Question

find the orthogonal trajectories for the family of curves of $y = kx^2$. sketch several members of each family.

the equation for the orthogonal trajectories is \boxed{}.
(type an equation.)

Explanation:

Step1: Differentiate the given equation

Given \(y = kx^{2}\), differentiate with respect to \(x\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), we get \(\frac{dy}{dx}=2kx\). From \(y = kx^{2}\), we can express \(k=\frac{y}{x^{2}}\). Substituting \(k\) into \(\frac{dy}{dx}=2kx\), we have \(\frac{dy}{dx}=\frac{2y}{x}\).

Step2: Find the slope of the orthogonal trajectory

The slope of the orthogonal trajectory \(m\) is the negative reciprocal of \(\frac{2y}{x}\). So \(m =-\frac{x}{2y}\).

Step3: Solve the differential equation for the orthogonal trajectory

We have the differential equation \(y\ dy=-\frac{1}{2}x\ dx\). Integrate both sides.
Integrating \(\int y\ dy=-\frac{1}{2}\int x\ dx\). Using the power - rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we get \(\frac{y^{2}}{2}=-\frac{1}{2}\times\frac{x^{2}}{2}+C\).

Step4: Simplify the equation

Multiply through by \(4\) to clear the fractions. We obtain \(2y^{2}+x^{2}=C_1\) (where \(C_1 = 4C\)).

Answer:

\(x^{2}+2y^{2}=C\) (where \(C\) is an arbitrary constant)