Question

How do you find an equation of the line tangent to the curve at the given point `x^3+y^3=36xy` given point (18,18)?

Answer 1

Use implicit differentiation

`3x^2+3y^2(dy)/dx=36(x(1)(dy)/dx+y(1))`

Simplify

`3x^2+3y^2(dy)/dx=36(x(dy)/dx+y)`

Distribute

`3x^2+3y^2(dy)/dx=36x(dy)/dx+36y`

Isolate `dy/dx`

`3y^2(dy)/dx-36x(dy)/dx=-3x^2+36y`

Factor out `dy/dx`

`(dy)/dx(3y^2-36x)=-3x^2+36y`

Divide to isolate `dy/dx`

`((dy)/dxcancel(3y^2-36x))/cancel((3y^2-36x))=(-3x^2+36y)/(3y^2-36x)`

Simplify

`(dy)/dx=(-3x^2+36y)/(3y^2-36x)`

Factor out a 3

`(dy)/dx=(3(-x^2+12y))/(3(y^2-12x))`

Simplify

`(dy)/dx=(cancel3(-x^2+12y))/(cancel3(y^2-12x))`

Simplify

`(dy)/dx=(12y-x^2)/(y^2-12x)`

Substitute in the point (18,18) to find the slope

`(dy)/dx=(12(18)-18^2)/(18^2-12(18))`

`(dy)/dx=-1`

Now we have the slope(m), -1, and a point (18,18)

Use the point slope formula, `y=m(x-x_1)+y_1`

`y=-1(x-18)+18`

Simplify

`y=-x+18+18`

`y=-x+36`

Click here to see a graph

https://www.desmos.com/calculator/ulvfbcnxtz

Click here to watch a similar example solved

and