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

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

Answers (2)

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. }

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

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=\sqrt{x^{3}}

\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}

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

\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,x=4

\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}

\text { The slope of the tangent at } x=4 \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-4}}

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

 

 

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