If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, find the rate at which the diameter decreases when the diameter is 10 cm. (Give your answer correct to 4 decimal places.)

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Answer:

The diameter decreases at the rate of 0.0796 cm/min when the diameter is 10 cm.

Step-by-step explanation:

A snowball is spherical, so it's area is given by the following formula:

[tex]A = 4\pi r^{2}[/tex]

The radius is half the diameter so:

[tex]A = 4\pi (\frac{d}{2})^{2}[/tex]

[tex]A = 4\pi \frac{d^{2}}{4}[/tex]

[tex]A = \pi d^{2}[/tex]

If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, find the rate at which the diameter decreases when the diameter is 10 cm.

We have to find [tex]\frac{dd}{dt}[/tex] when [tex]\frac{dA}{dt} = -5, d = 10[/tex]

[tex]A = \pi d^{2}[/tex]

Applying implicit differentiation:

We have to variables(A and d), so:

[tex]\frac{dA}{dt} = 2\pi d \frac{dd}{dt}[/tex]

[tex]-5 = 2\pi (10) \frac{dd}{dt}[/tex]

[tex]\frac{dd}{dt} = -\frac{5}{20\pi}[/tex]

[tex]\frac{dd}{dt} = -0.0796[/tex]

The diameter decreases at the rate of 0.0796 cm/min when the diameter is 10 cm.

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