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Name: Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 7 Maths Textbook Solution.
Uploaded: Thu, 2021-08-12 13:33
Description: 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:

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

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

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

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