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

Answer: $The \: Slope\: of \: the\: tangent \: is\: 3$

$The \: Slope\: of \: the\: normals \: is\: \frac{1}{3}$

Hint:

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

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

Hint:

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

$\inline The slope of normal is the normal to a curve at P=(x, y) is a line perpendicular to the tangent at P and passing through P.$

Given: $\inline y=\sqrt{x^{3}} \text { at } x=4$

Solution:

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

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

$\inline y=\sqrt{x^{3}}$

\inline \begin{aligned} &\therefore \sqrt[n]{x}=x^{\frac{1}{n}} \\ &y=\left(x^{3}\right)^{\frac{1}{2}}=(x)^{\frac{3}{2}} \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}

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

\inline \begin{aligned} &\frac{d y}{d x} \Rightarrow \frac{3}{2}(x)^{\frac{3}{2}-1} \\ &\frac{d y}{d x} \Rightarrow \frac{3}{2}(x)^{\frac{1}{2}} \end{aligned}

Since,$\inline x=4$

\inline \begin{aligned} &\left(\frac{d y}{d x}\right)_{x-4} \Rightarrow \frac{3}{2}(4)^{\frac{1}{2}} \\ &\left(\frac{d y}{d x}\right)_{x-4} \Rightarrow \frac{3}{2} \times \sqrt{4} \\ &\left(\frac{d y}{d x}\right)_{x-4} \Rightarrow \frac{3}{2} \times 2 \\ &\left(\frac{d y}{d x}\right)_{x-4} \Rightarrow 3 \end{aligned}

$\inline \text { The slope of the tangent at } x=4 \text { is } 3$

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

$\inline \text { The slope of the normal }=\frac{-1}{\left(\frac{d y}{d x}\right)_{x-4}}$

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