MHT CET · Maths · Application of Derivatives

Let \(x\) be the length of each of the equal sides of an isosceles triangle and \(\theta\) be the angle between these sides. If \(x\) is increasing at the rate \(\frac{1}{12} \mathrm{~m} /\) hour and \(\theta\) is increasing at the rate \(\frac{\pi}{180} \mathrm{rad} /\) hour, then the rate at which area of the triangle is increasing when \(x=12 \mathrm{~m}\) and \(\theta=\frac{\pi}{4}\) is

A \(\left(\frac{\pi}{5}+\frac{1}{2}\right) \mathrm{m}^2 /\) hour
B \(\sqrt{2}\left(\frac{\pi}{5}+\frac{1}{2}\right) \mathrm{m}^2 /\) hour
C \(2\left(\frac{\pi}{5}+\frac{1}{2}\right) \mathrm{m}^2 /\) hour
D \(\sqrt{3}\left(\frac{\pi}{5}+\frac{1}{2}\right) \mathrm{m}^2 /\) hour

Verified Solution

Answer & Solution

Correct Answer

(B) \(\sqrt{2}\left(\frac{\pi}{5}+\frac{1}{2}\right) \mathrm{m}^2 /\) hour

Step-by-step Solution

Detailed explanation

Area \( A = \frac{1}{2}x^2 \sin \theta \) \( \frac{dA}{dt} = \frac{1}{2} \left( 2x \frac{dx}{dt} \sin \theta + x^2 \cos \theta \frac{d\theta}{dt} \right) \)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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