Overview
Related rates problems involve two (or more) variables that change
at the same time, possibly at different rates. If we know how
the variables are related, and how fast one of them is changing,
then we can figure out how fast the other one is changing.
This usually involves writing an equation relating the two variables
and using implicit differentiation
to take the derivative of the equation with respect to time.
Example: A particle is moving clockwise around a circle of
radius 5 cm centered at the origin. As it passes through the
point $(3,4)$, its $x$ position is changing at a rate of 2 cm per
second. How fast is $y$ changing at that instant?
Solution: The rates of change (with respect to time
$t$) mentioned in the example are the rates of change of the
$x$-coordinate, $\frac{dx}{dt}$, and the $y$-coordinate,
$\frac{dy}{dt}$. Note that we will be trying to find
$\frac{dy}{dt}$ at a particular moment in time, which is the moment
that the particle is at the point $(3,4)$.
We have found that our
variables are $x$ and $y$, and we need to find a way to relate
them. In this problem, the relationship between $x$ and $y$ is
given by the fact that they are coordinates on the circle, and that
equation will relate them. We know that the equation for the
circle is $$x^2 + y^2 = 25.$$ To find the related
rates, i.e. to find a relationship
between the rates of change of $x$ and $y$ with respect to
time, we can implicitly differentiate the
equation above with respect to $t$. $$ 2 x \frac{dx}{dt} + 2 y
\frac{dy}{dt} = 0.$$ This is the general relationship between the
speed of $x$ and $y$. When the particle is passing $(3,4)$,
then its velocity is $\displaystyle{\frac{dx}{dt}\Big|_{(3,4)} =
2}$, $x=3$ and $y=4$, so we can solve for $\dfrac{dy}{dt}$ at this
precise moment: $$ 2(3)(2) + 2(4)\frac{dy}{dt}\Big|_{(3,4)} = 0.$$
We have $\dfrac{dy}{dt}\Big|_{(3,4)}=
-\frac{12}{8}=-\frac{3}{2}.$
We can now answer the question: $y$ is changing at the rate of
$-\frac{3}{2}$ cm/sec at that instant.