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:

x2+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 dydx.

Step by step differentiation:

x2+y2=(2x2+2y2x)2

2x+2y(dydx)=2(2x2+2y2x)(4x+4y(dydx)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 dydx terms to one side:
x+y(dydx)=8x3+8x2y(dydx)2x2+8xy2+8y3(dydx)2y24x24xy(dydx)+x

dydx=8x36x2+8xy22y2y8x2y+8y34xy

Equation of tangent to line is given by:

yy1=(dydx)x1,y1(xx1)

where (x1,y1)=(0,12)

Putting values of (x1,y1) in the equation we got for dydx:

(dydx)x1,y1=13

Putting it into the equation of tangent:

y12=(13)(x0)

Simplifying:

2x+6y3=0

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

Mar 19, 2018

Please see below.

Explanation:

ddx(x2+y2)=ddx((2x2+2y2x)2)

2x+2y(dydx)=2(2x2+2y2x)(4x+4y(dydx)1)

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

x=(0) and y=(12)

2(0)+2(12)(dydx)=2(2(0)2+2(12)2(0))(4(0)+4(12)(dydx)1)

dydx=2(12)(2dydx1)

dydx=2dydx1

dydx=1

So the tangent line has slope 1 and y-intercept 12.

y=x+12