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The radius of a cone is increasing at a rate of 3 centimeters per second and the height of the cone is decreasing at a rate of 4 centimeters per second.
At a certain instant, the radius is 8 centimeters and the height is 10 centimeters.
What is the rate of change of the volume of the cone at that instant (in cubic centimeters per second)?
Choose 1 answer:
(A) 
(224 pi)/(3)
(B) 
-(224 pi)/(3)
(C) 
-(736 pi)/(3)
(D) 
(736 pi)/(3)
The volume of 
a cone with radius 
r and height 
h is 
pir^(2)(h)/(3).

The radius of a cone is increasing at a rate of 33 centimeters per second and the height of the cone is decreasing at a rate of 44 centimeters per second.\newlineAt a certain instant, the radius is 88 centimeters and the height is 1010 centimeters.\newlineWhat is the rate of change of the volume of the cone at that instant (in cubic centimeters per second)?\newlineChoose 11 answer:\newline(A) 224π3 \frac{224 \pi}{3} \newline(B) 224π3 -\frac{224 \pi}{3} \newline(C) 736π3 -\frac{736 \pi}{3} \newline(D) 736π3 \frac{736 \pi}{3} \newlineThe volume of a a cone with radius r r and height h h is πr2h3 \pi r^{2} \frac{h}{3} .

Full solution

Q. The radius of a cone is increasing at a rate of 33 centimeters per second and the height of the cone is decreasing at a rate of 44 centimeters per second.\newlineAt a certain instant, the radius is 88 centimeters and the height is 1010 centimeters.\newlineWhat is the rate of change of the volume of the cone at that instant (in cubic centimeters per second)?\newlineChoose 11 answer:\newline(A) 224π3 \frac{224 \pi}{3} \newline(B) 224π3 -\frac{224 \pi}{3} \newline(C) 736π3 -\frac{736 \pi}{3} \newline(D) 736π3 \frac{736 \pi}{3} \newlineThe volume of a a cone with radius r r and height h h is πr2h3 \pi r^{2} \frac{h}{3} .
  1. Volume of Cone Formula: The formula for the volume of a cone is V=13πr2hV = \frac{1}{3}\pi r^2 h. We need to find dVdt\frac{dV}{dt}, the rate of change of the volume.
  2. Chain Rule Application: To find dVdt\frac{dV}{dt}, we use the chain rule: dVdt=dVdrdrdt+dVdhdhdt\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} + \frac{dV}{dh} \cdot \frac{dh}{dt}.
  3. Calculate dV/drdV/dr: First, we find dV/drdV/dr. Since V=(1/3)πr2hV = (1/3)\pi r^2 h, dV/dr=(2/3)πrhdV/dr = (2/3)\pi rh.
  4. Calculate dVdh\frac{dV}{dh}: Next, we find dVdh\frac{dV}{dh}. Since V=13πr2hV = \frac{1}{3}\pi r^2 h, dVdh=13πr2\frac{dV}{dh} = \frac{1}{3}\pi r^2.
  5. Find drdt\frac{dr}{dt} and dhdt\frac{dh}{dt}: Now we plug in the values for drdt\frac{dr}{dt} and dhdt\frac{dh}{dt}. We know drdt=3cm/s\frac{dr}{dt} = 3 \, \text{cm/s} and dhdt=4cm/s\frac{dh}{dt} = -4 \, \text{cm/s}.
  6. Substitute Values: We also plug in the values for rr and hh at the instant we are considering. r=8r = 8 cm and h=10h = 10 cm.
  7. Calculate dVdt\frac{dV}{dt}: Now we calculate dVdt=(23)πrhdrdt+(13)πr2dhdt\frac{dV}{dt} = \left(\frac{2}{3}\right)\pi rh \cdot \frac{dr}{dt} + \left(\frac{1}{3}\right)\pi r^2 \cdot \frac{dh}{dt}.
  8. Substitute Known Values: Substitute the known values: dVdt=(23)π(8cm)(10cm)×3cm/s+(13)π(8cm)2×(4cm/s)\frac{dV}{dt} = \left(\frac{2}{3}\right)\pi(8\,\text{cm})(10\,\text{cm}) \times 3\,\text{cm/s} + \left(\frac{1}{3}\right)\pi(8\,\text{cm})^2 \times (-4\,\text{cm/s}).
  9. Simplify Expression: Simplify the expression: dVdt=(23)π(80 cm2)×3 cm/s+(13)π(64 cm2)×(4 cm/s)\frac{dV}{dt} = \left(\frac{2}{3}\right)\pi(80 \text{ cm}^2) \times 3 \text{ cm/s} + \left(\frac{1}{3}\right)\pi(64 \text{ cm}^2) \times (-4 \text{ cm/s}).
  10. Perform Calculations: Perform the calculations: dVdt=(23)π(240 cm3/s)+(13)π(256 cm3/s)\frac{dV}{dt} = \left(\frac{2}{3}\right)\pi(240 \text{ cm}^3/s) + \left(\frac{1}{3}\right)\pi(-256 \text{ cm}^3/s).
  11. Combine Terms: Combine the terms: dVdt=(23)π(240 cm3/s)(13)π(256 cm3/s)\frac{dV}{dt} = \left(\frac{2}{3}\right)\pi(240 \text{ cm}^3/s) - \left(\frac{1}{3}\right)\pi(256 \text{ cm}^3/s).
  12. Final Result: dVdt=160πcm3/s85.333πcm3/s\frac{dV}{dt} = 160\pi \, \text{cm}^3/s - 85.333\pi \, \text{cm}^3/s.

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