Q.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.

Concept: Related Rates — we relate the rates of change of the sides of a right triangle using implicit differentiation.

Let $x$ be the horizontal distance (in metres) from the point directly below the kite, and let $s$ be the length of the string (in metres). The height is constant at $3$ m. By Pythagoras:

Differentiate both sides with respect to time $t$:

$2x \frac{dx}{dt} = 2s \frac{ds}{dt} \quad\Rightarrow\quad \frac{ds}{dt} = \frac{x}{s} \cdot \frac{dx}{dt}$ …

Why Related Rates Works Here

The problem gives you a rate of change (horizontal speed of the kite) and asks for another rate (how fast the string lengthens). Both quantities change with time, and they are linked by a geometric relationship — in this case, the right triangle formed by the height, horizontal distance, and the string.

The key insight: you don't need to know the actual horizontal distance at every moment. You only need the relationship between the rates at the instant when the string is 5 m long. That's the power of implicit differentiation with respect to time.

Step-by-Step Solution

1. Set up the variables and convert units

Let:

• $x$ = horizontal distance of the kite from the point directly below it (in metres)
• $y$ = height of the kite = 3 m (constant — the kite flies at a fixed height)
• $s$ = length of the string out (in metres)

We are given:

• $\frac{dx}{dt} = 200 \text{ cm/s} = 2 \text{ m/s}$ (since $200 \text{ cm} = 2 \text{ m}$)
• At the instant of interest: $s = 5 \text{ m}$, $y = 3 \text{ m}$

We need $\frac{ds}{dt}$ at that instant.

2. Write the geometric relation

The string, the height, and the horizontal distance form a right triangle:

Since $y = 3$ is constant, this becomes:

3. Differentiate implicitly with respect to time $t$

Differentiate both sides:

Divide through by 2:

Method: Solving a Related-Rates Problem via Implicit Differentiation

Use this method whenever two or more quantities are changing with time, one rate is given, and you need another rate at a specific instant — classic setups include a right triangle formed by a moving object (kite, ladder, shadow) or a growing/shrinking solid.

Steps

Step 1: Name every changing quantity and identify what is constant

Assign a variable to every quantity that changes with time $t$, and note which quantities in the picture stay fixed throughout. List the known rate(s) and the rate you must find, being careful to convert every length/rate to the same unit system before you substitute anything.

Step 2: Write an equation connecting the quantities

Use the geometry of the situation (commonly the Pythagorean theorem for a right-triangle setup, or a standard volume/area formula) to write a single equation relating the variables — this equation must hold true at every instant, not just the one you're asked about.

Step 3: Differentiate both sides with respect to time $t$ …

Common Mistakes

Mistake 1: Substituting numbers before differentiating

Plugging $s=5$ into the relation $x^2+9=s^2$ too early — treating $s$ as already fixed before differentiating — removes the variable whose rate you need and leaves no $ds/dt$ term. Differentiate the general relation with respect to $t$ first, then substitute the known numbers.

Mistake 2: Not converting units before combining rates …