Two curves are orthogonal if their tangent lines are...

Two curves are orthogonal if their tangent lines are perpendicular at each point of intersection. Show that the given families of curves are orthogonal trajectories of each other; that is, every curve in one family is orthogonal to every curve in the other family. Sketch both families of curves on the same axes. $y=a x^{3}, \quad x^{2}+3 y^{2}=b$

Key Concepts

Orthogonal Trajectories

Orthogonal trajectories are families of curves that intersect such that their tangent lines at each point of intersection are perpendicular. This concept is essential for understanding how one family of curves can be determined from another by ensuring the condition for perpendicularity between tangents, which is often translated into a differential relation between their slopes.

Tangent Lines and Perpendicularity

The analysis of tangent lines involves computing the derivative of a curve at any point, and perpendicularity is characterized by the condition that the product of the slopes of two intersecting curves is -1. This key concept ensures that, at every intersection point, the curves are intersecting at right angles.

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations that define y implicitly as a function of x. When curves are given in forms where y is not isolated, like in conic sections, applying implicit differentiation allows us to obtain the slope (dy/dx) necessary for analyzing the tangents and establishing orthogonality conditions.

Elimination of Parameters

When dealing with families of curves defined by an arbitrary constant, eliminating the parameter from the equation leads to a differential equation that characterizes the family. This process is crucial when comparing different families of curves to determine their relationships, such as verifying orthogonality, without the need for the specific parameter values.

Transcript

00:04 In this question, we're supposed to show that the given families of curves are orthogonal trajectories of each other.

00:12 And you have the two equations given.

00:17 And in order to do that, we need to find the derivatives of each of those.

00:24 And so the derivative of y would be y prime.

00:28 The derivative of a x cube would be 3ax squared.

00:34 And then the derivative of the other equation is going to be 2x plus 6 y, y prime, and b is just a constant, so it'll drop off.

00:53 And then if we solve that for y prime, we're going to get 6yy, y prime equals minus 2x.

01:00 So your y prime is equal to minus 2x over 6y.

01:08 We simplify a little bit, you'll get minus x over 3y.

01:15 And for the other equation, we don't want to have that a in there.

01:19 So we're going to take the original equation and solve it for a.

01:25 So a is going to equal y over x cubed.

01:30 We're going to take and plug it back in for our a there.

01:35 So y prime is going to equal 3 times y over x cubed times x cubed times x.

01:45 Squared.

01:46 So our x squared will cancel out with two of those, and we're going to get y prime is equal to 3y over x.

01:57 This is our first one.

01:59 This is our simple one.

02:01 So you notice our first derivative and our second first derivative are negative reciprocals to each other, which indicates that they are orthogonal, or if you take and multiply them together, you'll get negative 1 and that also indicates that they're orthogonal to each other.

02:20 And so then we need to look at graphing them.

02:39 So our first equation, y equals a x cubed.

02:49 If we start with letting a equal 1, then if we put x is 1 and y would be 1.

03:25 If we put x is 2, 2 cubed would be 8.

03:32 It's going to go off up there.

03:34 If you put in 0, you get 0.

03:35 If you put in negative 1, you get negative 1.

03:38 And if you get negative 2, you're going to get negative 8, which should be able to...