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.

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## Comments

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TopNewestkhanacademy.com also has very nice explanations of this.

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The problem says that the rate of filling is 3 cubic feet per minute, so why is \(\frac{dv}{dt} = 2\)?

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Oops...typo....Sorry for that....

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Nice explanation thanks

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