How do you find the equation of the tangent and normal line to the curve y=sqrtx at x=9?

Jan 6, 2017

eqn. tgt: $\text{ } x - 6 y + 27 = 0$

eqn. normal $\text{ } 6 x + y - 57 = 0$

Explanation:

To find both the equation of the tangent and normal we will use

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

where $m =$gradient; $\left({x}_{1} , {y}_{1}\right) =$ the coordinate in question

just considering the  +sqrt

$y = \sqrt{x} , {y}_{1} = \sqrt{9} = 3$

$\left({x}_{1} , {y}_{1}\right) = \left(9 , 3\right)$

the problem now is to find the gradients

tangent

${m}_{t} = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({x}^{\frac{1}{2}}\right) = \frac{1}{2} {x}^{- \frac{1}{2}}$

${m}_{t} \left(9\right) = {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{x = 9} = \frac{1}{2} \times \frac{1}{\sqrt{9}} = \frac{1}{6}$

eq tgt

$y - 3 = \frac{1}{6} \left(x - 9\right)$

$6 y - 18 = x - 9$

$6 y - x - 27 = 0$

$x - 6 y + 27 = 0$

normal

the normal is perpendicular to the tangent, by definition. So we can use the fact that the product of the gradients of perpendicular lines is $\left(- 1\right)$

${m}_{t} \times {m}_{n} = - 1$

$\frac{1}{6} \times {m}_{n} = - 1$

${m}_{n} = - 6$

eqn normal

$y - 3 = - 6 \left(x - 9\right)$

$y - 3 = - 6 x + 54$

$6 x + y - 57 = 0$