Related Rates

Related rates is just an application of the chain rule. Typically, we have a function $y = x$, where $x$ is implicitly dependent on another variable, usually time, $t$. Thus, if we differentiate the equation with respect to time, we have $\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}$. Related rates problems usually give su two of these differentials, and ask us to compute the third.

Example 1

Taken from practice exercise in Cracking the GRE Mathematics Subject Test, 4th Edition

The radius of a circle is decreasing at a rate of $0.5$ cm per second. At what rate in cm²/sec is the circle's area decreasing when the radius is $4$ cm?

The area of the circle is given by $A = \pi r^2$. Differentiating with respect to time, $t$, we have

$$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$$

We are given that $\frac{dr}{dt} = -0.5$, and that, at the refrence time $t_0$, the radius is $4$. Plugging these values in, we have

$$\frac{dA}{dt} = 2 \pi (4) (-0.5) = -4 \pi$$

Thus, the area is decreasing by $-4\pi$ cm²/sec.

Example 2

Taken from practice exercise in Calculus with Analytic Geometry, Third Edition

A board 5 feet long slides down a wall. At the instant the bottom end is 4 feet from the wall, the other end is moving down the wall at the rate of 2 feet per second. At that moment, how fast is the bottom end sliding along the ground?

If we let $y$ denote the height of the board against the wall, and $x$ the distance of the board from the wall, then we have the equation

$$5^2 = x^2 + y ^2$$

Differentiating with respect to time, we have

$$0 = 2x\frac{dx}{dt} + 2y \frac{dy}{dt}$$

We are given that, at the reference time, $t_0$, $\frac{dx}{dy} = -2$ and $x = 4$. Since, $x = 4$, we know that $y=3$ (it's a 3-4-5 triangle). Plugging these values into the above equation, we have

$$0 = 2(4)\frac{dx}{dt} + 2(3)(-2)$$

Which gives us $\frac{dx}{dt} = \frac{3}{2}$. Thus, the bottom of the board is sliding at a rate of 1.5 feet per second.