What is the equation of the normal line of f(x)=x^3/sqrt(16x^2-x) at x=2?

Feb 20, 2018

Equation of normal at $x = 2$ is $y - 1.016 = - 0.992253 \left(x - 2\right)$

Explanation:

For a function $y = f \left(x\right)$, the slope of tangent at $x = {x}_{0}$ i.e. at $\left({x}_{0.} f \left({x}_{0}\right)\right)$ is given by $f ' \left({x}_{0}\right)$, where $f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$

and hence slope of normal is $- \frac{1}{f ' \left({x}_{0}\right)}$ as normal is perpendicular to tangent.

and while equation of tangent is $y - f \left({x}_{0}\right) = f ' \left({x}_{0}\right) \left(x - {x}_{0}\right)$

and equation of normal is $y - f \left({x}_{0}\right) = - \frac{1}{f ' \left({x}_{0}\right)} \left(x - {x}_{0}\right)$

Here we have $f \left(x\right) = {x}^{3} / \sqrt{16 {x}^{2} - x}$ and at ${x}_{0} = 2$. As $f \left({x}_{0}\right) = \frac{8}{\sqrt{62}}$, hence we are seeking normal at point $\left(2 , \frac{8}{\sqrt{62}}\right)$ or $\left(2 , 1.016\right)$

Further using quotient rule $f ' \left(x\right) = \frac{3 {x}^{2} \sqrt{16 {x}^{2} - x} - {x}^{3} \left(\frac{32 x - 1}{2 \sqrt{16 {x}^{2} - x}}\right)}{16 {x}^{2} - x}$

and $f ' \left(2\right) = \frac{12 \sqrt{62} - 8 \left(\frac{63}{2 \sqrt{62}}\right)}{62}$

and slope of normal is $- \frac{62}{12 \sqrt{62} - 8 \left(\frac{63}{2 \sqrt{62}}\right)}$

= $- \frac{\sqrt{62}}{12 - \frac{504}{124}} = - \frac{\sqrt{62}}{12 - \frac{126}{31}}$

= $- \frac{31 \sqrt{62}}{372 - 126} = - \frac{31 \sqrt{62}}{246} = - 0.992253$

and equation of normal is $y - 1.016 = - 0.992253 \left(x - 2\right)$

graph{(y-1.016+0.992253(x-2))(x^3/sqrt(16x^2-x)-y)=0 [-10, 10, -5, 5]}