Chapter 3.5, Problem 26E

To determine

To find: The equation of the tangent line to the given equation at the point.

Answer to Problem 26E

The equation of the tangent line to the equation $x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$ at the point $\left(-3\sqrt{3},1\right)$ is $y=\frac{1}{\sqrt{3}}x+4$.

Explanation of Solution

Given:

The curve is $x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$.

The point is $\left(-3\sqrt{3},1\right)$.

Derivative rules: Chain rule

If $y=f\left(u\right)$ and $u=g\left(x\right)$  are both differentiable functions, then $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$.

Formula used:

The equation of the tangent line at $\left(x_1,y_1\right)$ is, $y-y_1=m\left(x-x_1\right)$ (1)

Where, m is the slope of the tangent line at $\left(x_1,y_1\right)$ and ${m=\frac{dy}{dx}|}_{x=x_1,y=y_1}$.

Calculation:

Consider the equation $x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$.

Differentiate the given equation implicitly with respect to x,

$\begin{array}{c}\frac{d}{dx}\left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)=\frac{d}{dx}\left(4\right)\\ \frac{d}{dx}\left(x^{\frac{2}{3}}\right)+\frac{d}{dx}\left(y^{\frac{2}{3}}\right)=0\\ \frac{2}{3}{x}^{\frac{2}{3}-1}+\frac{d}{dx}\left(y^{\frac{2}{3}}\right)=0\\ \frac{2}{3}{x}^{\frac{-1}{3}}+\frac{d}{dx}\left(y^{\frac{2}{3}}\right)=0\end{array}$

Apply the chain rule and simplify the terms,

$\begin{array}{l}\frac{2}{3}{x}^{\frac{-1}{3}}+\left[\frac{d}{dy}\left(y^{\frac{2}{3}}\right)\frac{dy}{dx}\right]=0\\ \frac{2}{3}{x}^{\frac{-1}{3}}+\left[\frac{2}{3}{y}^{\frac{-1}{3}}\frac{dy}{dx}\right]=0\\ \frac{2}{3}{x}^{\frac{-1}{3}}+\frac{2}{3}{y}^{\frac{-1}{3}}\frac{dy}{dx}=0\\ \frac{2}{3}{y}^{\frac{-1}{3}}\frac{dy}{dx}=-\frac{2}{3}{x}^{\frac{-1}{3}}\end{array}$

Separate $\frac{dy}{dx}$ to one side of the equation,

$\begin{array}{l}\frac{2}{3}{y}^{\frac{-1}{3}}\frac{dy}{dx}=-\frac{2}{3}{x}^{\frac{-1}{3}}\\ \frac{dy}{dx}=\frac{-\frac{2}{3}{x}^{\frac{-1}{3}}}{\frac{2}{3}{y}^{\frac{-1}{3}}}\\ \frac{dy}{dx}=\frac{-x^{\frac{-1}{3}}}{y^{\frac{-1}{3}}}\\ \frac{dy}{dx}=-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}\end{array}$

Therefore, the derivative of the equation is $\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}$.

The slope of the tangent line at the point $\left(-3\sqrt{3},1\right)$ is computed as follows,

$\begin{array}{c}{\frac{dy}{dx}|}_{\left(x,y\right)=\left(-3\sqrt{3},1\right)}=-\frac{\sqrt[3]{1}}{{\left(-3\sqrt{3}\right)}^{\frac{1}{3}}}\\ =-\frac{1}{{\left(\left(-1\right)\sqrt{3}\sqrt{3}\sqrt{3}\right)}^{\frac{1}{3}}}\\ =-\frac{1}{-{\left(\sqrt{3}\sqrt{3}\sqrt{3}\right)}^{\frac{1}{3}}}\\ =\frac{1}{\sqrt{3}}\end{array}$

Thus, the slope of the tangent line at $\left(-3\sqrt{3},1\right)$ is $m=\frac{1}{\sqrt{3}}$.

Substitute $\left(-3\sqrt{3},1\right)$ for $\left(x_1,y_1\right)$ and $m=\frac{1}{\sqrt{3}}$ in equation (1),

$\begin{array}{c}y-1=\frac{1}{\sqrt{3}}\left(x+3\sqrt{3}\right)\\ y-1=\frac{1}{\sqrt{3}}x+3\sqrt{3}\left(\frac{1}{\sqrt{3}}\right)\\ y-1=\frac{1}{\sqrt{3}}x+3\\ y=\frac{1}{\sqrt{3}}x+4\end{array}$

Therefore, the equation of the tangent line to the equation $x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$ at the point $\left(-3\sqrt{3},1\right)$ is $y=\frac{1}{\sqrt{3}}x+4$.

Graph:

The graph of the curve and the tangent line is shown below in Figure 1.

From Figure 1, it is observed that the line $y=\frac{1}{\sqrt{3}}x+4$ is tangent to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$ at the point $\left(-3\sqrt{3},1\right)$.