Question

How do you find the equation of the tangent and normal line to the curve `y=x^(1/2)` at x=1?

Answer 1

Read below.

Explanation:

We find the tangent line to a curve by finding the derivative first.

Power rule:

`d/dx(x^n)=nx^(n-1)` given that `n` is a constant.

`f(x)=x^1/2`

`=>f'(x)=1/2*x^(1-1/2)`

`=>f'(x)=1/2*x^(-1/2)`

`=>f'(x)=1/2*1/sqrtx`

`=>f'(x)=1/(2sqrtx)`

`=>f'(1)=1/(2sqrt1)`

`=>f'(1)=1/2`

Let's now find the y value when `x=1`

`=>f(1)=sqrt1`

`=>f(1)=1`

We can now use these information to find the equation of the tangent line using the formula `m(x-x_1)=y-y_1`

`=>1/2(x-1)=y-1`

`=>1/2x-1/2=y-1`

`=>1/2x+1/2=y`

That is the equation of the tangent line at the point `(1,1)`

A normal line is perpendicular to the tangent line.

We can find the slope of the normal line by finding the negative reciprocal of the slope of the tangent line.

`1/2=>-2`

We use the formula `m(x-x_1)=y-y_1` once again.

`=>-2(x-1)=y-1`

`=>-2x+2=y-1`

`=>-2x+3=y`

That is the equation of the normal line at the point `(1,1)`