Anonymous
Anonymous asked in Science & MathematicsMathematics · 4 months ago

a spherical snowball is melting symmetrically at the rate of 4pi cubic cm per hr. how fast is the diameter changing when it is 20 cm?

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  • 4 months ago
    Favorite Answer

    V = (4/3)πr³ 

    dV/dt = 4πr².dr/dt

    so, when the diameter is 20 cm we have:

    -4π = 4πr².dr/dt

    Hence, dr/dt = -1/r²

    i.e. dr/dt = -1/10² => -1/100

    Now, d = 2r

    so, d(d)/dt = 2.dr/dt

    => d(d)/dt = -1/50 cm/hour

    or, -0.2 mm/hour

    :)>

  • Ian H
    Lv 7
    4 months ago

    V = (4/3)πr^3 = (π/6)D^3

    dV/dt = dV/dD * dD/dt

    -4π = (π/2)D^2* dD/dt, (where V is in cubic cm)

    dD/dt = -8/D^2, and when D = 20

    dD/dt = -8/400 = -1/50 cm/hour

    dD/dt = - 0.2 mm/hour

  • 4 months ago

    r: radius  and  d: diammeter

    Volume, V = (4/3)πr³   ⇒   V = (4/3)π(d/2)³   ⇒   V = (1/6)πd³

    dV/dt = d[(1/6)πd³]/dt

    dV/dt = (1/2)πd²(dd/dt)

    Now, dV/dt = -4π cm³/hr  and  d = 20 cm:

    -4π =(1/2)π(20)²(dd/dt)

    dd/dt = -0.02 cm/hr

    The diameter DECREASES at a rate of 0.02 cm/hr = 0.2 mm/hr

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