with circular base of radius r and height h:
S = π r r 2 + h2 + π r 2
Surface Area S of a sphere of radius r :
S = 4π r 2
2
S = 2π r 2 + 2π r h
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A point is moving toward the origin on the parabola x + y 2 = 0 with
of
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Solution:
Step 1: Sketch:
2
1
0
-4
-3
-2
-1
x
0
1
-1
-2
Steps 2&3: The x-coordinate x and the y-coordinate y of the point and its distance
s from the origin are the only variables.
Step 4: Relations: x + y 2 = 0
and
s 2 = x2 + y 2.
Step 5: Identiﬁcation: The rates of change y of y and s of s are to be found, given
units
that x is changing at the rate of x = 2
.
ms
Step 6: Diﬀerentiate: x + 2yy = 0,
Step 7: Solve:
and
Insert the values x = −1, y = 1, x = 2
equations:
units
2
+ 2(1)y = 0,
ms
2ss = 2xx + 2yy .
units
into the diﬀerentiated
ms
2ss = 2(−1) 2 units
ms + 2(1)y .
to get y = −1 units
ms and
units .
2 (−1)2 + 12 s = 2(−1) 2 units
+
2(1)
−1
ms
ms
or
√
3 2 units
.
−
2 ms
√
6 units
2 2s = −6 units
ms , so s = − 2√2 ms =
Step 8: Translate:
The y coordinate is decreasing at the rate of one unit per mil√
3 2
lisecond, while the distance from the origin is decreasing at the rate of −
units per
2
millisecond.
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its x-coordinate increasing at the rate of 2 units/ms . What is the rate of change of s k
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the y-coordinate as it passes through the point (-1,1)? How fast is it approaching the
y
origin?
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parabola y = 8 − x 2 . If its height is decreasing at the rate of 4 units/sec, how fast are s k
y
its width and area changing when its height is 4 units?
9
8
Solution:
7
Step 1: Sketch:
6
5
4
3
2
1
-3
-2
0
-1 0
-1
x
1
2
3
-2
Steps 2&3: The width w and the height h of the rectangle and its area A are the only
variables.
Step 4: Relations: h = 8 − (w/2)2 = 8 − w 2 /4
and
A = wh.
Step 5: Identiﬁcation: The rates of change w of w and A of A are to be found, given
units
that h is changing at the rate of h = −4
.
sec
Step 6: Diﬀerentiate: h = −(w/2)w ,
Step 7: Solve:
Insert the values w = 4, h = 4, h = −4
equations:
−4
units
= −(4/2)w , gives us
sec
w = 2
A = wh + w h.
and
units
into the diﬀerentiated
sec
units
sec
and thus
units
A = (4units) −4 units
sec + (2units) sec 4 =
−8
units2
sec
.
Step 8: Translate: The width coordinate is increasing at the rate of two units per
second, while the area is decreasing at the rate of eight square units per second.
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A rectangle of has its base on the x-axis and is inscribed inside the
of
Example 2:
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Example 3:
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width changing when its area is twelve cm?
y
7
6
Solution:
5
Step 1: Sketch:
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9
x
Steps 2&3: The width w and the height h of the rectangle and its area A are the only
variables.
3
Step 4: Relations: h = 6 − w
4
3
A = wh = w 6 − w
4
and
= 6w −
3 2
w .
4
Step 5: Identiﬁcation: The rates of change w of w and h of h are to be found, given
cm2
.
that A is changing at the rate of A = 1
sec
3
Step 6: Diﬀerentiate: h = − w ,
and
A = 6w − 32 ww = (6 − 32 w)w .
4
Step 7: Solve: First we need to ﬁnd the value of w when A = 12.
must solve the quadratic equation 12 = 6w −
Insert the values w = 4cm, h = 3cm, A = 1
3
w2
4
To do this we
for w = 4
cm2
into the second diﬀerentiated equation:
sec
2
3
A = (6 − 32 w)w becomes 1 cm
sec = (6 − 2 4)w = 0, so w = h = 0!
Step 8: Translate: Something weird is happening here! Neither height nor width is
changing, but the area is!
Can you ﬁnd the division by zero?
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A rectangle is inscribed in a right triangle with legs of lengths 6 cm
and 8 cm. with two sides of the rectangle lie along the legs. If the area of the rectangle s
a
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is increasing at the rate of one square cm per second, how fast are the height and
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Problem 3.10.18.
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25
Solution:
30
25
Step 1: Sketch:
w
30
w
h
30
Step 2: Variables: The width of the horizontal sides of the two triangles and their
height will be needed to express the area of the “wet” end of the trough and the volume
of water in the trough.
