Question

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

Answer 1

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

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

When `x=9,`

`y=sqrtx`

`=sqrt9`

`=3`

\[\text { Now }, \]

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

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