MHT CET · Maths · Differentiation
If \(\mathrm{x}=\mathrm{e}^{\mathrm{t}}(\sin \mathrm{t}-\cos \mathrm{t})\) and \(\mathrm{y}=\mathrm{e}^{\mathrm{t}}(\sin \mathrm{t}+\cos \mathrm{t})\), then \(\frac{\mathrm{dy}}{\mathrm{dx}}\) at \(\mathrm{t}=\frac{\pi}{3}\) is
- A \(\sqrt{3}\)
- B \(\frac{1}{\sqrt{3}}\)
- C \(\frac{\sqrt{3}}{2}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & x=e^t(\sin t-\cos t) \text { and } y=e^t(\sin t+\cos t) \\ & \therefore \quad \frac{d x}{d t}=e^t(\sin t-\cos t)+e^t(\cos t+\sin t)=2 e^t \sin t \\ & \frac{d y}{d t}=e^t(\sin t+\cos t)+e^t(\cos t-\sin t)=2 e^t \sin t \\ & \therefore \quad \frac{d y}{d x}=\frac{2 e^t \cos t}{2 e^t \sin t} \Rightarrow\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{3}}=\cot \left(\frac{\pi}{3}\right)=\frac{1}{\sqrt{3}} \\ & \end{aligned}\)
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