AP EAMCET · Maths · Application of Derivatives
The area (in sq units) of the triangle formed by the normal drawn at the point \((1,0)\) on the curve \(\mathrm{x}=\mathrm{e}^{\sin \mathrm{y}}\) with the coordinate axes is
- A \(1\)
- B \(\frac{1}{4}\)
- C \(\frac{1}{2}\)
- D \(\frac{3}{8}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(\because x=e^{\sin y} \Rightarrow \log x=\sin y\) \(\Rightarrow \frac{1}{x}=\cos y \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{x \cos y}\) Slope of normal at \((1,0)=-\frac{1}{\left(\frac{d y}{d x}\right)_{(1,0)}}\) \(=-\left.x \cos y\right|_{(1,0)}\)…
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