Blog #32

Well, I feel as though I have reviewed just about everything we’ve learned several times over…except word problems. The free response questions on the practice AP tests that we’ve been doing are essentially just word problems. I’ll go over how to work through a simple related rate problem.

There are steps that you can take to decipher related rate word problems and make them easier to understand. First, you need to identify all of the information the word problem gives, such as variables and what you’re actually trying to find. Once you do that, then you need to find a formula that connects all of your variables and your shape together (such as the area of a triangle or the volume of a cube). After this, solve for your desired rate with respect to t (such as dA/dt for the rate of area changing or dV/dt for the rate of volume changing). And lastly, plug in all of your variables to get your answer.
*Don’t forget to include your units (such as cm/min or ft^3/sec).

Some key words you need to know are:
“rate of change” = derivative
“speed” = derivative
“with respect to” = what’s the bottom variable of the derivative (ex. rate of changing area A with respect to time t : dA/dt)
“area” or “volume” = formula
“cube” or “sphere” or “square” or “triangle” or “circle” = shape
“radius” or “diameter” or “height” or “depth” = possibly what you need to take the derivate of or important components of the formula
“increasing” or “decreasing” = whether the derivative is positive or negative
and any form of units (especially squared or cubic units) = squared means area; cubic means volume

Ex. The radius r of a circle is increasing at a rate of 4 centimeters per minute. Find the rate of change of the area when r = 8 cm.

First of all, we know the following info: r = 8 & dr/dt = 4 cm/min
We also know we are looking for: dA/dt = ?
Second, because we are looking for the rate of change of the AREA, we know where using the area of a circle formula: A = π(r^2)
Now, solve for dA/dt by taking the derivative of the area formula: dA/dt = 2π r dr/dt
**Remember to put dr/dt after you take the derivative of r because it is an implicit derivative.
Finally, plug in your variables: dA/dt = 2π (8)(4) = 64π cm^2/min
***cm is squared because area is always squared, just like volume is always cubed.