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The radius of the base of a cone is increasing at a rate of 10 meters per second. The height of the cone is fixed at 6 meters. At a certain instant, the radius is 1 meter. What is the rate of change of the volume of the cone at that instant (in cubic meters per second)? Choose 1 answer: (A) 20 pi (B) 40 pi (C) 400 pi (D) 200 pi The volume of a cone with radius r and height h is pir^(2)(h)/(3).

The radius of the base of a cone is increasing at a rate of 1010 meters per second.\newlineThe height of the cone is fixed at 66 meters.\newlineAt a certain instant, the radius is 11 meter.\newlineWhat is the rate of change of the volume of the cone at that instant (in cubic meters per second)?\newlineChoose 11 answer:\newline(A) 20π 20 \pi \newline(B) 40π 40 \pi \newline(C) 400π 400 \pi \newline(D) 200π 200 \pi \newlineThe volume of a cone with radius r r and height h h is πr2h3 \pi r^{2} \frac{h}{3} .

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Q. The radius of the base of a cone is increasing at a rate of 1010 meters per second.\newlineThe height of the cone is fixed at 66 meters.\newlineAt a certain instant, the radius is 11 meter.\newlineWhat is the rate of change of the volume of the cone at that instant (in cubic meters per second)?\newlineChoose 11 answer:\newline(A) 20π 20 \pi \newline(B) 40π 40 \pi \newline(C) 400π 400 \pi \newline(D) 200π 200 \pi \newlineThe volume of a cone with radius r r and height h h is πr2h3 \pi r^{2} \frac{h}{3} .
  1. Volume Formula Derivation: The formula for the volume of a cone is V=(13)πr2hV = (\frac{1}{3}) \pi r^2 h. We need to find dVdt\frac{dV}{dt}, the rate of change of volume with respect to time.
  2. Given Information: Given: drdt=10 m/s\frac{dr}{dt} = 10 \, \text{m/s} (rate at which radius increases), h=6 mh = 6 \, \text{m} (height of the cone is constant), and r=1 mr = 1 \, \text{m} (radius at the instant we are considering).
  3. Differentiation with Respect to Time: Differentiate the volume formula with respect to time tt to find dVdt\frac{dV}{dt}. dVdt=13πd(r2)dth\frac{dV}{dt} = \frac{1}{3} \pi \frac{d(r^2)}{dt} h.
  4. Substitution and Calculation: Since r2r^2 differentiates to 2r⋅drdt2r \cdot \frac{dr}{dt}, we substitute drdt=10 m/s\frac{dr}{dt} = 10 \, \text{m/s} and r=1 mr = 1 \, \text{m} into the equation. dVdt=(13)⋅π⋅2⋅1 m⋅10 m/s⋅6 m.\frac{dV}{dt} = \left(\frac{1}{3}\right) \cdot \pi \cdot 2 \cdot 1 \, \text{m} \cdot 10 \, \text{m/s} \cdot 6 \, \text{m}.
  5. Final Simplification: Simplify the expression: dVdt=(13)⋅π⋅2⋅10⋅6=40π\frac{dV}{dt} = \left(\frac{1}{3}\right) \cdot \pi \cdot 2 \cdot 10 \cdot 6 = 40 \pi cubic meters per second.

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