Chapter 3.3, Problem 41E

To determine

The tangents to Newton’s serpentine at the origin and the point $\left(1,2\right)$ .

Answer to Problem 41E

The required tangents are $y=4x$ and $y=2$ .

Explanation of Solution

Given information:

The Newton’s serpentine: $y=\frac{4x}{x^2+1}$

Formula used:

Differentiation formula:

  $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-v^{\prime }u}{v^2}$

Point-slope form of a line:

  $y-y_1=m\left(x-x_1\right)$

Calculation:

The given Newton’s serpentine is $y=\frac{4x}{x^2+1}$ and the given point is $\left(1,2\right)$ .

Differentiate the given equation using the formula $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-v^{\prime }u}{v^2}$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(\frac{4x}{x^2+1}\right)\\ =\frac{\left(x^2+1\right)\frac{d}{dx}\left(4x\right)-4x\frac{d}{dx}\left(x^2+1\right)}{{\left(x^2+1\right)}^2}\\ =\frac{\left(x^2+1\right)\left(4\right)-4x\left(2x\right)}{{\left(x^2+1\right)}^2}\\ =\frac{4x^2+4-8x^2}{{\left(x^2+1\right)}^2}\end{array}$

Simplify the above derivative further.

  $\begin{array}{c}\frac{dy}{dx}=\frac{4x^2+4-8x^2}{{\left(x^2+1\right)}^2}\\ \frac{dy}{dx}=\frac{-4x^2+4}{{\left(x^2+1\right)}^2}\end{array}$

To find the slope of the tangent at the point $\left(0,0\right)$ , substitute $0$ for x in the above derivative.

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

Thus, the slope of the line is 4.

Substitute $0$ for $x_1$ , $0$ for $y_1$ and 4 for m in the point-slope form of the equation of the line that is, $y-y_1=m\left(x-x_1\right)$ .

  $\begin{array}{c}y-0=4\left(x-0\right)\\ y=4x\end{array}$

Now, to find the slope of the tangent at the point $\left(1,2\right)$ , substitute $1$ for x in the above derivative.

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

Thus, the slope of the line is 0.

Now, substitute $1$ for $x_1$ , $2$ for $y_1$ and 0 for m in the point-slope form of the equation of the line that is, $y-y_1=m\left(x-x_1\right)$ .

  $\begin{array}{c}y-2=0\left(x-1\right)\\ y-2=0\\ y=2\end{array}$

Hence, the required tangents are $y=4x$ and $y=2$ .