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Calculus IChapter 4: Application of DerivativesSection 4-1

Rates of Change — Question 3

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Question 3

Problem:

Water is being poured into a cylindrical tank at a rate of 500cm3/s500 \, \text{cm}^3/\text{s}. The tank has a radius of 10cm10 \, \text{cm}. How fast is the water level rising when the water is 20cm20 \, \text{cm} deep?

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Question 3 - Solution

The volume of a cylinder is: V=πr2hV = \pi r^2 h

Given: dVdt=500\frac{dV}{dt} = 500 , r=10r = 10 , h=20h = 20

We differentiate with respect to time: dVdt=πr2dhdt⇒500=π(10)2dhdt⇒500=100πdhdt⇒dhdt=500100π=5π\frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \Rightarrow 500 = \pi (10)^2 \frac{dh}{dt} \Rightarrow 500 = 100\pi \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{500}{100\pi} = \frac{5}{\pi}

Answer: dhdt=5π cm/sec\boxed{\frac{dh}{dt} = \frac{5}{\pi} \text{ cm/sec}}

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