Question

1. (15 points) find the equation of the tangent line to the curve $x^2 + y^2 = (2x^2 + 2y^2 - x)^2$ at point $(0, \frac{1}{2})$.

Explanation:

Step1: Implicit Differentiation

Differentiate both sides of \(x^{2}+y^{2}=(2x^{2}+2y^{2}-x)^{2}\) with respect to \(x\).
Left side: \(\frac{d}{dx}(x^{2}+y^{2}) = 2x + 2y\frac{dy}{dx}\)
Right side: Use chain rule. Let \(u = 2x^{2}+2y^{2}-x\), then \(\frac{d}{dx}(u^{2}) = 2u\frac{du}{dx}\).
Compute \(\frac{du}{dx}=\frac{d}{dx}(2x^{2}+2y^{2}-x)=4x + 4y\frac{dy}{dx}-1\)
So right side: \(2(2x^{2}+2y^{2}-x)(4x + 4y\frac{dy}{dx}-1)\)
Now we have:
\(2x + 2y\frac{dy}{dx}=2(2x^{2}+2y^{2}-x)(4x + 4y\frac{dy}{dx}-1)\)

Step2: Substitute Point \((0,\frac{1}{2})\)

Substitute \(x = 0\), \(y=\frac{1}{2}\) into the equation.
Left side: \(2(0)+2(\frac{1}{2})\frac{dy}{dx}=\frac{dy}{dx}\)
Right side: First, \(2x^{2}+2y^{2}-x=2(0)+2(\frac{1}{4})-0=\frac{1}{2}\)
\(4x + 4y\frac{dy}{dx}-1=4(0)+4(\frac{1}{2})\frac{dy}{dx}-1 = 2\frac{dy}{dx}-1\)
So right side: \(2(\frac{1}{2})(2\frac{dy}{dx}-1)=2\frac{dy}{dx}-1\)
Now the equation becomes:
\(\frac{dy}{dx}=2\frac{dy}{dx}-1\)

Step3: Solve for \(\frac{dy}{dx}\) (Slope)

\(\frac{dy}{dx}-2\frac{dy}{dx}=-1\)
\(-\frac{dy}{dx}=-1\)
\(\frac{dy}{dx}=1\)

Step4: Equation of Tangent Line

Use point - slope form \(y - y_{1}=m(x - x_{1})\), where \(m = 1\), \(x_{1}=0\), \(y_{1}=\frac{1}{2}\)
\(y-\frac{1}{2}=1(x - 0)\)
Simplify: \(y=x+\frac{1}{2}\) or \(x - y+\frac{1}{2}=0\) or \(2x - 2y + 1 = 0\)

Answer:

The equation of the tangent line is \(y=x+\frac{1}{2}\) (or \(2x - 2y+1 = 0\))