Question

Find the equation of the tangent line to the curve x^2 + y^2 = (2x^2 + 2y^2 -x)^2 at the point 0, 1/2?

I dont understand how to use implicit differentiation

Answer 1

2x + 6y - 3 = 0

Explanation:

`x^2+y^2 = (2x^2 + 2y^2 - x)^2`

Differentiating term by term w.r.t. x

That means simple x terms differentiate normally but while differentiating those with y; since you are differentiating with x; you'll have to multiply those with `dy/dx`.

Step by step differentiation:

`x^2+y^2 = (2x^2 + 2y^2 - x)^2`

`2x+2y (dy/dx) = 2 (2x^2 + 2y^2 -x)(4x + 4y(dy/dx) - 1)`

`x + y(dy/dx) = (2x^2 + 2y^2 - x)(4x + 4y(dy/dx) - 1)`

I hope you understood what I did on the RHS of the second to last step; I treated the whole term first as a huge x term that I had to differentiate and then following the chain rule I differentiated the inside and multiplied it.

Next, expanding and separating out the `dy/dx` terms to one side:
`x+y(dy/dx) = 8x^3 + 8x^2y(dy/dx) - 2x^2 +8xy^2 + 8y^3(dy/dx) - 2y^2 - 4x^2 - 4xy(dy/dx) +x`

`dy/dx = (8x^3 - 6x^2 +8xy^2 - 2y^2)/(y-8x^2y+8y^3-4xy)`

Equation of tangent to line is given by:

`y-y_1 = (dy/dx)_(x_1,y_1) (x-x_1)`

where `(x_1,y_1) = (0,1/2)`

Putting values of `(x_1,y_1)` in the equation we got for `dy/dx`:

`(dy/dx)_(x_1,y_1) = -1/3`

Putting it into the equation of tangent:

`y - 1/2 = (-1/3)(x-0)`

Simplifying:

`2x + 6y - 3 = 0`

And there you have it! It was a pretty simple question. Hope it helped!

Answer 2

Please see below.

Explanation:

`d/dx(x^2+y^2) = d/dx((2x^2 + 2y^2 - x)^2)`

`2x+2y (dy/dx) = 2 (2x^2 + 2y^2 -x)(4x + 4y(dy/dx) - 1)`

We were not asked for a general formula for `dy/dx` only for its value at `(0,1/2)`. So let's plug in the numbers and solve for `dy/dx`

`x = (0)` and `y = (1/2)`

`2(0)+2(1/2)(dy/dx) = 2 (2(0)^2 + 2(1/2)^2 -(0))(4(0) + 4(1/2)(dy/dx) - 1)`

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

`dy/dx = 2dy/dx-1`

`dy/dx = 1`

So the tangent line has slope `@1` and `y`-intercept `1/2`.

`y = x+1/2`