Process

Here are steps to help you solve a related rates problem.

DO:  For each step, see if we used it in the previous example (copied below for your convenience).  This will help you visualize and remember the steps.  Did we skip any steps?

1. Draw a picture!!  This always helps.


2. Assign variables to all changing values. List everything you know: Which variables do you know about?  Which variable are you trying to figure out?


3. Label your picture with variables and constants (to avoid confusion, do not assign names to constant values).


4. Figure out how the variables are related, and write down an equation.  This is where you bring in knowledge from outside of calculus, typically geometry or physics.  If the variables represent two sides of a right triangle, use the Pythagorean theorem.  If the variables represent the radius and volume of a sphere, use the formula for volume in terms of radius.  You don't have to express one of the variables in terms of the other -- just find an equation that involves both.


5. If there are more than two variables, find a relationship to eliminate one.


6. Differentiate your equation(s) with respect to time.


7. Solve for the rate of change of the variable you want in terms of the rate of change of the variable you already understand.


8. If a variable assumes a specific value under some conditions (for example the velocity changes, but it equals 2 mph at 4 PM), replace it at this time. Remember to plug-in after differentiating.  (Why?)  Convert that information into an answer to your question.

Previous 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.