Find the slopes of the tangent and the normal to the curve `y=sqrt(x^3)` at `x=4`

Answer

Slop of the tangent `m = 3`

Slop of the normal `=frac{-1}{m}=frac{-1}{3}`

Explanation

`y=\sqrt{x^{3}} \quad` [Given `]`

`\Rightarrow y=x^{\frac{3}{2}}`

`\Rightarrow \frac{d y}{d x}=\frac{3}{2} x^{\frac{1}{2}}`

[Differentiating both sides ]

`\therefore \frac{d y}{d x}=\frac{3}{2} \sqrt{x}`

When `x=4`,

`\Rightarrow y=\sqrt{x^{3}}`

`=\sqrt{(4)^{3}}`

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