If a snowball melts so that its surface area decreases at a rate of 3 cm2/min, find the rate at which the diameter decreases when the diameter is 10 cm.

Answer :

Answer:

The diameter decreases at a rate of -0.0477 cm/min when it is 10 cm.

Step-by-step explanation:

Surface area of a snowball:

A snowball has a spheric format, which means that it's surface area is given by:

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

In which r is the radius, which is half the diameter. In function of the diameter, the area is given by:

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

Solving the question:

To solve this question, we have to implicitly derivate the area in function of t. So

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

Snowball melts so that its surface area decreases at a rate of 3 cm2/min

This means that [tex]\frac{dA}{dt} = -3[/tex]

Find the rate at which the diameter decreases when the diameter is 10 cm.

This is [tex]\frac{dd}{dt}[/tex] when [tex]d = 10[/tex]. So

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

[tex]-3 = 20\pi\frac{dd}{dt}[/tex]

[tex]\frac{dd}{dt} = -\frac{3}{20\pi}[/tex]

[tex]\frac{dd}{dt} = -0.0477[/tex]

The diameter decreases at a rate of -0.0477 cm/min when it is 10 cm.

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