Step 3: Symbols: w and h will be as depicted.
wT and wB will denote the widths of the top and bottom of the blue trapezoid, and A
will be its area.
V will denote the volume of water in the trough.
Step 4: Relations:
wB =
3
,
10
wT =
3
10
V = 10A,
+ 2w =
3
10
A=
wB + wT
h,
2
w
25
1
=
= ,
h
50
2
so h = 2w.
+ h, so
3
3
+ h+
10 h = 5(h + 3 )h = 5h2 + 3h, or
V = 10A = 10 10
5
2
V = 5h2 + 3h
Step 5: Identiﬁcation: The rate of change h of h is to be found, given that V is chang1 m3
ing at the rate of
5 min
Step 6: Diﬀerentiate: V = 10hh + 3h = (10h + 3)h .
Step 7: Solve:
Insert the values h =
V = (10h + 3)h :
3
1 m3
= (10
+ 3)h = 6h to get
5 min
10
Step 8: Translate:
minute.
3
,
10
V =
h =
1
30
1 m3
into the diﬀerentiated equation
5 min
m
min
The water level is increasing at the rate of 1/30 of a metre per
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A trough is ten metres long and its ends have the shape of
isosceles trapezoids that are 80 cm across at the top and 30 cm across at the bottom, s
m3 , k a t c h e w a
and has a height of 50 cm.
If the trough is ﬁlled with water at a rate of 0.2 min
how fast is the water level rising when the water is 30 cm deep?
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3.10.28.
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Solution:
P
Step 1: Sketch:
x
z
12
A
w
y
B
Q
Steps 2&3: Let w and x be the distance of A from Q and P respectively, and let y and
z be the distance of B from Q and P respectively.
Step 4: Relations: There are three obvious relations:
w 2 + 122 = x 2 , x + y = 39, and z2 + 122 = y 2 .
Step 5: Identiﬁcation: We must ﬁnd
when w = 5ft.
Step 6: Diﬀerentiate:
dz
,
dt
given that
2ww = 2xx ,
dw
ft
=2
, at the point in time
dt
sec
x + y = 0,
and
2zz = 2yy ,
which we may reduce to:
z =
y y ,
z
y = −x , and
x =
w w ,
x
or
z = −
yw w .
z x
Step 7: Solve: We now ﬁnd the values of x, y, and z when w = 5:
√
√
x = 144 + 25 = 169 = 13, y = 39 − x = 39 − 13 = 26,
√
√
z = y 2 − 122 = 262 − 122 = 2 133.
ft
26
5
Thus z = − √
2
=
13
sec
133
10 ft
.
−√
133 sec
Step 8: Translate: Cart B is moving towards Q at the rate of
10
√
0.87feet per second
133
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Two carts, A and B, are connected by a rope 39 feet long that passes
over a pulley P. The point Q is on the ﬂoor 12 feet directly beneath P and between the s
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carts A and B. Cart A is being pulled away from Q at a speed of 2 sec . How fast is cart
B moving toward Q at the instant when cart A is 5 feet from Q?
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Problem 3.10.34.
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Solution:
Step 1: Sketch:
y
z
x
Steps 2&3: Let x be the distance of the ﬁrst person from the initial point at time t,
let y be the distance of the second person from the initial point at time t, and let z
be the distance between the two at time t.
Step 4: Relations: By the Law of Cosines,
z2 = x 2 + y 2 − 2xy cos π4 = x 2 + y 2 −
√
2xy.
given that x = 3 miles and
hour
Step 5: Identiﬁcation: We must ﬁnd z ,
when t = 0.25hour.
Step 6: Diﬀerentiate: 2zz = 2xx + 2yy − 2(x y + xy ) or
√
2xx + 2yy − 2(x y + xy )
z =
2z
1
3
Step 7: Solve: After 15 minutes, x = mile, y = mile, so
4
2
2 2 1
3
1
3
z=
mile=
+
− 2
4
2
4
2
√
6
9
13 − 6 2
4
+
− 2
mile=
mile.
16 16
16
4
1 3 3
1
2
3+2
2− 2 3
+
2
miles
4
2 2
4
Thus z =
=
√
hour
13 − 6 2
2
4
miles
13 − 6 2
hour
Step 8: Translate: The two people are separating at the rate of
13 − 6 2 2.1 miles per hour.
8
miles
y = 2
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Two people start from the same point. One walks east at
miles . How fast is the distance between the
3 miles and the other walks northeast at 2
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people changing after 15 minutes?