Blog #7

This week, we learned about word problems that have to do with rate of change (aka derivatives). There are steps that you can take to decipher the word problems and make the 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).

Ex. The radius r of a circle is increasing at a rate of 4 centimeters per minute. Find the rates 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.