Sunday, September 9, 2018

Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 39

Find the points where the tangent is horizontal on the lemniscate having a function of 2(x2+y2)2=25(x2y2)

Taking the derivative of both sides of the equation implicitly we have


4(x2+y2)(2x+2yy)=25(2x2yy)8x3+8x2yy+8xy2+8y3y=50x50yy8x2yy+8y3y+50yy=50x8x38xy2y=x[508x28y2]8x2y+8y3+50y


The equation of the horizontal tangent can be determined where y=0 so..


0=x[508x28y2]8x2+8y2=50x2+y2=254Equation 1


From the equation of the lemniscate,


2(254)2=25(x2y2)x2y2=258Equation 2


Use Equations 1 and 2 to locate the points where the horizontal tangent is .


x2+y2=254x2y2=258


Solving for x and y simultaneously, we have..

x=±534 and y=±54

Therefore, we have four points at (±534,±54)

No comments:

Post a Comment

Why is the fact that the Americans are helping the Russians important?

In the late author Tom Clancy’s first novel, The Hunt for Red October, the assistance rendered to the Russians by the United States is impor...