CBSE Board Examination 2023 · Section E · 4 marks

Q37: Case study: water draining from a conical tank

Board 2023Ch 6 · Application of Derivatives

The question

A tank, as shown in the figure below, formed using a combination of a cylinder and a cone, offers better drainage as compared to a flat bottomed tank.

The tank: a cylinder on top of a cone, with a tap at the bottom of the cone.

A tap is connected to such a tank whose conical part is full of water. Water is dripping out from a tap at the bottom at the uniform rate of 2 cm³/s. The semi-vertical angle of the conical tank is 45°.

On the basis of given information, answer the following questions :

(i) Find the volume of water in the tank in terms of its radius r. (1 mark)

(ii) Find rate of change of radius at an instant when $r = 2\sqrt2$ cm. (1 mark)

(iii) (a) Find the rate at which the wet surface of the conical tank is decreasing at an instant when radius $r = 2\sqrt2$ cm. (2 marks)

(iii) (b) Find the rate of change of height ‘h’ at an instant when slant height is 4 cm. (2 marks)

Exam answer

Let r, h and l be the radius, height and slant height of the water cone at time t.

(i) The semi-vertical angle is $45^\circ$ ⁠, so $\tan 45^\circ = \frac{r}{h} = 1$ ⁠, i.e. $r = h$ ⁠.

Why?

The semi-vertical angle is the angle between the axis of the cone and its slant side. The water fills a smaller cone with the same angle.

In the right triangle formed by the height $h$ along the axis and a radius $r$ ⁠, the side opposite the angle is $r$ ⁠. So $\tan 45^\circ = \frac{r}{h}$ ⁠, and $\tan 45^\circ = 1$ ⁠.

$\therefore\ V = \frac13\pi r^2h = \frac13\pi r^3$

Why?

Volume of a cone:

V = \frac13\pi r^2h

Put $h = r$ ⁠, so that $V$ depends on $r$ alone.

\begin{aligned} V &= \frac13\pi r^2(r) \\ &= \frac13\pi r^3 \end{aligned}

(ii) Water drips out at 2 cm³/s, so $\frac{dV}{dt} = -2$ ⁠.

Why?

The volume of water is going down, so its rate of change is negative.

\frac{dV}{dt} = \pi r^2\frac{dr}{dt}

Why?

Differentiate $V = \frac13\pi r^3$ with respect to $t$ ⁠. $r$ changes with time, so by the chain rule

\frac{d}{dt}\left(r^3\right) = 3r^2\frac{dr}{dt}

and $\frac13\pi\cdot 3r^2 = \pi r^2$ ⁠.

\Rightarrow\ -2 = \pi\left(2\sqrt2\right)^2\frac{dr}{dt} = 8\pi\frac{dr}{dt}

$\Rightarrow\ \left(\frac{dr}{dt}\right)_{r = 2\sqrt2} = -\frac{1}{4\pi}$ cm/s

Why?

$\left(2\sqrt2\right)^2 = 4 \times 2 = 8$ ⁠. Then divide by $8\pi$ ⁠:

\begin{aligned} -2 &= 8\pi\frac{dr}{dt} \\ \frac{dr}{dt} &= \frac{-2}{8\pi} \\ &= -\frac{2 \div 2}{8\pi \div 2} = -\frac{1}{4\pi} \end{aligned}

The minus sign says the radius is getting smaller.

Watch out: put $r = 2\sqrt2$ only after differentiating.

(iii) (a) $l = \sqrt{r^2 + h^2} = \sqrt2\,r$ ⁠. The wet surface is the curved surface of the cone:

C = \pi rl = \pi r\cdot\sqrt2\,r = \sqrt2\,\pi r^2

Why?

By Pythagoras, $l^2 = r^2 + h^2$ ⁠. With $h = r$ ⁠, $l^2 = 2r^2$ ⁠, so $l = \sqrt2\,r$ ⁠.

\begin{aligned} l^2 &= r^2 + r^2 = 2r^2 \\ l &= \sqrt{2r^2} = \sqrt2\,r \end{aligned}

The wet surface is the part of the cone in contact with the water, which is its curved surface. Its area is $\pi rl$ ⁠.

\frac{dC}{dt} = \sqrt2\,\pi\cdot 2r\,\frac{dr}{dt}

Why?

Differentiate with respect to $t$ ⁠, using the chain rule:

\frac{d}{dt}\left(r^2\right) = 2r\frac{dr}{dt}

The value of $\frac{dr}{dt}$ at $r = 2\sqrt2$ comes from part (ii).

\begin{aligned} \frac{dC}{dt} &= \sqrt2\,\pi\cdot 2r\,\frac{dr}{dt} \\ &= 2\sqrt2\,\pi r\,\frac{dr}{dt} \end{aligned}

Now put $r = 2\sqrt2$ and $\frac{dr}{dt} = -\frac{1}{4\pi}$ ⁠.

\begin{aligned} &\left(\frac{dC}{dt}\right)_{r = 2\sqrt2} \\ &= 2\sqrt2\,\pi\left(2\sqrt2\right)\left(-\frac{1}{4\pi}\right) \\ &= -2 \end{aligned}

Why?

$2\sqrt2 \times 2\sqrt2 = 8$ ⁠, so this is

\begin{aligned} 2\sqrt2 \times 2\sqrt2 &= (2 \times 2)(\sqrt2 \times \sqrt2) \\ &= 4 \times 2 = 8 \end{aligned}

\begin{aligned} 8\pi \times \left(-\frac{1}{4\pi}\right) &= -\frac{8\pi}{4\pi} \\ &= -2 \end{aligned}

The negative rate means the wet surface is shrinking. The question asks how fast it is decreasing, so the answer is the positive number 2.

$\therefore$ The wet surface is decreasing at the rate of 2 cm²/s.

(iii) (b) $l^2 = h^2 + r^2 = 2h^2$ (as $r = h$ ⁠).

$l = 4 \Rightarrow r = h = 2\sqrt2$ ⁠.

Why?

With $l = 4$ ⁠, $2h^2 = 16$ ⁠, so $h^2 = 8$ and $h = \sqrt8 = 2\sqrt2$ (a height is positive). As $r = h$ ⁠, also $r = 2\sqrt2$ ⁠.

\begin{aligned} 2h^2 &= 4^2 = 16 \\ h^2 &= \frac{16}{2} = 8 \\ h &= \sqrt8 = \sqrt{4 \times 2} \\ &= \sqrt4 \times \sqrt2 = 2\sqrt2 \end{aligned}

$h = r \Rightarrow \frac{dh}{dt} = \frac{dr}{dt}$ ⁠, and by (ii) at $r = 2\sqrt2$ ⁠,

Why?

$h$ and $r$ are equal at every moment, so they change at the same rate.

The moment when $l = 4$ is the moment when $r = 2\sqrt2$ ⁠, the same moment as in part (ii). So the rate found there applies.

$\therefore\ \dfrac{dh}{dt} = -\dfrac{1}{4\pi}$ cm/s.

Where this comes from

Chapter 6, Application of Derivatives: Rates of change: draining conical tank.

Question type: Rates in time

How to spot it: click to see

Type 2. Rates in time: how fast is it changing now?

Spot it: words like "per second", "how fast", "at the rate of", and one moment, such as "when the radius is 10 cm".

Here: Water drips out at 2 cm³/s, so $\frac{dV}{dt} = -2$ ⁠. With $r = h$ ⁠, $V = \frac13\pi r^3$ ⁠: differentiate with respect to $t$ ⁠, then put in $r = 2\sqrt2$ for (ii) and (iii)(a), or $h = 2\sqrt2$ for (iii)(b).

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