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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Answer 1 · 2016-12-10T22:29:29

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

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

$A = 4pir^2 = 4pi(D/2)^2 = piD^2$

Differentiating wrt $r$ we have:

$(dA)/(dD) = 2piD$

We are told that $(dA)/dt=-3$ and we want to find $(dD)/dt$

By the chain rule we have:

$(dA)/(dD) = (dA)/(dt)*(dt)/(dD) = ((dA)/(dt)) / ((dD)/dt)$
$:. 2piD = -3/ ((dD)/dt)$
$:. (dD)/dt = -3/ (2piD)$

When $D=12 => (dD)/dt = -3/ (24pi) = -1/(8pi)$ ( $~~0.039788...$ )
The sign confirms that $D$ is decreasing