Chapter 3.5, Problem 43E

To determine

To show: One family of the curves are orthogonal trajectories to the other family.

Explanation of Solution

Given:

The curves $y=c{x}^2$ and $x^2+2y^2=k$.

Derivative rules:

Chain rule: $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$

Proof:

Obtain the slope of the curves $y=c{x}^2$.

$\begin{array}{c}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(c{x}^2\right)\\ \frac{dy}{dx}=c\frac{d}{dx}\left(x^2\right)\\ =2cx\end{array}$

Thus the slope of the tangent to the curve $y=c{x}^2$ is $m_1=2cx$.

Obtain the slope of the curves $x^2+y^2=k$.

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

Thus, the slope of the equation $x^2+y^2=k$ is $m_2=\frac{-x}{2y}$.

The product slope is computed as follows,

$\begin{array}{c}{m}_1{m}_2=2cx\left(-\frac{x}{2y}\right)\\ =\frac{-2c{x}^2}{2y}\\ =\frac{-2y}{2y}\left[Qy=c{x}^2\right]\\ =-1\end{array}$

If $c\ne 0$ then the family of the curves orthogonal trajectories to another family of the curves.

If $c=0$, then

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

Substitute the value $y=0$ in the equation $x^2+y^2=k$.

$\begin{array}{l}{x}^2+{\left(0\right)}^2=k\\ x^2=k\\ x=\pm \sqrt{k}\end{array}$

The line $y=0$ is horizontal line and it intersect $x^2+2y^2=k$ at the points $\left(-\sqrt{k},0\right)\text{ and }\left(\sqrt{k},0\right)$. That is, the curve $x^2+2y^2=k$ has vertical tangent at the points $\left(-\sqrt{k},0\right)\text{ and }\left(\sqrt{k},0\right)$.

Therefore, the curves $y=c{x}^2$ and $x^2+y^2=k$ are orthogonal trajectories to each other is proved.

Graph:

The family of the curves $y=c{x}^2$ and $x^2+y^2=k$ as shown below in Figure 1.

From the graph, it is observed that the family of the curves $y=c{x}^2$ and $x^2+y^2=k$ are orthogonal trajectories to each other.