A spherical snowball is melting in such a way that it maintains its shape. The snowball is decreasing in volume at a constant rate of 8 cubi

Question

A spherical snowball is melting in such a way that it maintains its shape. The snowball is decreasing in volume at a constant rate of 8 cubic centimeters per hour. At what rate, in centimeters per hour, is the radius of the snowball decreasing at the instant when the radius is 10 centimeters?

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Ruby 6 days 2021-09-10T23:03:23+00:00 2 Answers 0

Answers ( )

    0
    2021-09-10T23:04:32+00:00

    Answer:

    0.0628cm/hr

    Step-by-step explanation:

    the snowball decreases in volume at a constant rate of 8 cubic cm per hour

    Volume of a sphere = 4/3πr^3

    Differentiate volume with respect to time(t)

    dV/dt = 3(4/3πr^3)

    = 4πr^2 . dr/dt

    From the question, dV/dt = 8, r= 10

    Substituting,

    8 = 4π(10)^2 . dr/dt

    8 = 400π . dr/dt

    dr/dt = 8/400π

    dr/dt =1/50π

    dr/dt = 0.0628cm/hr

    0
    2021-09-10T23:04:36+00:00

    Answer:

    0.006366 cm/h

    Step-by-step explanation:

    First we need to know that the volume of a sphere is calculated with the following formula:

    V = (4/3)*pi*r^3

    Where V is the volume and r is its radius.

    Then, to find the rate of change, we need to derivate the equation with respect to the radius:

    dV/dr = (4/3)*pi*(3r^2) = 4*pi*r^2

    Then, we can write dV/dr as being (dV/dt)*(dt/dr), where dt is the change of time (variable t)

    (dV/dt)*(dt/dr) = 4*pi*r^2

    (dV/dt) = 4*pi*r^2*(dr/dt)

    The rate of change of the volume is 8 cm3/h when the radius is 10, so using these values, we can find the rate of change of the radius (dr/dt):

    8 = 4*pi*10^2*(dr/dt)

    8 = 400*pi*(dr/dt)

    dr/dt = 8/(400*pi)

    dr/dt = 0.006366 cm/h

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