Calculus AB Related Rates

A spherical snowball is melting in such a way that its volume is decreasing at a rate of 1cm^3/min. At what rate is the diameter decreasing when the diameter is 10 cm?

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Recall: Volume of a Sphere = (4/3)(Pi)r^3

Since we are concerned with matters involving the diameter and not the radius, let's re-write our formula in terms of the diameter...

Recall: r = d/2

Thus...

Volume of a Sphere = (4/3)(Pi)(d/2)^3

= (4/3)(Pi)(d^3)/(2^3)

= (4/3)(Pi)(d^3)/8

V = (Pi/6)d^3

Now, let's differentiate with respect to t...

dV/dt = (Pi/6)3d^2 dd/dt

dV/dt = (Pi/2)d^2 dd/dt

We were given the constant dV/dt = 1cm^3/min and asked to find the rate of change of the diameter, i.e., dd/dt, when d = 10 cm; therefore, let's plug in the values for dV/dt and d and solve for dd/dt...

1cm^3/min = (Pi/2)(10cm)^2 dd/dt

1cm^3/min = (Pi/2)10^2cm^2 dd/dt

1cm^3/min = (Pi/2)100cm^2 dd/dt

1cm^3/min = 50(Pi)cm^2 dd/dt

1cm^3/min = 50(Pi)cm^2 dd/dt

1cm/min = 50Pi dd/dt

1/(50Pi)cm/min = 50Pi/(50Pi) dd/dt

dd/dt = 1/(50Pi)cm/min

dd/dt =~ 0.0063661977 cm/min

dd/dt =~ 6.4 mm/min

Volume of a sphere` (V) = 4/3*pi*(d^3/8)`

Where d is the diameter.

It is given that rate of decreasing volume is 1cm^3/min.

`(dV)/dt = 4/3*pi*3*d^2/8*(dd)/dt`

`(dV)/dt = 1/2*pi*r^2*(dd)/dt`

`(dV)/dt =` rate of volume decreasing with time

`(dr)/dt =` rate of diameter decreasing with time

When the diameter is 10cm

`(dV)/dt = 1/2*pi*r^2*(dd)/dt`

`1 = 1/2*pi*10^2*(dd)/dt`

`(dd)/dt = 0.0064`

So the rate of decreasing diameter is 0.0064cm/min or 6.4mm/min.

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