Question

"What is the rate at which the volume ((V)) and the surface area ((S)) of a shape are changing when the two radii are 15 centimeters and 25 centimeters, and the height is 10 centimeters, given that (\frac{dV}{dt}) is in (\mathrm{cm}^{3}/\mathrm{min}) and (\frac{dS}{dt}) is in…?"

Answer 1

To solve the problem, we first need to find out the surface area (S) and the volume (V) of the given shape. Given the radii r1 and r2 are 15 cm and 25 cm respectively and the height h is 10 cm, we use the following formulas:

Surface Area (S): S = 2 * pi * r1 * r2 + pi * (r1^2 + r2^2)
Volume (V): V = (pi * h * (r1^2 + r2^2)) / 2

Calculating these with the given values, S equals 5026.55 cm^2, and V equals 13351.77 cm^3.

Next, we need to determine how the volume V and surface area S of the shape are changing over time (dV/dt and dS/dt). We use the rule d(uv)/dt = u'v + uv' to find the rates of change. Given that dV/dt is 1 cm^3/min, we have:

Rate of change of volume (dV/dt): dV/dt = pi * h * (2r1 + 2r2)
Rate of change of surface area (dS/dt): dS/dt = 2 * pi * r1 * 1 + 2 * pi * r2 * 1 + 2 * pi * r1 * 1 + 2 * pi * r2 * 1

Plugging the given values into these equations, dV/dt evaluated equals 2513.27, and dS/dt evaluated equals 502.65 (both rates per minute).

Therefore, given the specified radii and height, the volume of the shape is increasing at a rate of 2513.27 cm^3/min, and the surface area is increasing at a rate of 502.65 cm^2/min.