Question

use implicit differentiation to find an equation of the tangent line to the curve at the given point. $x^{2}+y^{2}=(3x^{2}+4y^{2}-x)^{2}$, $(0,\frac{1}{4})$ (cardioid)

Explanation:

Step1: Differentiate both sides

Differentiate $x^{2}+y^{2}=(3x^{2}+4y^{2}-x)^{2}$ with respect to $x$.
Using the chain - rule and sum - rule, the left - hand side is $2x + 2y\frac{dy}{dx}$. For the right - hand side, let $u = 3x^{2}+4y^{2}-x$, so $(3x^{2}+4y^{2}-x)^{2}=u^{2}$. Then $\frac{d}{dx}(u^{2}) = 2u\frac{du}{dx}$, and $\frac{du}{dx}=6x + 8y\frac{dy}{dx}-1$. So the right - hand side is $2(3x^{2}+4y^{2}-x)(6x + 8y\frac{dy}{dx}-1)$.
So we have $2x + 2y\frac{dy}{dx}=2(3x^{2}+4y^{2}-x)(6x + 8y\frac{dy}{dx}-1)$.

Step2: Substitute the point $(0,\frac{1}{4})$

Substitute $x = 0$ and $y=\frac{1}{4}$ into the above equation.
First, when $x = 0$ and $y=\frac{1}{4}$, $3x^{2}+4y^{2}-x=4\times(\frac{1}{4})^{2}=\frac{1}{4}$.
The left - hand side of the differentiated equation becomes $2\times0+2\times\frac{1}{4}\frac{dy}{dx}=\frac{1}{2}\frac{dy}{dx}$.
The right - hand side becomes $2\times\frac{1}{4}(0 + 8\times\frac{1}{4}\frac{dy}{dx}-1)=\frac{1}{2}(2\frac{dy}{dx}-1)=\frac{dy}{dx}-\frac{1}{2}$.
So $\frac{1}{2}\frac{dy}{dx}=\frac{dy}{dx}-\frac{1}{2}$.

Step3: Solve for $\frac{dy}{dx}$

Subtract $\frac{1}{2}\frac{dy}{dx}$ from both sides: $0=\frac{dy}{dx}-\frac{1}{2}\frac{dy}{dx}-\frac{1}{2}$, which simplifies to $\frac{1}{2}\frac{dy}{dx}=\frac{1}{2}$, so $\frac{dy}{dx}=1$.

Step4: Find the equation of the tangent line

Use the point - slope form of a line $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(0,\frac{1}{4})$ and $m = 1$.
The equation is $y-\frac{1}{4}=1\times(x - 0)$, which simplifies to $y=x+\frac{1}{4}$.

Answer:

$y=x+\frac{1}{4}$