Question

Suppose the radius of the cylinder is decreasing at a rate of 2 cm/sec, but the volume is constant. How fast is the height changing when the radius is 10 cm and the height is 35 cm? Include units with your answer. Hint: If the volume is constant, what do you know about the rate of change of the volume?

Answer 1

The height changing at "14 cm/s".

According to the question,

• r = 10 cm
• h = 35 cm

• $\frac{dr}{dt} = -2 \ cm/s$

As we know the formula,

• $V = \pi r^2 h$

then,

→ $\frac{dv}{dt} = \pi [2r\frac{dr}{dt} h+ r^2 \frac{dh}{dt} ]$

→   $0 = \pi [2 \pi h\frac{dr}{dt} + r^2 \frac{dh}{dt} ]$

By substituting the given values,

→ $0 = 2\times 10\times 35\times (-2)+(10)^2\times \frac{dh}{dt}$

  $\frac{dh}{dt} = \frac{2\times 10\times 35\times 2}{10\times 10}$

  $\frac{dh}{dt} = \frac{140}{10}$

  $\frac{dh}{dt} = 4 \ cm/s$                                  

Thus the answer above is right.

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