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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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 $\frac{3}{24 π}$ $3/(24pi)$ ( $0.040$ $0.040$ (2sf)) ${cm}^{2} /min$ $"cm"^2"/min"$

Explanation:

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

$A = 4 π r^{2} = 4 π {(\frac{D}{2})}^{2} = π D^{2}$ $A = 4pir^2 = 4pi(D/2)^2 = piD^2$

Differentiating wrt $r$ $r$ we have:

$\frac{d A}{d D} = 2 π D$ $(dA)/(dD) = 2piD$

We are told that $\frac{d A}{d t} = - 3$ $(dA)/dt=-3$ and we want to find $\frac{d D}{d t}$ $(dD)/dt$

By the chain rule we have:

$\frac{d A}{d D} = \frac{d A}{d t} \cdot \frac{d t}{d D} = \frac{\frac{d A}{d t}}{\frac{d D}{d t}}$ $(dA)/(dD) = (dA)/(dt)*(dt)/(dD) = ((dA)/(dt)) / ((dD)/dt)$
$:. 2piD = -3/ ((dD)/dt)$
$:. (dD)/dt = -3/ (2piD)$

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

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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?

1 Answer

Explanation:

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