Thursday, April 11, 2019

Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 48

Show that the family curve y=ax3,x2+3y2=b 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.

Taking the derivative of the function y=ax3 implicitly we have

dydx=3ax2

So the slope of the tangent line to this curve at point (x1,y1) is


dydx=3a(x1)2dydx=3ax21


Similarly, on the other curve x2+3y2=b


2x+3(2y)dydx=0dydx=x3y


So the slope at point (x1,y1) is

dydx=x13y1

but y1=ax31, so the slope of the second curve is

dydx=x13(ax31)=13ax21

We know that if the two curves are orthogonal to each other, their tangent lines are perpendicular at each point of intersection, that is, the product of their slopes is equal to 1. Multiplying the slopes we get..

(\cancel3ax21)(1\cancel3ax21)=1


Therefore, the curve must be orthagonal.

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