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  • Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 6 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 6 Maths Textbook Solution.

Answers (1)

Answer:

\text { The slope of the tangent is }-1

\text { The slope of the normal is } 1

Hint

\text { The slope of tangent is the limit of } \frac{\Delta y}{\Delta x} \text { as } \Delta x \text { approaches zero. }

\text { The slope of normal is the normal to a curve at } P=(x, y) \text { is a line perpendicular to the tangent }

\text { at } P \text { and passing through } P \text { . }

Given:x=a \cos ^{3} \theta, y=a \sin ^{3} \theta \text { at } \theta=\frac{\pi}{4}

Solution:

\text { Here to find } \frac{d y}{d x}, \text { we have to find } \frac{d y}{d \theta} \& \frac{d x}{d \theta} \text { and }

\text { divide } \frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \text { and we get desired } \frac{d y}{d x}

\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\ &\frac{d}{d x}(\cos x)=-\sin x \\ &x=a \cos ^{3} \theta \\ &\frac{d x}{d \theta}=a\left\{\frac{d }{d \theta}\left(\cos ^{3} \theta\right)\right\} \end{aligned}

\frac{d x}{d \theta}=a\left(3 \cos ^{3-1} \theta \times-\sin \theta\right)

\frac{d x}{d \theta}=-3 a \cos ^{2} \theta \sin \theta \quad \rightarrow(1)

y=a \sin ^{3} \theta

\frac{d y}{d \theta}=a\left\{\frac{d}{d \theta}\left(\sin ^{3} \theta\right)\right\}

\frac{d}{d x}(\sin x)=\cos x

\frac{d y}{d \theta}=a\left(3 \sin ^{3-1} \theta \times \cos \theta\right)

\frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta \quad \rightarrow(2)

\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \Rightarrow \frac{3 a \sin ^{2} \theta \cos \theta}{-3 a \cos ^{2} \theta \sin \theta}

\frac{d y}{d x}=\frac{\sin \theta}{-\cos \theta}

\frac{d y}{d x}=-\tan \theta

\text { The slope of the tangent is }-\tan \theta

\text { Since, } \theta=\frac{\pi}{4}

\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{4}} \Rightarrow-\tan \frac{\pi}{4}=-1

\text { The slope of the tangent at } \theta=\frac{\pi}{4} \text { is }-1

\text { The slope of the normal }=\frac{-1}{\text { The slope of the tangent }}

\text { The slope of the normal }=\frac{-1}{\left(\frac{d y}{d x}\right)_{\theta=\frac{-\pi}{2}}}

\text { The slope of the normal }=\frac{-1}{-1}=1

 

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