A particle travels the circumference of the equation x^2 + y^2 = 1 counterclockwise. At what points in the circle does the ordinate(y) decrease with the same speed that the abscissa(x) grows? Use derivates
1 Answer
Jun 10, 2018
See below
Explanation:
For the unit circle, differentiate wrt time:
#2 x \ dot x + 2 y \ dot y = 0#
The condition [" ordinate(y) decrease with the same speed [ sic ] that the abscissa(x) grows"] might be interpreted as:
#dot y = - dot x#
So:
-
#2 x dot x - 2 y dot x = 0# -
#dot x ( x - y ) = 0#
So that leaves solutions along:
#y = x# .
The points on the unit circle are:
#pm( 1/sqrt2, 1/sqrt2)#
But it's easily possible to construe the question as requiring solution along:
#y = - x#