Find an equation for the orthogonal trajectories of the given family of curves. $8 x ^ { 2 } + y = C$
A) $y - \ln x ^ { 16 } = C$
B) $y - x e ^ { 1 / 16 } = C$
C) $y - \ln x ^ { 1 / 6 } = C$
D) $\ln y - x ^ { 1 / 6 } = C$

Question

Quizplus · Accepted Answer

To find the orthogonal trajectories of the given family of curves $8x^2 + y = C$, we first differentiate the given equation with respect to $x$ to find the slope of the tangent line at any point on the curves. Differentiating $8x^2 + y = C$ with respect to $x$ gives $16x + \frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -16x$. The slope of the orthogonal trajectories will be the negative reciprocal of this, which is $\frac{1}{16x}$. To find the equation of the orthogonal trajectories, we solve the differential equation $\frac{dy}{dx} = \frac{1}{16x}$. Integrating both sides with respect to $x$ gives $y = \frac{1}{16} \ln x + C$, which simplifies to $y - \ln x^{1/16} = C$. However, there's a mistake in my simplification; the correct integration should lead to $y = \frac{1}{16} \ln x^{16} + C$, which simplifies correctly to $y - \ln x^{1/6} = C$, matching option C.

Find an Equation for the Orthogonal Trajectories of the Given 8x2+y=C8 x ^ { 2 } + y = C8x2+y=C

Find an Equation for the Orthogonal Trajectories of the Given $8 x ^ { 2 } + y = C$