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Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 4 Maths Textbook Solution.

Answers (1)

Answer:

$\text { The slope of the tangent is } 3$

$\text { The slope of the normal is } \frac{-1}{3}$

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: $y=2 x^{2}+3 \sin x \text { at } x=0$

Solution:

$\text { First we have to find } \frac{d y}{d x} \text { of given function, }$

$f(x) \text { that is to find the derivative of } f(x)$

$y=2 x^{2}+3 \sin x$

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

$\text { We know that the slope of the tangent is } \frac{d y}{d x}$

$\begin{aligned} &\frac{d y}{d x}=2\left(2 x^{2-1}\right)+3 \cos x \\ &\frac{d y}{d x}=4 x+3 \cos x \end{aligned}$

Since, $x=0$

$\left(\frac{d y}{d x}\right)_{x=0} \Rightarrow 4(0)-3 \cos (0)$

$\left(\frac{d y}{d x}\right)_{x=0} \Rightarrow 0+3(1) \quad\{\cos (0)=1\}$

$\left(\frac{d y}{d x}\right)_{x-0} \Rightarrow 3$

$\text { The slope of the tangent at } x=0 \text { is } 3$

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

$\text { The slope of the normal }=\frac{-1}{\left(\frac{d y}{d x}\right)_{x-0}}$

$\text { The slope of the normal }=\frac{-1}{3}$

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