If the radius of a sphere is increasing at a constant rate of 4 cm/sec, then the volume is increasing at a rate of ???? cm^3/sec when the radius is 5 cm.
v = 4/3 pi r^3
dv/dt = 4 pi r^2 dr/dt
now just plug in your numbers ...
I got 264586.8
by calculating
268.08x 986.96
where did I go wrong?
Thanks steve
where ever did you get those numbers?
4 pi r^2 dr/dt
= 4 * pi * 5^2 * 4
= 400 pi cm^3/s
To find the rate at which the volume of a sphere is increasing, we need to use the formula for the volume of a sphere:
V = (4/3)πr^3
Where V is the volume and r is the radius of the sphere.
Given that the radius is increasing at a rate of 4 cm/sec, we can determine the rate at which the volume is increasing by taking the derivative of the volume formula with respect to time:
dV/dt = (4/3)π * 3r^2 * dr/dt
Where dV/dt represents the rate of change of the volume with respect to time, and dr/dt represents the rate of change of the radius with respect to time.
Since we know that the radius is increasing at a constant rate of 4 cm/sec, dr/dt is constant and equal to 4 cm/sec. Additionally, we are asked to find the rate of change of the volume when the radius is 5 cm.
Substituting the given values into the formula, we have:
dV/dt = (4/3)π * 3(5^2) * 4
Simplifying further:
dV/dt = (4/3)π * 3 * 25 * 4
dV/dt = 400π cm^3/sec
Therefore, the volume is increasing at a rate of 400π cm^3/sec when the radius is 5 cm.