Question

How do you find the equation of the tangent line to the curve at `(81, 9)` of `y=sqrtx`?

Answer 1

See below

Explanation:

Given equation
`y^2 = x`

This is an equation of a parabola. graph{y^2 = x [-10, 10, -5, 5]}
For finding the tangent to the equation, we first differentiate it.

The goal here is to find the value of `dy/dx` which will give the slope of the tangent to the parabola.

So,
`d/dy(y^2) = dx/dy`

`2y = dx/dy`

`1/(2y) = dy/dx`

Now, we want to evaluate the slope at the given point. So substituting y we get,

`1/(18) = dy/dx`

This is the slope of the tangent.

It passes through (81,9) [as stated in the question]

The general equation of a line is
`y = mx +c`

m is the slope of the line.

For the line/ tangent that we got `m = dy/dx`.

`y = x/(18) + c`

The point should satisfy the equation of the line/ tangent. So,

`9 = 81/(18) + c`

So, `c = 9/2`

Finally substituting the value in our equation of the line/tangent,

`y = x/(18) + 9/2`

`18y = x + 81`

Answer 2

`y=1/18x+9/2`

Explanation:

`•color(white)(x)m_(color(red)"tangent")=dy/dx" at x "= 81`

`y=sqrtx=x^(1/2)`

`rArrdy/dx=1/2x^(-1/2)=1/(2sqrtx)`

`x=81tody/dx=1/(2sqrt81)=1/18`

`"equation of tangent in point-slope form is"`

`y-9=1/18(x-81)`

`rArry=1/18x+9/2`