Sample 2024-25 Class 12 Maths Q26 Solution: Rate at which a kite string is being released

The question

A kite is flying at a height of 3 metres and 5 metres of string is out. If the kite is moving away horizontally at the rate of 200 cm/s, find the rate at which the string is being released.

Exam answer

Let the kite K be at height 3 m above the point M, let O be the person, $OM = x$ m and the string $OK = y$ m.

x^2 + 3^2 = y^2

Why?

Triangle OMK has a right angle at M, so Pythagoras' theorem gives $OM^2 + MK^2 = OK^2$ ⁠.

The kite moves horizontally, so its height $MK = 3$ m stays fixed. Only $x$ and $y$ change with time $t$ ⁠.

When $y = 5$ ⁠, $x^2 = 25 - 9 = 16$ ⁠, so $x = 4$ ⁠.

Why?

$x$ is a distance, so we take the positive root of $x^2 = 16$ ⁠. The triangle at this instant is the 3, 4, 5 right triangle.

\begin{aligned} x^2 + 3^2 &= 5^2 \\ x^2 + 9 &= 25 \\ x^2 &= 25 - 9 = 16 \\ x &= \sqrt{16} = 4 \end{aligned}

Differentiating w.r.t. $t$ ⁠,

2x\frac{dx}{dt} = 2y\frac{dy}{dt}

Why?

Both $x$ and $y$ depend on $t$ ⁠, so by the chain rule

\frac{d}{dt}(x^2) = 2x\frac{dx}{dt}

and in the same way for $y^2$ ⁠. The constant $3^2 = 9$ has derivative 0. Dividing both sides by 2 gives the next line.

\Rightarrow x\frac{dx}{dt} = y\frac{dy}{dt}

Putting $x = 4$ ⁠, $y = 5$ ⁠, $\dfrac{dx}{dt} = 200$ cm/s,

4(200) = 5\frac{dy}{dt} \Rightarrow \frac{dy}{dt} = 160

Why?

The rate at which the string is released is the rate at which its length $y$ grows, that is $\frac{dy}{dt}$ ⁠.

\begin{aligned} 4(200) &= 5\frac{dy}{dt} \\ 800 &= 5\frac{dy}{dt} \\ \frac{dy}{dt} &= \frac{800}{5} \quad (\div\,5) \\ &= 160 \end{aligned}

The lengths are in metres and the speed in cm/s. That is fine: the equation only uses the ratio $x : y = 4 : 5$ ⁠, so $\frac{dy}{dt}$ comes out in cm/s, like $\frac{dx}{dt}$ ⁠.

$\therefore$ The string is being released at the rate of 160 cm/s.

Where this comes from

Chapter 6, Application of Derivatives: Rate of change: kite string being released.

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: The kite moves at 200 cm/s, so $\frac{dx}{dt} = 200$ ⁠, and the moment is when 5 m of string is out. Differentiate $x^2 + 3^2 = y^2$ with respect to $t$ ⁠, then put in $x = 4$ and $y = 5$ ⁠.

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Q26: Rate at which a kite string is being released