find an equation of the tangent line to the astroid at the (-3√3, 1). x^2/3 + y^2/3 = 4 y = need help? read…

Explanation:

Step1: Differentiate implicitly

Differentiate $x^{\frac{2}{3}}+y^{\frac{2}{3}} = 4$ with respect to $x$. Using the power - rule $(u^n)^\prime=nu^{n - 1}u^\prime$, we have $\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}y^\prime=0$.

Step2: Solve for $y^\prime$

First, subtract $\frac{2}{3}x^{-\frac{1}{3}}$ from both sides: $\frac{2}{3}y^{-\frac{1}{3}}y^\prime=-\frac{2}{3}x^{-\frac{1}{3}}$. Then, divide both sides by $\frac{2}{3}y^{-\frac{1}{3}}$ to get $y^\prime=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}$.

Step3: Find the slope at the given point

Substitute $x = - 3\sqrt{3}$ and $y = 1$ into $y^\prime$. $y^\prime\big|_{x=-3\sqrt{3},y = 1}=-\left(\frac{1}{-3\sqrt{3}}\right)^{\frac{1}{3}}=\left(\frac{1}{3\sqrt{3}}\right)^{\frac{1}{3}}=\frac{1}{\sqrt{3}}$.

Step4: Use the point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-3\sqrt{3},1)$ and $m=\frac{1}{\sqrt{3}}$. $y - 1=\frac{1}{\sqrt{3}}(x + 3\sqrt{3})$. Expand the right - hand side: $y - 1=\frac{1}{\sqrt{3}}x+3$. Add 1 to both sides to get the equation of the tangent line: $y=\frac{1}{\sqrt{3}}x + 4$.

Answer:

$y=\frac{1}{\sqrt{3}}x + 4$