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Find the Slope of the Tangent and the Normal to the Following Curve at the Indicted Point Y = √ X 3 at X = 4 ? - Mathematics

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प्रश्न

Find the slope of the tangent and the normal to the following curve at the indicted point \[y = \sqrt{x^3} \text { at } x = 4\] ?

योग

उत्तर

\[y = \sqrt{x^3} = x^\frac{3}{2} \]

\[ \Rightarrow \frac{dy}{dx} = \frac{3}{2} x^\frac{1}{2} = \frac{3}{2}\sqrt{x}\]

When `x=4,`

`y=sqrt(x^3)`

`=sqrt(4^3)`

`=sqrt64`

`=8`

\[\text { Now,} \]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( 4, 8 \right) =\frac{3}{2}\sqrt{4} = 3\]

\[\text { Slope of the normal }=\frac{- 1}{\left( \frac{dy}{dx} \right)_\left( 4, 8 \right)}=\frac{- 1}{3}\]

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अध्याय 16: Tangents and Normals - Exercise 16.1 [पृष्ठ १०]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 16 Tangents and Normals
Exercise 16.1 | Q 1.01 | पृष्ठ १०

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