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Question
Find the slope of the tangent and the normal to the following curve at the indicted point \[y = \sqrt{x^3} \text { at } x = 4\] ?
Solution
\[y = \sqrt{x^3} = x^\frac{3}{2} \]
\[ \Rightarrow \frac{dy}{dx} = \frac{3}{2} x^\frac{1}{2} = \frac{3}{2}\sqrt{x}\]
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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