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The radius of the base of a cylinder is increasing at a rate of 1 meter per hour and the height of the cylinder is decreasing at a rate of 4 meters per hour.
At a certain instant, the base radius is 5 meters and the height is 8 meters.
What is the rate of change of the volume of the cylinder at that instant (in cubic meters per hour)?
Choose 1 answer:
(A) 
180 pi
(B) 
20 pi
(C) 
-20 pi
(D) 
-180 pi
The volume of a cylinder with base radius 
r and height 
h is 
pir^(2)h.

The radius of the base of a cylinder is increasing at a rate of 11 meter per hour and the height of the cylinder is decreasing at a rate of 44 meters per hour.\newlineAt a certain instant, the base radius is 55 meters and the height is 88 meters.\newlineWhat is the rate of change of the volume of the cylinder at that instant (in cubic meters per hour)?\newlineChoose 11 answer:\newline(A) 180π 180 \pi \newline(B) 20π 20 \pi \newline(C) 20π -20 \pi \newline(D) 180π -180 \pi \newlineThe volume of a cylinder with base radius r r and height h h is πr2h \pi r^{2} h .

Full solution

Q. The radius of the base of a cylinder is increasing at a rate of 11 meter per hour and the height of the cylinder is decreasing at a rate of 44 meters per hour.\newlineAt a certain instant, the base radius is 55 meters and the height is 88 meters.\newlineWhat is the rate of change of the volume of the cylinder at that instant (in cubic meters per hour)?\newlineChoose 11 answer:\newline(A) 180π 180 \pi \newline(B) 20π 20 \pi \newline(C) 20π -20 \pi \newline(D) 180π -180 \pi \newlineThe volume of a cylinder with base radius r r and height h h is πr2h \pi r^{2} h .
  1. Write Formula: First, write down the formula for the volume of a cylinder, which is V=πr2hV = \pi r^2 h.
  2. Find Rate of Change: Next, we need to find the rate of change of the volume, which is dVdt\frac{dV}{dt}. To do this, we'll use the chain rule from calculus, since VV is a function of both rr and hh, which are both functions of time tt.
  3. Apply Chain Rule: The chain rule gives us dVdt=(dVdr)(drdt)+(dVdh)(dhdt)\frac{dV}{dt} = \left(\frac{dV}{dr}\right)\left(\frac{dr}{dt}\right) + \left(\frac{dV}{dh}\right)\left(\frac{dh}{dt}\right). We know drdt=1m/hr\frac{dr}{dt} = 1 \, \text{m/hr} (since the radius is increasing at 11 meter per hour) and dhdt=4m/hr\frac{dh}{dt} = -4 \, \text{m/hr} (since the height is decreasing at 44 meters per hour).
  4. Calculate Partial Derivatives: Now, calculate the partial derivatives dV/drdV/dr and dV/dhdV/dh. The partial derivative of VV with respect to rr is dV/dr=2πrhdV/dr = 2\pi rh, and the partial derivative of VV with respect to hh is dV/dh=πr2dV/dh = \pi r^2.
  5. Plug in Values: Plug in the values of rr and hh into the partial derivatives. We have r=5r = 5 meters and h=8h = 8 meters, so dVdr=2π(5)(8)\frac{dV}{dr} = 2\pi(5)(8) and dVdh=π(5)2\frac{dV}{dh} = \pi(5)^2.
  6. Calculate Derivatives: Now, calculate the actual values of the partial derivatives. dV/dr=2π(5)(8)=80πdV/dr = 2\pi(5)(8) = 80\pi and dV/dh=π(5)2=25πdV/dh = \pi(5)^2 = 25\pi.
  7. Find dV/dtdV/dt: Finally, plug in all the known values into the chain rule equation to find dV/dtdV/dt. dV/dt=(80π)(1)+(25π)(4)dV/dt = (80\pi)(1) + (25\pi)(-4).
  8. Final Calculation: Calculate dVdt\frac{dV}{dt}. dVdt=80π100π=20π\frac{dV}{dt} = 80\pi - 100\pi = -20\pi cubic meters per hour.

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