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If a snowball melts so that its surface area decreases at a rate of 9 cm2/min, find the rate (in cm/min) at which the diameter decreases when the diameter is 10 cm. (Round your answer to three decimal places.)

Sagot :

Answer:

The diameter decreases at a rate of 0.143cm/min when it is of 10 cm.

Step-by-step explanation:

Surface area of an snowball:

An snowball has spherical format. The surface area of an sphere is given by:

[tex]S = d^2\pi[/tex]

In which d is the diameter of the sphere.

In this question:

We need to differentiate S implicitly in function of time. So

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

Surface area decreases at a rate of 9 cm2/min

This means that [tex]\frac{dS}{dt} = -9[/tex]

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

[tex]-9 = 2(10)\pi\frac{dd}{dt}[/tex]

[tex]\frac{dd}{dt} = -\frac{9}{20\pi}[/tex]

[tex]\frac{dd}{dt} = -0.143[/tex]

Area in cm², so diameter in cm.

The diameter decreases at a rate of 0.143cm/min when it is of 10 cm.

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