A tumor is assumed to be roughly spherical. If the radius R is found to be increasing at the rate of 0.5 mm/day when the radius is R=2 mm, how fast is the volume changing at this time?

V = (4/3)πr^3

given dr/dt = .5 when r = 2
dV/dt = 4πr^2 dr/dt
dV/dt = 4π(,25)(2)
= 2π mm^3/day

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To find how fast the volume of the tumor is changing, we need to calculate the derivative of the volume with respect to time.

The volume of a sphere can be calculated using the formula V = (4/3)πR^3, where V is the volume and R is the radius of the sphere.

To find how fast the volume is changing, we take the derivative of the volume with respect to time t:

dV/dt = d/dt (4/3)πR^3

To evaluate this derivative, we need to apply the chain rule. The chain rule states that if we have a function of the form f(g(t)), then the derivative of f(g(t)) with respect to t can be calculated as:

df/dt = (df/dg)(dg/dt)

Applying the chain rule to our equation, we have:

dV/dt = (d/dR (4/3)πR^3) * (dR/dt)

The first derivative (d/dR) of (4/3)πR^3 with respect to R is:

(d/dR) (4/3)πR^3 = 4πR^2

And the derivative of R with respect to t is given in the problem as 0.5 mm/day.

Now, plugging these values into the equation:

dV/dt = (4πR^2) * (dR/dt)

At the given time when the radius is R = 2 mm, we can substitute this value into the equation:

dV/dt = (4π(2)^2) * 0.5

Simplifying:

dV/dt = 8π * 0.5

dV/dt = 4π

Therefore, at the time when the radius is 2 mm, the volume is changing at a rate of approximately 4π cubic millimeters per day.