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

2 Answers
Jan 17, 2018

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

Jan 17, 2018

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