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Zhenghao was given this problem:
The radius 
r(t) of a sphere is decreasing at a rate of 1 meter per hour. At a certain instant 
t_(0), the radius is 4 meters. What is the rate of change of the volume 
V(t) of the sphere at that instant?
Which equation should Zhenghao use to solve the problem?
Choose 1 answer:
(A) 
V(t)=2pi*r(t)
(B) 
V(t)=pi[r(t)]^(2)
(C) 
V(t)=4pi[r(t)]^(2)
(D) 
V(t)=(4)/(3)pi[r(t)]^(3)

Zhenghao was given this problem:\newlineThe radius r(t) r(t) of a sphere is decreasing at a rate of 11 meter per hour. At a certain instant t0 t_{0} , the radius is 44 meters. What is the rate of change of the volume V(t) V(t) of the sphere at that instant?\newlineWhich equation should Zhenghao use to solve the problem?\newlineChoose 11 answer:\newline(A) V(t)=2πr(t) V(t)=2 \pi \cdot r(t) \newline(B) V(t)=π[r(t)]2 V(t)=\pi[r(t)]^{2} \newline(C) V(t)=4π[r(t)]2 V(t)=4 \pi[r(t)]^{2} \newline(D) V(t)=43π[r(t)]3 V(t)=\frac{4}{3} \pi[r(t)]^{3}

Full solution

Q. Zhenghao was given this problem:\newlineThe radius r(t) r(t) of a sphere is decreasing at a rate of 11 meter per hour. At a certain instant t0 t_{0} , the radius is 44 meters. What is the rate of change of the volume V(t) V(t) of the sphere at that instant?\newlineWhich equation should Zhenghao use to solve the problem?\newlineChoose 11 answer:\newline(A) V(t)=2πr(t) V(t)=2 \pi \cdot r(t) \newline(B) V(t)=π[r(t)]2 V(t)=\pi[r(t)]^{2} \newline(C) V(t)=4π[r(t)]2 V(t)=4 \pi[r(t)]^{2} \newline(D) V(t)=43π[r(t)]3 V(t)=\frac{4}{3} \pi[r(t)]^{3}
  1. Volume Formula: To find the rate of change of the volume of the sphere, we need to use the formula for the volume of a sphere. The correct formula for the volume of a sphere is given by V(t)=43π[r(t)]3V(t) = \frac{4}{3}\pi[r(t)]^3, where r(t)r(t) is the radius of the sphere as a function of time tt.
  2. Differentiate Volume Formula: We need to differentiate the volume formula with respect to time to find the rate of change of the volume. This is because the rate of change of the volume is the derivative of the volume with respect to time, denoted as dVdt\frac{dV}{dt}.
  3. Substitute Given Values: The derivative of V(t)V(t) with respect to tt is given by dVdt=4π[r(t)]2drdt\frac{dV}{dt} = 4\pi[r(t)]^2 \frac{dr}{dt}, where drdt\frac{dr}{dt} is the rate of change of the radius with respect to time. We are given that the radius is decreasing at a rate of 1-1 meter per hour, so drdt=1\frac{dr}{dt} = -1 meter/hour.
  4. Calculate Derivative: Now we substitute the given values into the derivative formula. At the instant t0t_0, the radius r(t)r(t) is 44 meters, so we have dVdt=4π(4 meters)2(1 meter/hour)\frac{dV}{dt} = 4\pi(4 \text{ meters})^2 * (-1 \text{ meter/hour}).
  5. Final Rate of Change: Calculating the derivative, we get dVdt=4π(16 meters2)×(1 meter/hour)=64π meters3/hour\frac{dV}{dt} = 4\pi(16 \text{ meters}^2) \times (-1 \text{ meter/hour}) = -64\pi \text{ meters}^3/\text{hour}. This is the rate of change of the volume of the sphere at the instant when the radius is 4 meters4 \text{ meters}.