Differentiation problem solving

Question

I have a question which I'm unable to provide much working for as I'm not sure how to start it.

A metal sphere is placed in seawater to study the corrosive effect of seawater. If the surface area decreases at a rate of $35cm^2/year$ due to corrosion, how fast is the radius changing when the radius is $7cm$? The surface area of a sphere $r$ is given by $$S=4πr^2$$

All I can decipher is that $\frac{d}{dt}s$ = $35cm/year$ where t is time

I understand that the relation between S and r is through the equation and the rate of change of r will be proportional to the rate of change of S (I think) and that I need to use the chain rule somewhere. Could someone please provide a little bit of help as to where to go from here? Thanks

Answer 1

You noted that $\frac{d}{dt}s=35$cm${}^2$/year, which is very close to correct. While the surface area is changing by that much, the derivative doesn't just measure change, but the direction of change. Since surface area is decreasing, $\frac{d}{dt}s=-35$cm${}^2$/year.

The next step is to differentiate the constraint equation:

$$\frac{d}{dt}S=\frac{d}{dt}(4\pi r^2)=8\pi r \frac{d}{dt}r.$$

Since $\frac{d}{dt}s=-35$cm${}^2$/year and we are examining when $r=7$cm, we have at that point

$$-35\text{cm}{}^2/\text{year}=56\pi \text{cm} \frac{d}{dt}r.$$

From here it is straightforward to solve for $\frac{d}{dt}r.$

Answer 2

Notice that $$\frac{d}{dt}S=\frac{d}{dt}(4\pi r^2) = 8\pi r\frac{dr}{dt}$$ thus from $\frac{d}{dt}S=35$ you get $$\frac{dr}{dt}=\frac{35}{8\pi r}$$ For $r=7$ $$\frac{dr}{dt}=\frac{35}{7\cdot8\pi} = \frac{5}{8\pi}$$ Through all this the units of length are $cm$, of area $cm^2$ and of time $years$.

Answer 3

$\frac{dS}{dt} = 35$

$S = 4\pi R^2$

$\frac{d(4\pi R^2)}{dt}= 35$

$8\pi R\frac{dR}{dt} = 35$

$\frac{dR}{dt} =35/(7*8\pi )= 5/(8*\pi)$