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.
