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# Related Rates

This note along with the problem set is based on basic applications of differentiation to real-world problems.To understand this note you need to have basic ideas of Differential Calculus.

This note is about problems on related rates-which is the study of change of one variable with respect to another provided that we find a constant parameter providing a relationship between those two variables.AN important point for approaching problems like these is to find that constant variable providing the link between the variables we want to evaluate.

$$\textbf{Problem.}$$Consider a conical tank. Its radius at the top is 5 feet, and it’s 12 feet high. It’s being filled with water at the rate of 3 cubic feet per minute. How fast is the water level rising when it is 7 feet high? (See the given figure)

$$\textbf{Ans.}$$

From the second figure by triangle similarity we have $\frac{r}{5} = \frac{h}{12}$ $r = 5 \frac{h}{12}$

Now we need to find the constant parameter.Clearly it is the rate of change of volume!!Since given $$\dfrac{\text{d}V}{\text{d}t} = 2$$. We have $V = \frac{1}{3}\pi r^{2} h$

We put the value of $$r$$ in terms of $$h$$ to get

$V = \frac{1}{3}\pi (5 \frac{h}{12})^{2} h$

$V = \frac{1}{3}\pi (5 \frac{h}{12})^{2} h$

$V = \frac{1}{3}\pi (25 \frac{h^3}{144})$

Now we differentiate both sides to get:

$\dfrac{\text{d}V}{\text{d}t} = \frac{1}{3}\pi (25*3 \frac{h^2}{144})\dfrac{\text{d}h}{\text{d}t}$

.Now we simply put the values of $$h$$ and $$\dfrac{\text{d}V}{\text{d}t}$$ to get the correct answer.

Now that you have learnt how to approach these situations try your hand at the problems.They are not that hard......After that you'll see derivatives everywhere.

2 years, 10 months ago

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The problem says that the rate of filling is 3 cubic feet per minute, so why is $$\frac{dv}{dt} = 2$$? · 2 years, 10 months ago

Oops...typo....Sorry for that.... · 2 years, 10 months ago

khanacademy.com also has very nice explanations of this. · 2 years, 10 months ago