# If a snowball melts so that its surface area decreases at a rate of 3 cm2/min, how do you find the rate at which the diameter decreases when the diameter is 12 cm?

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#### Explanation:

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Steve M Share
Dec 10, 2016

diameter is decreasing at rate of $\frac{3}{24 \pi}$ ($0.040$ (2sf)) $\text{cm"^2"/min}$

#### Explanation:

Assuming the snowball is a perfect sphere, then if $A$ denotes the surface area and $D$ the diameter then:

$A = 4 \pi {r}^{2} = 4 \pi {\left(\frac{D}{2}\right)}^{2} = \pi {D}^{2}$

Differentiating wrt $r$ we have:

$\frac{\mathrm{dA}}{\mathrm{dD}} = 2 \pi D$

We are told that $\frac{\mathrm{dA}}{\mathrm{dt}} = - 3$ and we want to find $\frac{\mathrm{dD}}{\mathrm{dt}}$

By the chain rule we have:

$\frac{\mathrm{dA}}{\mathrm{dD}} = \frac{\mathrm{dA}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dD}} = \frac{\frac{\mathrm{dA}}{\mathrm{dt}}}{\frac{\mathrm{dD}}{\mathrm{dt}}}$
$\therefore 2 \pi D = - \frac{3}{\frac{\mathrm{dD}}{\mathrm{dt}}}$
$\therefore \frac{\mathrm{dD}}{\mathrm{dt}} = - \frac{3}{2 \pi D}$

When $D = 12 \implies \frac{\mathrm{dD}}{\mathrm{dt}} = - \frac{3}{24 \pi} = - \frac{1}{8 \pi}$ ($\approx 0.039788 \ldots$)
The sign confirms that $D$ is decreasing

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