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  • Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 7 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 7 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(\theta-\sin \theta) y=a(1-\cos \theta) \text { at } \theta=\frac{\pi}{2}

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}(\sin x)=\cos x \end{aligned}

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

y=a(1-\cos \theta)

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

\frac{d}{d x}(\text { constant })=0

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

\frac{d y}{d \theta}=a(0-(-\sin \theta))

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

\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \Rightarrow \frac{a \sin \theta}{a(1-\cos \theta)}

\frac{d y}{d x}=\frac{\sin \theta}{(1-\cos \theta)}

\text { Since, } \theta=\frac{\pi}{2}

\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{2}}=\frac{\sin \frac{\pi}{2}}{\left(1-\cos \frac{\pi}{2}\right)}

\text { We know that } \cos \left(\frac{\pi}{2}\right)=0 \text { and } \sin \left(\frac{\pi}{2}\right)=1

\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{2}}=\frac{1}{(1-0)}=1

\text { The slope of the tangent at } \theta=\frac{\pi}{2} \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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