Related Rates

Next, give the variables names. The important thing to figure out is which variables are changing.
dD At D = 50, x = 40. (We know this because it’s a 3-4-5 right triangle.) In addition, = D? =
dt
?80. D? is negative because the car is moving in the ?x direction. Don’t plug in the value for D
yet! D is changing, and it depends on x.
The Pythagorean theorem says
302 + x2 = D2
Differentiate this equation with respect to time (implicit differentiation:
d ? 2 ? 2DD?
302 + x = D2 = 2xx? = 2DD? = x? =
dt
? ? 2x
Now, plug in the instantaneous numerical values:
50 feet
x? =
40
(?80) = ?100
s
This exceeds the speed limit of 95 feet per second; you are, in fact, speeding.
1
?
There is another, longer, way of solving this problem. Start with
D = 302 + x2 = (302 + x2)1/2
d 1 dx
D = (302 + x2)?1/2(2x )
dt 2 dt
Plug in the values:
1 dx
?80 = (302 + 402)?1/2(2)(40)
2 dt
and solve to find
dx feet
= ?100
dt s
(A third strategy is to differentiate x =
?
D2 ? 302). It is easiest to differentiate the equation in its
simplest algebraic form 302 + x2 = D2, our first approach.
The general strategy for these types of problems is:
1. Draw a picture. Set up variables and equations.
2. Take derivatives.
3. Plug in the given values. Don’t plug the values in until after taking the derivatives.
Example 2. Consider a conical tank. Its radius at the top is 4 feet, and it’s 10 feet high. It’s being
filled with water at the rate of 2 cubic feet per minute. How fast is the water level rising when it is
5 feet high?
h