Question

find the orthogonal trajectories for the family of curves of y = kx². sketch several members of each family. the equation for the orthogonal trajectories is (type an equation )

Explanation:

Step1: Differentiate the given family of curves

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

Step2: Find the slope of the orthogonal trajectories

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

Step3: Solve the differential equation for the orthogonal trajectories

We have the differential equation \(\frac{dy}{dx}=-\frac{x}{2y}\). Cross - multiply to get \(2y\;dy=-x\;dx\).
Integrate both sides:
Integrate \(\int 2y\;dy\) and \(\int - x\;dx\). Using the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have \(\int 2y\;dy=y^{2}+C_1\) and \(\int - x\;dx=-\frac{x^{2}}{2}+C_2\).
So \(y^{2}=-\frac{x^{2}}{2}+C\) (where \(C = C_2 - C_1\)).

Answer:

\(x^{2}+2y^{2}=C\)