Question

find the orthogonal trajectories of the family of curves of $y = ce^{8x}$. sketch at least two members of each family and at least two of the orthogonal trajectories. what is the equation of the orthogonal trajectories? (type an equation )

Explanation:

Step1: Differentiate the given family of curves

Given \(y = ce^{8x}\). Differentiate with respect to \(x\) using the chain - rule \((e^{u})^\prime=e^{u}\cdot u^\prime\) (where \(u = 8x\)).
\(y^\prime=\frac{dy}{dx}=8ce^{8x}\). But from \(y = ce^{8x}\), we can substitute \(ce^{8x}=y\). So \(\frac{dy}{dx}=8y\).

Step2: Find the slope of the orthogonal trajectories

If two curves are orthogonal, the product of their slopes is \(- 1\). Let \(m_1=\frac{dy}{dx}\) be the slope of the given family of curves and \(m_2\) be the slope of the orthogonal trajectories. Then \(m_1\cdot m_2=-1\).
Since \(m_1 = 8y\), then \(m_2=-\frac{1}{8y}\). So the differential equation for the orthogonal trajectories is \(\frac{dy}{dx}=-\frac{1}{8y}\).

Step3: Solve the differential equation for the orthogonal trajectories

Separate the variables: \(8y\ dy=-dx\).
Integrate both sides.
Integrate \(\int8y\ dy\) using the power - rule \(\int y^n\ dy=\frac{y^{n + 1}}{n+1}+C\) (\(n = 1\)) and \(\int - dx\).
\(\int8y\ dy=8\times\frac{y^{2}}{2}=4y^{2}\) and \(\int - dx=-x + C\).
So \(4y^{2}=-x + C\).

Answer:

\(4y^{2}=-x + C\)