avissidep

2021-10-05

Suppose you are working for a manufacturing company that is analyzing a newer product using the following:
$A=\pi {r}^{2}$
$\frac{dA}{dt}=2\pi r\left\{\left(\frac{dr}{dt}\right\}\right)$ Explain what the product engineers may be examining/testing/analyzing. Why is the factor $\left\{dr\right\}\left\{dt\right\}$ needed?

Bentley Leach

Step 1
We know that a circle of radius r, the area is $\pi {r}^{2}$
Hence $A=\pi {r}^{2}$ is pretty straightforward. It tells us that the engineers are analysing a product that has a circular part that changes in size while retaining the circular shape. We can also understand that area A is a function of the radius r.
Step 2
Now, differentiating A with respect to time t, gives us $\frac{dA}{dt}=\pi \frac{d{r}^{2}}{dr}\cdot \frac{dr}{dt}=2\pi r\frac{dr}{dt}$
This gives us the rate of change of area w.r.t time. We can see that it is dependent on the current radius r as well as the rate of change of radius w.r.t. time, i.e. $\frac{dr}{dt}$
It appears that the engineers are testing the rate of change of area using the radius and the rate of change of radius.

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