Question

Related rates(calculus)

Answer 1

Answer:

The volume increases at a rate of $125663.71 mm^ 3 / s$

Step-by-step explanation:

The rate of change of the radius with respect to time is 4 mm / s.

So:

$\frac{dr}{dt} = 4mm / s$

Now we must find a relationship between the volume of a sphere and its radius.

The equation of the volume of a sphere is:

$V = \frac{4}{3}\pi r^ 3$

So:

$\frac{dV}{dt} = \frac{4}{3}\pi * 3r ^ 2 * \frac{dr}{dt}\\\frac{dV}{dt} = 4\pi r ^ 2 * 4\\\frac{dV}{dt} = 16\pi r ^ 2$

The diameter of the sphere is 100 mm. Therefore its radius is 100/2 = 50 mm.

So:

$\frac{dV}{dt} = 16\pi (50) ^ 2\\\frac{dV}{dt} = 40000\pi mm^3 / s$

The volume increases at a rate of :

40000 π mm ^ 3 / s = 125663.71 mm^3/s