Chapter 3.5, Problem 24E

To determine

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

Answer to Problem 24E

The equation of the tangent line to the equation $x^2+2xy+4y^2=12$ at $\left(2,1\right)$ is $y=\frac{-x}{2}+2$.

Explanation of Solution

Given:

The curve is $x^2+2xy+4y^2=12$.

The point is $\left(2,1\right)$.

Derivative rules:

(1) Chain rule: If $y=f\left(u\right)$ and $u=g\left(x\right)$  are both differentiable function, then

$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$.

(2) Product rule: If $y=f\left(u\right)$ and $u=g\left(x\right)$  are both differentiable function, then

$\frac{d}{dx}\left(f\left(x\right)+g\left(x\right)\right)=\frac{d}{dx}\left(f\left(x\right)\right)+\frac{d}{dx}\left(g\left(x\right)\right)$.

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 $x^2+2xy+4y^2=12$.

Differentiate the above equation implicitly with respect to x,

$\begin{array}{l}\frac{d}{dx}\left(x^2+2xy+4y^2\right)=\frac{d}{dx}\left(12\right)\\ \frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2xy\right)+\frac{d}{dx}\left(4y^2\right)=0\\ \frac{d}{dx}\left(x^2\right)+2\frac{d}{dx}\left(xy\right)+4\frac{d}{dx}\left(y^2\right)=0\end{array}$

Apply the product rule (2),

$\begin{array}{l}2x+2\left[x\frac{d}{dx}\left(y\right)+y\frac{d}{dx}\left(x\right)\right]+4\frac{d}{dx}\left(y^2\right)=0\\ 2x+2\left[x\frac{dy}{dx}+y\left(1\right)\right]+4\frac{d}{dx}\left(y^2\right)=0\end{array}$

Apply the chain rule (1) and simplify the terms,

$\begin{array}{l}2x+2x\frac{dy}{dx}+2y+4\left[\frac{d}{dy}\left(y^2\right)\frac{dy}{dx}\right]=0\\ 2x+2x\frac{dy}{dx}+2y+4\left[2y\frac{dy}{dx}\right]=0\\ 2x+2x\frac{dy}{dx}+2y+8y\frac{dy}{dx}=0\end{array}$

Combine the terms $\frac{dy}{dx}$,

$\begin{array}{l}8y\frac{dy}{dx}+2x\frac{dy}{dx}=-2x-2y\\ \frac{dy}{dx}\left(8y+2x\right)=-2x-2y\\ \frac{dy}{dx}=-\frac{2x+2y}{8y+2x}\\ \frac{dy}{dx}=-\frac{x+y}{4y+x}\end{array}$

Therefore, the derivative of the equation is $\frac{dy}{dx}=-\frac{x+y}{4y+x}$.

The slope of the tangent line at $\left(2,1\right)$ is computed as follows,

$\begin{array}{c}m={\frac{dy}{dx}|}_{\left(x_1,y_1\right)=\left(2,1\right)}\\ =-\frac{2+1}{4\left(1\right)+2}\\ =\frac{-3}{6}\\ =\frac{-1}{2}\end{array}$

Thus, the slope of the tangent line at $\left(2,1\right)$ is $m=\frac{-1}{2}$.

Substitute $\left(2,1\right)$ for $\left(x_1,y_1\right)$ and $m=\frac{-1}{2}$ in equation (1),

$\begin{array}{c}y-1=\frac{-1}{2}\left(x-2\right)\\ y-1=\frac{-x}{2}+1\\ y=\frac{-x}{2}+1+1\\ y=\frac{-x}{2}+2\end{array}$

Therefore, the equation of the tangent line to the equation $x^2+2xy+4y^2=12$ at $\left(2,1\right)$ is $y=\frac{-x}{2}+2$.