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The radius of the base of a cylinder is decreasing at a rate of 9 millimeters per hour and the height of the cylinder is increasing at a rate of 2 millimeters per hour. At a certain instant, the base radius is 8 millimeters and the height is 3 millimeters. What is the rate of change of the surface area of the cylinder at that instant (in square millimeters per hour)? Choose 1 answer: (A) 310 pi (B) -310 pi (C) 155 pi (D) -155 pi The surface of a cylinder with base radius r and height h is 2pi rh+2pir^(2)". "

The radius of the base of a cylinder is decreasing at a rate of 99 millimeters per hour and the height of the cylinder is increasing at a rate of 22 millimeters per hour.\newlineAt a certain instant, the base radius is 88 millimeters and the height is 33 millimeters.\newlineWhat is the rate of change of the surface area of the cylinder at that instant (in square millimeters per hour)?\newlineChoose 11 answer:\newline(A) 310π 310 \pi \newline(B) −310π -310 \pi \newline(C) 155π 155 \pi \newline(D) −155π -155 \pi \newlineThe surface of a cylinder with base radius r r and height h h is 2πrh+2πr2.  2 \pi r h+2 \pi r^{2} \text {. }

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Q. The radius of the base of a cylinder is decreasing at a rate of 99 millimeters per hour and the height of the cylinder is increasing at a rate of 22 millimeters per hour.\newlineAt a certain instant, the base radius is 88 millimeters and the height is 33 millimeters.\newlineWhat is the rate of change of the surface area of the cylinder at that instant (in square millimeters per hour)?\newlineChoose 11 answer:\newline(A) 310π 310 \pi \newline(B) −310π -310 \pi \newline(C) 155π 155 \pi \newline(D) −155π -155 \pi \newlineThe surface of a cylinder with base radius r r and height h h is 2πrh+2πr2.  2 \pi r h+2 \pi r^{2} \text {. }
  1. Surface Area Formula Derivative: The formula for the surface area of a cylinder is S=2πrh+2πr2S = 2\pi rh + 2\pi r^2. We need to find the derivative of this formula with respect to time to find the rate of change of the surface area.
  2. Rate of Change Variables: Let's denote the rate of change of the radius as drdt=−9 mm/h\frac{dr}{dt} = -9 \, \text{mm/h} (since the radius is decreasing) and the rate of change of the height as dhdt=2 mm/h\frac{dh}{dt} = 2 \, \text{mm/h} (since the height is increasing).
  3. Derivative Calculation: Now we take the derivative of the surface area with respect to time: dSdt=2π(d(rh)dt)+2π(d(r2)dt)\frac{dS}{dt} = 2\pi\left(\frac{d(rh)}{dt}\right) + 2\pi\left(\frac{d(r^2)}{dt}\right).
  4. Product Rule Application: Using the product rule for d(rh)dt\frac{d(rh)}{dt}, we get d(rh)dt=r(dhdt)+h(drdt)\frac{d(rh)}{dt} = r\left(\frac{dh}{dt}\right) + h\left(\frac{dr}{dt}\right).
  5. Substitute Values: Substitute the values of rr, hh, rac{dr}{dt}, and rac{dh}{dt} into the equation: rac{d(rh)}{dt} = 8(2) + 3(-9).
  6. Calculate d(rh)dt\frac{d(rh)}{dt}: Calculate d(rh)dt\frac{d(rh)}{dt}: d(rh)dt=16−27=−11 mm2/h\frac{d(rh)}{dt} = 16 - 27 = -11 \, \text{mm}^2/\text{h}.
  7. Calculate d(r2)dt\frac{d(r^2)}{dt}: Now, calculate d(r2)dt\frac{d(r^2)}{dt}: d(r2)dt=2rdrdt=2(8)(−9)\frac{d(r^2)}{dt} = 2r\frac{dr}{dt} = 2(8)(-9).
  8. Calculate dSdt\frac{dS}{dt}: Calculate d(r2)dt\frac{d(r^2)}{dt}: d(r2)dt=−144 mm2/h\frac{d(r^2)}{dt} = -144 \, \text{mm}^2/\text{h}.
  9. Final Calculation: Now we have all the parts to find dSdt\frac{dS}{dt}: dSdt=2π(−11)+2π(−144)\frac{dS}{dt} = 2\pi(-11) + 2\pi(-144).
  10. Conclusion: Calculate dSdt\frac{dS}{dt}: dSdt=−22π−288π=−310π\frac{dS}{dt} = -22\pi - 288\pi = -310\pi mm2^2/h.
  11. Conclusion: Calculate dSdt\frac{dS}{dt}: dSdt=−22π−288π=−310π\frac{dS}{dt} = -22\pi - 288\pi = -310\pi mm2^2/h. The rate of change of the surface area of the cylinder at that instant is −310π-310\pi square millimeters per hour, which corresponds to answer choice (B).

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