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Question
Find the slope of the tangent and the normal to the following curve at the indicted point \[y = \sqrt{x} \text { at }x = 9\] ?
Solution
\[ y = \sqrt{x} = x^\frac{1}{2} \]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2} x^\frac{- 1}{2} = \frac{1}{2\sqrt{x}}\]
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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