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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) 155 pi (B) 310 pi (C) -155 pi (D) -310 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) 155π 155 \pi \newline(B) 310π 310 \pi \newline(C) −155π -155 \pi \newline(D) −310π -310 \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) 155π 155 \pi \newline(B) 310π 310 \pi \newline(C) −155π -155 \pi \newline(D) −310π -310 \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. Write Given Information: First, let's write down what we know:\newlineThe radius rr is decreasing at 99 mm/hr, so drdt=−9\frac{dr}{dt} = -9 mm/hr.\newlineThe height hh is increasing at 22 mm/hr, so dhdt=2\frac{dh}{dt} = 2 mm/hr.\newlineAt the instant we're considering, r=8r = 8 mm and h=3h = 3 mm.\newlineThe formula for the surface area AA of a cylinder is A=2πrh+2πr2A = 2\pi rh + 2\pi r^2.
  2. Find Rate of Change: Now, we need to find the rate of change of the surface area, dAdt\frac{dA}{dt}. To do this, we'll use the formula for the derivative of the surface area with respect to time: dAdt=ddt(2πrh+2πr2)\frac{dA}{dt} = \frac{d}{dt}(2\pi rh + 2\pi r^2).
  3. Apply Derivative Formula: Using the product rule and chain rule for differentiation, we get: dAdt=2π(rdhdt+hdrdt)+4πrdrdt\frac{dA}{dt} = 2\pi(r\frac{dh}{dt} + h\frac{dr}{dt}) + 4\pi r\frac{dr}{dt}.
  4. Substitute Known Values: Plug in the values we know: dAdt=2π(8 mm×2 mm/hr+3 mm×−9 mm/hr)+4π(8 mm×−9 mm/hr)\frac{dA}{dt} = 2\pi(8 \, \text{mm} \times 2 \, \text{mm/hr} + 3 \, \text{mm} \times -9 \, \text{mm/hr}) + 4\pi(8 \, \text{mm} \times -9 \, \text{mm/hr}).
  5. Perform Calculations: Now, let's do the math:\newlinedAdt=2π(16 mm2/hr−27 mm2/hr)−4π(72 mm2/hr)\frac{dA}{dt} = 2\pi(16 \text{ mm}^2/\text{hr} - 27 \text{ mm}^2/\text{hr}) - 4\pi(72 \text{ mm}^2/\text{hr}).
  6. Simplify Expression: Simplify the expression: dAdt=2π(−11 mm2/hr)−288π mm2/hr\frac{dA}{dt} = 2\pi(-11 \text{ mm}^2/\text{hr}) - 288\pi \text{ mm}^2/\text{hr}.
  7. Combine Terms: Combine the terms: dAdt=−22π mm2/hr−288π mm2/hr\frac{dA}{dt} = -22\pi \, \text{mm}^2/\text{hr} - 288\pi \, \text{mm}^2/\text{hr}.
  8. Add Together: Add the two terms together: dAdt=−310π mm2/hr.\frac{dA}{dt} = -310\pi \, \text{mm}^2/\text{hr}.

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