If a snowball melts so that its surface area decreases at a rate of 7 cm2/min, find the rate at which the diameter decreases when the diamet

Question

If a snowball melts so that its surface area decreases at a rate of 7 cm2/min, find the rate at which the diameter decreases when the diameter is 8 cm. cm/min

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Bơ 5 years 2021-07-20T14:52:39+00:00 1 Answers 23 views 0

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  1. Nick
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    2021-07-20T14:53:49+00:00
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    Answer:

    7/16π

    Step-by-step explanation:

    The surface area of the snow ball = 4πr^2

    Let the surface area= y

    Then y= 4πr^2

    The derivative can be taken with respect to time as

    dy/dt= (8πr)dr/dt………..eqn(1)

    We were told that its surface area decreases at a rate of 7 cm2/min.

    Then dy/dt=-7

    diameter = 8 cm. cm/min, the radius r= 4cm/min

    r= 4

    Substitute the values of r into eqn(1)

    dy/dt=(8π×4)dr/dt

    -7=(32π) dr/dr

    dr/dt= -7/32π…………….eqn(**)

    But we know that diameter d= 2× radius

    d=2r…………eqn(*)

    We can find the derivatives with respect to t as

    dd/dt=2dr/dt

    Then substitute en(**) we have

    dd/dt=2(7/32π)

    dd/dt= -14/32π

    Divide both numerator and denominator by 2

    dd/dt= -7/16π

    Hence, the diameter decreases by the rate of 7/16π

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