If a snowball melts so that its surface area decreases at a rate of $3~\frac{\text{cm}^2}{\text{min}}$, find the rate at which the diameter decreases when the diameter is $8$ cm.
My answer is wrong. Can some one help me solve?
Given: The rate of decrease of the surface area is $3~\frac{\text{cm}^2}{\text{min}}$. I f we let $t$ be time in (in minutes) and $s$ be the surface area (in cm$^2$), then we are given that $\frac{ds}{dt} = -3~\text{cm}^3$
Unknown: The rate of decrease of the diameter when the diameter is $8$ cm
If we let $x$ be the diameter, then we want to find $\frac{dx}{dt}$ when $x = 8$ cm. If the radius is $r$ and the diameter is $x = 2r$ then $r = \frac{1}{2}x$ and $S = 4\pi r^2$, $x=2r$, then $r=\frac{1}{2}x$
$S=\pi x^2 =-\frac{3}{2}\pi 8$
My answer is wrong. Can some one help me understand?
Best Answer
We have that the surface area formula is:
$$SA = 4\pi r^2$$
If we take the derivative with respect with time, we get:
$$\frac{dSA}{dt} = 8\pi r \frac{dr}{dt}$$
Now, from the problem we are given that $\frac{dSA}{dt} = -3$, and $r = 4$
Now we can find $\frac{dr}{dt}$:
$$ 32\pi \frac{dr}{dt} = -3,\ \frac{dr}{dt} = -\frac{3}{32\pi}$$
Now, we can also make the relation that:
$$d = 2r,\ \frac{dd}{dt} = 2\frac{dr}{dt} = 2*\left(-\frac{3}{32\pi}\right) = -\frac{6}{32\pi}$$
Therefore our diameter decreases at a rate of $\frac{6}{32\pi}$