Monday, June 16, 2014

Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 49

Find the points at which the ellipse x2xy+y2=3 crosses the x-axis and show that the tangent lines at these points are parallel.

The ellipse crosses the x-axis at y=0 so...


x2xy+y2=3x2x(0)+(0)2=3x2=3x=±3


Taking the derivative of the curve implicitly we have,

x2xy+y2=3


ddx(x2)ddx(xy)+ddx(y2)=ddx(3)2x[ddx(x)y+xddx(y)]+ddy(y2)dydx=02x[y+xdydx]+2ydydx=02xyxdydx+2ydydx=0dydx=2xyx2y


@ x=3


dydx=2(3)032(0)dydx=2\cancel3\cancel3dydx=2


@ x=3


dydx=2(3)032(0)dydx=2(\cancel3)\cancel3dydx=2


The slopes are equal. Therefore, the tangent lines are parallel.

No comments:

Post a Comment