134.webp" data-orsrc="https://uploads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/d8e5e29f4134.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0px auto;max-height:40vh;vertical-align:baseline" alt="MHT CET 2023 12th May Morning Shift Mathematics - Application of Derivatives Question 224 English Explanation">

According to the figure, $$x^2+y^2=25$$ ..... (i)

Note that $$\cos \theta=\frac{\mathrm{OB}}{\mathrm{AB}}=\frac{x}{5}$$

$$\begin{aligned} & \therefore \quad x=5 \cos \theta \\ & \therefore \quad \text { (i) } \Rightarrow 25 \cos ^2 \theta+y^2=25 \end{aligned}$$

Differentiating w.r.t. 't', we get

$$\begin{aligned} & \quad-50 \cos \theta \sin \theta \frac{\mathrm{d} \theta}{\mathrm{dt}}+2 y \frac{\mathrm{d} y}{\mathrm{dt}}=0 \\ & 25 \sin \theta \cos \theta \frac{\mathrm{d} \theta}{\mathrm{dt}}=y \frac{\mathrm{d} y}{\mathrm{dt}} \\ & \therefore \quad 25 \sin \theta \cos \theta \frac{\mathrm{d} \theta}{\mathrm{dt}}=y(-0.1) \\ & \quad \ldots\left[\because \frac{\mathrm{d} y}{\mathrm{~d} x}=-10 \mathrm{~cm} / \mathrm{s}=-0.1 \mathrm{~m} / \mathrm{s}\right] \\ & \therefore \quad 25 \sin \theta \cos \theta \frac{\mathrm{d} \theta}{\mathrm{dt}}=-(0.1) y \quad \ldots \text { (ii) } \\ & \quad \text { at } x=4, \cos \theta=\frac{4}{5}, \sin \theta=\frac{3}{5} \quad \text { and } y=3 \\ & \therefore \quad \text { (ii) } \Rightarrow 25 \times \frac{3}{5} \times \frac{4}{5} \times \frac{\mathrm{d} \theta}{\mathrm{dt}}=-0.3 \\ & \Rightarrow \frac{\mathrm{d} \theta}{\mathrm{dt}}=-0.025 \end{aligned}$$

i.e., the angle is decreasing at the rate of $$0.025 \mathrm{~rad} / \mathrm{s}$$

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