Question

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

Answer 1

eqn. tgt: `" "x-6y+27=0`

eqn. normal `" "6x+y-57=0`

Explanation:

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

`y-y_1=m(x-x_1)`

where `m=`gradient; `(x_1,y_1)=` the coordinate in question

just considering the `+sqrt`

`y=sqrtx, y_1=sqrt9=3`

`(x_1,y_1)=(9,3)`

the problem now is to find the gradients

tangent

`m_t=(dy)/(dx)=d/(dx)(x^(1/2))=1/2x^(-1/2)`

`m_t(9)=((dy)/(dx))_(x=9)=1/2xx1/sqrt9=1/6`

eq tgt

`y-3=1/6(x-9)`

`6y-18=x-9`

`6y-x-27=0`

`x-6y+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 `(-1)`

`m_txxm_n=-1`

`1/6xxm_n=-1`

`m_n=-6`

eqn normal

`y-3=-6(x-9)`

`y-3=-6x+54`

`6x+y-57=0`