Question

A conical tank (with vertex down) is $$10$$ feet across the top and $$12$$ feet deep. Water is flowing into the tank at a rate of $$5$$ cubic feet per minute. Find the rate of change of the depth of the water when the water is $$4$$ feet deep.

Answer 1

Hint: In this question,we need to find the rate of change of the depth of the water when the water is $$4$$ feet deep. Given that the radius of the tank is $$10$$ feet. From this, we can find the radius of the tank. Also the height of the tank is $$12$$ feet and the water is $$8$$ feet deep. The rate of change of the depth of the water can be found by differentiating the volume of the cone. The differentiation is nothing but a rate of change of function with respect to an independent variable given in the function. First let us consider the radius of the tank to be $$r$$ and $$h$$ be the height of the tank . We need to substitute the value of $$r$$ in the volume formula. Then on differentiating we can find the rate of change of the depth of the water when the water is $$4$$ feet deep.

Formula used:
The volume of the cone ,
$$V={\displaystyle \frac{1}{3}}\pi r^{2}h$$
Where $$r$$ is the radius of the cone and $$h$$ is the height of the cone.
Derivative rule used :
$${\displaystyle \frac{d}{dx}}{x}^n=n{x}^{n–1}$$

Complete step-by-step answer:

Let us consider the radius of the tank to be $$r$$ and $$h$$ be the height of the tank .
Given, the conical tank is $$10$$ feet across and $$12$$ feet deep, that is the radius of the tank is $$10$$ feet and height is $$12$$ feet.
$$r=10$$
By relating the radius $$r$$ to the height $$h$$ ,
$$\Rightarrow {\displaystyle \frac{r}{h}}={\displaystyle \frac{10}{12}}$$
Thus $$r={\displaystyle \frac{5}{6}}h$$
We know volume of the cone is $${\displaystyle \frac{1}{3}}\pi r^{2}h$$
$$V={\displaystyle \frac{1}{3}}\pi r^{2}h$$
By substituting $$r={\displaystyle \frac{5}{6}}h$$
We get,
$$V={\displaystyle \frac{1}{3}}\pi {({\displaystyle \frac{5}{6}}h)}^{2}h$$
On simplifying,
We get,
$$V={\displaystyle \frac{25}{108}}\pi h^3$$
On differentiating $$V$$ with respect to time $$t$$ ,
We get,
$${\displaystyle \frac{dV}{dt}}={\displaystyle \frac{25}{108}}\pi \left(3h^2{\displaystyle \frac{dh}{dt}}\right)$$
Also given the rate change of volume is $$5$$ cubic feet per minute.
By substituting the value of volume rate,
We get,
$$5={\displaystyle \frac{25}{108}}\pi \left(3h^2{\displaystyle \frac{dh}{dt}}\right)$$
We need to find the rate of change of the depth of the water when the water is $$4$$ feet deep,
By substituting the value of $$h=4$$ and $$\pi =3.14$$ ,
We get,
$$5={\displaystyle \frac{25}{108}}\left(3.14\right)\left(3{\left(4\right)}^2{\displaystyle \frac{dh}{dt}}\right)$$
On simplifying,
We get,
$$5=0.719((48){\displaystyle \frac{dh}{dx}})$$
On multiplying the term inside,
We get,
$$5=34.5{\displaystyle \frac{dh}{dt}}$$
By rewriting the terms,
We get,
$${\displaystyle \frac{dh}{dx}}={\displaystyle \frac{5}{34.5}}$$
On simplifying,
We get $${\displaystyle \frac{dh}{dt}}\approx 0.146$$ feet/min
Thus we get the rate of change of the depth of the water is approximately $$0.146$$ feet/min.
Final answer :
The rate of change of the depth of the water is approximately $$0.146$$ feet/min.

Note: Mathematically , Derivative helps in solving the problems in calculus and in differential equations. The derivative of $$y$$ with respect to $$x$$ is represented as $${\displaystyle \frac{dy}{dx}}$$ . Here the notation $${\displaystyle \frac{dy}{dx}}$$ is known as Leibniz's notation . In derivation, there are two types of derivative namely first order derivative and second order derivative. A simple example for a derivative is the derivative of $$x^3$$ is $$3x$$ . Derivative is applicable in trigonometric functions also .