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If a snowball melts so that its surface area decreases at a rate of 3 cm2/min, find the rate (in cm/min) at which the diameter decreases when the diameter is 9 cm. (Round your answer to three decimal places.)

Sagot :

Answer:

The diameter decreases at a rate of 0.053 cm/min.

Step-by-step explanation:

Surface area of an snowball

The surface area of an snowball has the following equation:

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

In which d is the diameter.

Implicit differentiation:

To solve this question, we differentiate the equation for the surface area implictly, in function of t. So

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

Surface area decreases at a rate of 3 cm2/min

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

Tind the rate (in cm/min) at which the diameter decreases when the diameter is 9 cm.

This is [tex]\frac{dd}{dt}[/tex] when [tex]d = 9[/tex]. So

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

[tex]-3 = 2*9\pi\frac{dd}{dt}[/tex]

[tex]\frac{dd}{dt} = -\frac{3}{18\pi}[/tex]

[tex]\frac{dd}{dt} = -0.053[/tex]

The diameter decreases at a rate of 0.053 cm/min.