Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Aiden was given this problem:
The radius 
r(t) of the base of a cylinder is increasing at a rate of 1 meter per hour and the height 
h(t) of the cylinder is decreasing at a rate of 4 meters per hour. At a certain instant 
t_(0), the base radius is 5 meters and the height is 8 meters. What is the rate of change of the volume 
V(t) of the cylinder at that instant?
Which equation should Aiden use to solve the problem?
Choose 1 answer:
(A) 
V(t)=pi[r(t)]^(2)+2pi*r(t)*h(t)
(B) 
V(t)=pi[r(t)]^(2)*h(t)
(C) 
V(t)=pi[r(t)]^(2)+pi*r(t)sqrt([r(t)]^(2)+[h(t)]^(2))
(D) 
V(t)=(pi[r(t)]^(2)*h(t))/(3)

Aiden was given this problem:\newlineThe radius r(t) r(t) of the base of a cylinder is increasing at a rate of 11 meter per hour and the height h(t) h(t) of the cylinder is decreasing at a rate of 44 meters per hour. At a certain instant t0 t_{0} , the base radius is 55 meters and the height is 88 meters. What is the rate of change of the volume V(t) V(t) of the cylinder at that instant?\newlineWhich equation should Aiden use to solve the problem?\newlineChoose 11 answer:\newline(A) V(t)=π[r(t)]2+2πr(t)h(t) V(t)=\pi[r(t)]^{2}+2 \pi \cdot r(t) \cdot h(t) \newline(B) V(t)=π[r(t)]2h(t) V(t)=\pi[r(t)]^{2} \cdot h(t) \newline(C) V(t)=π[r(t)]2+πr(t)[r(t)]2+[h(t)]2 V(t)=\pi[r(t)]^{2}+\pi \cdot r(t) \sqrt{[r(t)]^{2}+[h(t)]^{2}} \newline(D) V(t)=π[r(t)]2h(t)3 V(t)=\frac{\pi[r(t)]^{2} \cdot h(t)}{3}

Full solution

Q. Aiden was given this problem:\newlineThe radius r(t) r(t) of the base of a cylinder is increasing at a rate of 11 meter per hour and the height h(t) h(t) of the cylinder is decreasing at a rate of 44 meters per hour. At a certain instant t0 t_{0} , the base radius is 55 meters and the height is 88 meters. What is the rate of change of the volume V(t) V(t) of the cylinder at that instant?\newlineWhich equation should Aiden use to solve the problem?\newlineChoose 11 answer:\newline(A) V(t)=π[r(t)]2+2πr(t)h(t) V(t)=\pi[r(t)]^{2}+2 \pi \cdot r(t) \cdot h(t) \newline(B) V(t)=π[r(t)]2h(t) V(t)=\pi[r(t)]^{2} \cdot h(t) \newline(C) V(t)=π[r(t)]2+πr(t)[r(t)]2+[h(t)]2 V(t)=\pi[r(t)]^{2}+\pi \cdot r(t) \sqrt{[r(t)]^{2}+[h(t)]^{2}} \newline(D) V(t)=π[r(t)]2h(t)3 V(t)=\frac{\pi[r(t)]^{2} \cdot h(t)}{3}
  1. Define Volume Formula: To find the rate of change of the volume of the cylinder, we need to use the formula for the volume of a cylinder, which is the product of the area of the base (a circle) and the height of the cylinder.
  2. Identify Correct Formula: The area of the base of a cylinder (a circle) is given by π\pi times the radius squared, πr2\pi r^2. The volume of a cylinder is this area multiplied by its height, hh. Therefore, the correct formula for the volume of a cylinder is V(t)=π[r(t)]2h(t)V(t) = \pi[r(t)]^2 \cdot h(t).
  3. Differentiate Volume Formula: Looking at the options given, option (B) V(t)=π[r(t)]2h(t)V(t) = \pi[r(t)]^2 \cdot h(t) is the correct formula for the volume of a cylinder. Options (A), (C), and (D) are incorrect because they either add unnecessary terms or divide by an incorrect value.
  4. Substitute Given Values: Now that we have the correct formula, we can differentiate it with respect to time to find the rate of change of the volume. The derivative of V(t)V(t) with respect to time tt is dVdt=π[2r(t)drdth(t)+r(t)2dhdt]\frac{dV}{dt} = \pi[2r(t) \cdot \frac{dr}{dt} \cdot h(t) + r(t)^2 \cdot \frac{dh}{dt}], where drdt\frac{dr}{dt} is the rate of change of the radius and dhdt\frac{dh}{dt} is the rate of change of the height.
  5. Calculate Derivative: Given that the radius is increasing at a rate of 11 meter per hour (drdt=1m/hr)(\frac{dr}{dt} = 1 \, \text{m/hr}) and the height is decreasing at a rate of 44 meters per hour (dhdt=4m/hr)(\frac{dh}{dt} = -4 \, \text{m/hr}), we can substitute these values into the derivative formula along with the given radius and height at the instant t0t_0 (r(t0)=5m,h(t0)=8m)(r(t_0) = 5 \, \text{m}, h(t_0) = 8 \, \text{m}).
  6. Perform Calculations: Substituting the given values, we get dVdt=π[2×5m×1m/hr×8m+5m2×(4m/hr)]\frac{dV}{dt} = \pi[2 \times 5 \, \text{m} \times 1 \, \text{m/hr} \times 8 \, \text{m} + 5 \, \text{m}^2 \times (-4 \, \text{m/hr})].
  7. Final Rate of Change: Performing the calculations, we have dVdt=π[2×5×1×8+52×(4)]=π[80100]=π[20]\frac{dV}{dt} = \pi[2 \times 5 \times 1 \times 8 + 5^2 \times (-4)] = \pi[80 - 100] = \pi[-20] m3/hr\text{m}^3/\text{hr}.
  8. Final Rate of Change: Performing the calculations, we have dVdt=π[2×5×1×8+52×(4)]=π[80100]=π[20]\frac{dV}{dt} = \pi[2 \times 5 \times 1 \times 8 + 5^2 \times (-4)] = \pi[80 - 100] = \pi[-20] m3/hrm^3/hr.The rate of change of the volume of the cylinder at the instant t0t_0 is therefore 20π-20\pi cubic meters per hour.

More problems from Solve linear equations with variables on both sides: word problems