Final answer:
The rate at which the sphere's volume is being dissolved with respect to time is 28π cm^3/hr. The rate at which the radius changes with respect to time is 25/(2π) cm/hr when r = 1 meter and 25/π cm/hr when r = 1/2 meter.
Step-by-step explanation:
(a) To determine the rate at which the sphere's volume is being dissolved with respect to time, we need to use the formula for the volume of a sphere, which is V = (4/3)πr^3, where V is the volume and r is the radius. We are given the rate of change of the radius, which is 7 cm/hour. To find the rate of change of the volume, we differentiate the volume formula with respect to time t, and substitute the given rate of change of the radius. The derivative of the volume with respect to time is dV/dt = 4πr^2(dr/dt). Substituting the given values, we get dV/dt = 4π(1)^2(7) = 28π cm^3/hr.
(b) To find the rate at which the radius changes with respect to time, we need to use the formula for the surface area of a sphere, which is A = 4πr^2. We are given the rate of dissolving of the material, which is 1 m^3/hr. To find the rate of change of the radius, we differentiate the surface area formula with respect to time t, and solve for dr/dt. The derivative of the surface area with respect to time is dA/dt = 8πr(dr/dt). Solving for dr/dt, we get dr/dt = (dA/dt)/(8πr). Substituting the given values, we get dr/dt = (1)/(8π(1)) = 1/(8π) = 1/8π m/hr = (1/8π)(100) = 25/(2π) cm/hr when r = 1 meter, and dr/dt = (1)/(8π(1/2)) = 1/(4π) = 1/4π m/hr = (1/4π)(100) = 25/π cm/hr when r = 1/2 meter.