Question

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

Answer 1

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

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

When `x=4,`

`y=sqrt(x^3)`

`=sqrt(4^3)`

`=sqrt64`

`=8`

\[\text { Now,} \]

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

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