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Degger [83]
2 years ago
4

The radius of a cone is decreasing at a constant rate of 5 centimeters per minute, and the volume is decreasing at a rate of 148

cubic centimeters per minute. At the instant when the radius of the cone is 222 centimeters and the volume is 21 cubic centimeters, what is the rate of change of the height? The volume of a cone can be found with the equation V=1/3πr^2h. Round your answer to three decimal places.
Mathematics
1 answer:
Maslowich2 years ago
6 0

Answer:

The rate of change for the height of the cone is -2.849\times 10^{-3} centimeters per minute.

Step-by-step explanation:

We derive an expression for the rate of change for the height of the cone by differentiating the volume formula given on statement:

\dot V = \frac{\pi}{3}\cdot (2\cdot r\cdot h \cdot \dot r + r^{2}\cdot \dot h) (1)

Where:

\dot V - Rate of change for the volume of the cone, measured in cubic centimeters per minute.

r - Radius of the cone, measured in centimeters.

h - Height of the cone, measured in centimeters.

\dot r - Rate of change for the radius of the cone, measured in centimeters per minute.

\dot h - Rate of change for the height of the cone, measured in centimeters per minute.

If we know that \dot V = -148\,\frac{cm^{2}}{min}, r = 222\,cm, V = 21\,cm^{3}, \dot r = -5\,\frac{cm}{min}, then the rate of change for the height is:

h = \frac{3\cdot V}{\pi\cdot r^{2}} (2)

Where V is the volume of the cone, measured in cubic centimeters.

h = \frac{3\cdot (21\,cm^{3})}{\pi\cdot (222\,cm)^{2}}

h \approx 4.069\times 10^{-4}\,cm

From (1):

\frac{3\cdot \dot V}{\pi} = 2\cdot r\cdot h\cdot \dot r + r^{2}\cdot \dot h

r^{2}\cdot \dot h = \frac{3\cdot \dot V}{\pi}-2\cdot r\cdot h\cdot \dot r

\dot h = \frac{3\cdot \dot V}{\pi\cdot r^{2}}-\frac{2\cdot h\cdot \dot r}{r}

\dot h = \frac{3\cdot \left(-148\,\frac{cm^{3}}{min} \right)}{\pi\cdot (222\,cm)^{2}} -\frac{2\cdot (4.069\times 10^{-4}\,cm)\cdot \left(-5\,\frac{cm}{min} \right)}{222\,cm}

\dot h = -2.849\times 10^{-3}\,\frac{cm}{min}

The rate of change for the height of the cone is -2.849\times 10^{-3} centimeters per minute.

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