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  • Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 5 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 5 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}

\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}

Solution:

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

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

y=a(1+\cos \theta)

\frac{d y}{d \theta}=a\left\{\frac{d }{d \theta}(1)+\frac{d }{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)

\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 { The slope of the tangent is } \frac{-\sin \theta}{(1-\cos \theta)}

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

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

\begin{aligned} &\left(\frac{d y}{d x}\right)_{\theta_{-\frac{-\pi}{2}}}{ } \Rightarrow \frac{-(-1)}{(1-(-0))} \\ &\left(\frac{d y}{d x}\right)_{\theta_{-\frac{-\pi}{2}}} \Rightarrow \frac{1}{1-0} \Rightarrow 1 \end{aligned}

\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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