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Find the Slope of the Tangent and the Normal to the Following Curve at the Indicted Point Y = √ X at X = 9 ? - Mathematics

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प्रश्न

Find the slope of the tangent and the normal to the following curve at the indicted point \[y = \sqrt{x} \text { at }x = 9\] ?

योग

उत्तर

\[ y = \sqrt{x} = x^\frac{1}{2} \]

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

When `x=9,`

`y=sqrtx`

`=sqrt9`

`=3`

\[\text { Now }, \]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( 9, 3 \right) =\frac{1}{2\sqrt{9}}=\frac{1}{6}\]

\[\text { Slope of the normal }=\frac{- 1}{\left( \frac{dy}{dx} \right)_\left( 9, 3 \right)}=\frac{- 1}{\left( \frac{1}{6} \right)}=-6\]

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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 1.02 | पृष्ठ १०

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