Question

Orthogonal Trajectories use a graphing utility to sketch the intersecting graphs of the equations and show that they are orthogonal. [Two graphs are orthogonal if at their point(s) of intersection their tangent lines are perpendicular to each other.] $$ \begin{array}{l} y^{2}=x^{3} \\ 2 x^{2}+3 y^{2}=5 \end{array} $$

Short answer

Since the result is -1, the two curves are confirmed to be orthogonal trajectories.

Step-by-step solution

Find the First Derivative of Both Equations

The first step involves finding the first derivative of both equations. For the first equation, both sides need to be differentiated with respect to x. For the second equation to be differentiated implicitly, it implies that every y term will have to involve a y' (dy/dx). This will yield: \[2y \frac{dy}{dx} = 3x^2\] for the first equation and \[4x + 6 y \frac{dy}{dx} = 0\] for the second equation.

Solving for Dy/dx

Solving the derivative equations for dy/dx gives the slope of the tangent lines. For the first equation, we have: \[\frac{dy}{dx} = \frac{3x^2}{2y}\] And for the second equation, we get: \[\frac{dy}{dx} = -\frac{2x}{3y}\]

Checking for Orthogonality

Two lines are orthogonal if the product of their slopes is -1. Therefore, if the slopes found above when multiplied yield -1, then the curves will be orthogonal. Let's check: \[\left(-\frac{2x}{3y}\right) \left(\frac{3x^2}{2y}\right) = -1\] Which simplifies to -1 proving the curves are orthogonal trajectories.