Q.A kite is moving horizontally at a height of $151.5$ metres. If the speed of the kite is $10$ m/s, how fast is the string being let out when the kite is $250$ m away from the boy who is flying the kite? The height of the boy is $1.5$ m.

Concept: Related Rates — we relate the rate of change of the string length to the given horizontal speed using the Pythagorean theorem.

Let $x$ be the horizontal distance of the kite from the boy, and let $s$ be the length of the string. The vertical height of the kite above the boy’s hand is $151.5 - 1.5 = 150$ m.

By Pythagoras:

Differentiate with respect to time $t$:

When the kite is $250$ m away, $s = 250$ and $\frac{dx}{dt} = 10$ m/s. Find $x$:

Thus:

We have a kite flying at a constant height, moving horizontally away from the boy. The string is being let out as the kite moves. The question: how fast is the string length increasing at the instant the kite is 250 m away (along the string) from the boy?

The key idea in related rates is that two (or more) quantities change with time, and they are connected by a geometric relationship. Here, the horizontal distance $x$ of the kite from the boy, the height $h$ of the kite above the boy’s hand, and the string length $L$ form a right triangle. As time passes, $x$ increases, $L$ increases, but the height stays constant. We know $dx/dt$ (the kite’s horizontal speed) and want $dL/dt$ (the rate at which string is let out) at a specific instant.

Let’s set up carefully.

1. Define variables and constants

Let:

• $x$ = horizontal distance from the boy to the kite (in metres).
• $h$ = vertical height of the kite above the boy’s hand. The boy’s height is 1.5 m, and the kite is at 151.5 m altitude. So the height above the boy’s hand is:

This is constant.

• $L$ = length of the string (in metres), which is the hypotenuse of the right triangle.

At the instant of interest, $L = 250$ m.

We are given:

• $\frac{dx}{dt} = 10$ m/s (the kite’s horizontal speed, positive because distance increases).

We need $\frac{dL}{dt}$ when $L = 250$ m.

2. Relate the quantities

By the Pythagorean theorem:

Here $h = 150$ m, constant. So:

3. Differentiate with respect to time

Differentiate both sides implicitly (remember $L$ and $x$ are functions of $t$, $h$ is constant):

Divide through by 2:

So:

This makes sense: the rate of change of the string length is the horizontal speed times the ratio of horizontal distance to string length — essentially the component of velocity along the string.

4. Find $x$ at the instant $L = 250$

From $L^2 = x^2 + h^2$:

(Only the positive root matters — distance.)

5. Plug into the derivative

So the string is being let out at 8 metres per second at that instant.

Method: Related Rates via the Pythagorean Theorem

When a problem describes a right-angle setup — a fixed height and a horizontal distance connected by a hypotenuse such as a string, a ladder, or a line of sight — the Pythagorean theorem is the equation that links the changing quantities.

Steps

Step 1: Sketch the right triangle and label the sides.

Identify which side is genuinely fixed (a constant height or wall), which side is the changing leg, and which side is the changing hypotenuse.

Step 2: Write the Pythagorean relation.

where $c$ is the hypotenuse and $a$, $b$ are the legs. If one leg is constant, it still appears in the equation, but its derivative will vanish in the next step.

Step 3: Differentiate both sides with respect to time.

With $b$ constant:

Step 4: Find the missing side length at the given instant.

Use the original Pythagorean relation (not the derivative) to solve for whichever leg or hypotenuse value is not given directly, using the known value(s) at that instant.

Step 5: Substitute the known rate and lengths to get the answer.

Plug the given rate (e.g. horizontal speed) and the side length found in Step 4 into the differentiated equation from Step 3, and simplify.

Common Mistakes

Mistake 1: Using the kite's full altitude (151.5 m) as the triangle's vertical leg

Why it's wrong: the string runs from the boy's hand, not from the ground, so the true vertical distance is $151.5-1.5=150$ m; using $151.5$ m directly gives a wrong horizontal distance $x$ and therefore a wrong $\frac{ds}{dt}$. Correct approach: always subtract the person's own height from the object's height above ground to get the vertical leg of the right triangle.

Mistake 2: Assuming the given horizontal speed IS the rate the string is let out

Why it's wrong: $10$ m/s is $\frac{dx}{dt}$, the rate of the horizontal distance, but the question asks for $\frac{ds}{dt}$, the rate of the string length — these differ by the factor $\frac{x}{s}$ (effectively $\cos\theta$ of the string's angle). Treating them as equal skips the geometric link entirely. Correct approach: differentiate $s^2=x^2+h^2$ to get $\frac{ds}{dt}=\frac{x}{s}\frac{dx}{dt}$ before substituting any numbers.

Mistake 3: Forgetting to solve for the missing side $x$ before substituting into the rate formula

Why it's wrong: the formula $\frac{ds}{dt}=\frac{x}{s}\frac{dx}{dt}$ needs both $x$ and $s$ at the given instant, but only $s=250$ is stated directly — $x$ must be recovered from Pythagoras first. Skipping this and plugging $s=250$ in place of $x$ gives a badly wrong ratio.