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

2 Answers
Mar 19, 2018

2x + 6y - 3 = 0

Explanation:

x^2+y^2 = (2x^2 + 2y^2 - x)^2x2+y2=(2x2+2y2x)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/dxdydx.

Step by step differentiation:

x^2+y^2 = (2x^2 + 2y^2 - x)^2x2+y2=(2x2+2y2x)2

2x+2y (dy/dx) = 2 (2x^2 + 2y^2 -x)(4x + 4y(dy/dx) - 1)2x+2y(dydx)=2(2x2+2y2x)(4x+4y(dydx)1)

x + y(dy/dx) = (2x^2 + 2y^2 - x)(4x + 4y(dy/dx) - 1)x+y(dydx)=(2x2+2y2x)(4x+4y(dydx)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/dxdydx 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) +xx+y(dydx)=8x3+8x2y(dydx)2x2+8xy2+8y3(dydx)2y24x24xy(dydx)+x

dy/dx = (8x^3 - 6x^2 +8xy^2 - 2y^2)/(y-8x^2y+8y^3-4xy)dydx=8x36x2+8xy22y2y8x2y+8y34xy

Equation of tangent to line is given by:

y-y_1 = (dy/dx)_(x_1,y_1) (x-x_1)yy1=(dydx)x1,y1(xx1)

where (x_1,y_1) = (0,1/2)(x1,y1)=(0,12)

Putting values of (x_1,y_1)(x1,y1) in the equation we got for dy/dxdydx:

(dy/dx)_(x_1,y_1) = -1/3(dydx)x1,y1=13

Putting it into the equation of tangent:

y - 1/2 = (-1/3)(x-0)y12=(13)(x0)

Simplifying:

2x + 6y - 3 = 02x+6y3=0

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

Mar 19, 2018

Please see below.

Explanation:

d/dx(x^2+y^2) = d/dx((2x^2 + 2y^2 - x)^2)ddx(x2+y2)=ddx((2x2+2y2x)2)

2x+2y (dy/dx) = 2 (2x^2 + 2y^2 -x)(4x + 4y(dy/dx) - 1)2x+2y(dydx)=2(2x2+2y2x)(4x+4y(dydx)1)

We were not asked for a general formula for dy/dxdydx only for its value at (0,1/2)(0,12). So let's plug in the numbers and solve for dy/dxdydx

x = (0)x=(0) and y = (1/2)y=(12)

2(0)+2(1/2)(dy/dx) = 2 (2(0)^2 + 2(1/2)^2 -(0))(4(0) + 4(1/2)(dy/dx) - 1)2(0)+2(12)(dydx)=2(2(0)2+2(12)2(0))(4(0)+4(12)(dydx)1)

dy/dx = 2(1/2)(2dy/dx-1)dydx=2(12)(2dydx1)

dy/dx = 2dy/dx-1dydx=2dydx1

dy/dx = 1dydx=1

So the tangent line has slope @11 and yy-intercept 1/212.

y = x+1/2y=x+12