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Applications of Differentiation: Related Rates Problems - Prof. Jon W. Lamb, Study notes of Mathematics

Solutions to various related rates problems using differentiation. It covers topics such as computing the rate of change of one quantity in terms of another, differentiating implicitly with respect to time, and using equations based on similar triangles. Examples and exercises with answers.

Typology: Study notes

Pre 2010

Uploaded on 08/18/2009

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bg1
dx 1ft/sec
dt =
d(x y) ?
dt
+
=
dy ?
dt =
6
dx 5ft/sec
dt =
APPLICATIONS OF DIFFERENTIATION
4.1 RELATED RATES
I. Compute the rate of change of one quantity in terms of the rate of change of another quantity.
II. Normally differentiate all quantities with respect to time.
III. Example 2 on p. 266
A. Make a diagram:
y 10 The ladder is always 10 ft long
x The change in x is positive
because x is increasing.
B. Label diagram: variables, rates of change (derivatives), numerical values which do not change
C. Write an equation which relates all the variables: 22 2
xy10+=
D. Differentiate implicitly with respect to time:dx dy dx dy
2x 2y 0 x y 0
dt dt dt dt
+
=⇒ + =
E. Evaluate “when”: When x = 6, we can used the Pythagorean Theorem to get y = 8.
F. dx dy dy dy 3
xy06(1)80
dt dt dt dt 4
+=+==
G. Answer question(s) completely, include units and “when”: The top of the ladder is sliding down
(because rate of change of y is negative) the wall at a rate of ft per second when the foot of
3
4
the ladder is 6 ft from the wall and the foot is sliding away from the wall at a rate of 1 ft/second.
IV. Exercise 8 on p. 269
A. Diagram:
B. Label: 15
x y
C. Equation: Based on similar triangles: 15 x y 2
6x 6y 15y y x
6y 3
+
=⇒+==
D. Differentiate implicitly with respect to time: dd2d55dx
(x y) x x x
dt dt 3 dt 3 3 dt

+= + = =


E. Evaluate “when”: When x = 40: this is actually irrelevant because there is no x in our
differentiated equation!
F. 5dx 5 25
(5)
3dt 3 3
==
G. Answer: The tip of his shadow is always moving at a rate of ft per second whenever the
1
83
man is walking away from the light pole at a rate of 5 ft per second.
pf2

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dx 1 ft / sec dt

d (x y)? dt

dy ? dt

dx 5 ft / sec dt

APPLICATIONS OF DIFFERENTIATION

4.1 RELATED RATES

I. Compute the rate of change of one quantity in terms of the rate of change of another quantity.

II. Normally differentiate all quantities with respect to time.

III. Example 2 on p. 266

A. Make a diagram:

y 10 The ladder is always 10 ft long

x The change in x is positive

because x is increasing.

B. Label diagram: variables, rates of change (derivatives), numerical values which do not change

C. Write an equation which relates all the variables:

2 2 2 x + y = 10

D. Differentiate implicitly with respect to time :

dx dy dx dy 2x 2y 0 x y 0 dt dt dt dt

E. Evaluate “when”: When x = 6, we can used the Pythagorean Theorem to get y = 8.

F.

dx dy dy dy 3 x y 0 6(1) 8 0 dt dt dt dt 4

G. Answer question(s) completely, include units and “when”: The top of the ladder is sliding down

(because rate of change of y is negative) the wall at a rate of ft per second when the foot of

the ladder is 6 ft from the wall and the foot is sliding away from the wall at a rate of 1 ft/second.

IV. Exercise 8 on p. 269

A. Diagram:

B. Label: 15

x y

C. Equation: Based on similar triangles:

15 x y 2 6x 6y 15y y x 6 y 3

D. Differentiate implicitly with respect to time:

d d 2 d 5 5 dx (x y) x x x dt dt 3 dt 3 3 dt

E. Evaluate “when”: When x = 40: this is actually irrelevant because there is no x in our

differentiated equation!

F.

5 dx 5 25 (5) 3 dt 3 3

G. Answer: The tip of his shadow is always moving at a rate of ft per second whenever the

man is walking away from the light pole at a rate of 5 ft per second.

d (^) o 2 / min dt

V. Exercise 24 on p. 270

A. Diagram: 15 x

B. Label:

C. Equation: Use Law of Cosines:

2 2 2

x = 12 + 15 −2(12)(15) cos θ

D. Differentiate implicitly:

dx d dx d 2x 360( sin ) x 180 sin dt dt dt dt

θ θ = − − θ ⇒ = θ

E. Evaluate “when”:

o 2 1

60 x 369 360 189 x 189 3 21

F. Must use

dx d dx o dx 3 1

x 180 sin 3 21 180 sin 60.

dt dt dt 90 dt 3 21 3 7

θ π π π

radians!

G. Answer: The length of the third side is increasing by approximately .396 square meters per

minute when the angle between the given sides has measure and is increasing at a rate of

o 60

per minute.

o 2