CBSE 2025, Set 1

Question

**Case Study - 2**
A ladder of fixed length 'h' is to be placed along the wall such that it is free to move along the height of the wall.
![Diagram Ladder](https://www.edvaya.com/images/cbse/mathematics/2025/set1/q37-ladder.webp)

Based upon the above information, answer the following question:
(iii) (b) If the foot of the ladder whose length is $5 	ext{ m}$, is being pulled towards the wall such that the rate of decrease of distance ($y$) is $2 	ext{ m/s}$. then at what rate is the height on the wall ($x$) increasing, when the foot of the ladder is $3 	ext{ m}$ away from the wall?

Edvaya · Accepted Answer

The correct answer is Subjective. Given: Length of ladder $h = 5 	ext{ m}$. The foot is being pulled *towards* the wall, so $y$ is decreasing. Rate of decrease $\frac{dy}{dt} = -2 	ext{ m/s}$. We need to find $\frac{dx}{dt}$ when $y = 3 	ext{ m}$. First, find $x$ when $y = 3$ using Pythagoras theorem: $x^2 + y^2 = h^2$ $x^2 + (3)^2 = (5)^2$ $x^2 + 9 = 25 \implies x^2 = 16 \implies x = 4 	ext{ m}$ Now, differentiate the relation $x^2 + y^2 = h^2$ with respect to time $t$: $\frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(h^2)$ $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$ (since $h$ is constant, its derivative is $0$) Divide by $2$: $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$ Substitute the known values ($x = 4$, $y = 3$, $\frac{dy}{dt} = -2$): $4 \frac{dx}{dt} + 3(-2) = 0$ $4 \frac{dx}{dt} - 6 = 0$ $4 \frac{dx}{dt} = 6 \implies \frac{dx}{dt} = \frac{6}{4} = 1.5 	ext{ m/s}$. The height on the wall is increasing at a rate of $1.5 	ext{ m/s}$.

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Mathematics > Calculus > Application Of Derivatives

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