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  • Provide solution for RD Sharma maths class 12 Chapter 15 Tangents and Normals Exercise Very Short Answer Question 9 maths text book solution.

Provide solution for RD Sharma maths class 12 Chapter 15 Tangents and Normals Exercise Very Short Answer Question 9 maths text book solution.

Answers (1)

Answer : \frac{\pi }{2}

Hint : Differentiating both equation and find \frac{dy}{dx}

Given :

Given that the curve,

              x=e^{t} \cos t \text { and } y=e^{t} \sin t

We have to find the angle made by the tangent to the given curve.

Solution :

Here,

       x=e^{t} \cos t \text { and } y=e^{t} \sin t

Differentiating both equation with respect to 't'

We have,

\frac{d x}{d t}=e^{t} \cos t-e^{t} \sin t

\begin{aligned} &\text { - }\\ &\begin{aligned} &\frac{d y}{d t}=e^{t} \sin t+e^{t} \cos t \\ &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}\\ &=\frac{e^{t} \sin t+e^{t} \cos t}{e^{t} \cos t-e^{t} \sin t} \end{aligned}

Now, slope of tangent =\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{4}}

\begin{aligned} &=\frac{e^{t}\left(\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right)}{e^{t}\left(\cos \frac{\pi}{4}-\sin \frac{\pi}{4}\right)} \\ &=\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}}=\frac{\frac{2}{\sqrt{2}}}{0} \\ &\frac{d y}{d x}=\infty \end{aligned}

Let \theta be the angle made by the tangent with the x-axis.

\begin{array}{ll} \therefore & \tan \theta=\infty \\ \Rightarrow & \theta=\frac{\pi}{2} \end{array}

Hence, the angle made by the tangent to the given curve is \frac{\pi }{2}.

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