Question

A particle moves along the curve y=2/3x^3+1.find the point on the curve at wch y-coordinate is changing twice as fast as x-coordinate.

Answer 1

y-coordinate is changing twice as fast as x-coordinate means

2*dx = dy     ..eq1
 
differentiating the given equation will give

dy =  2/3 * 3x^2*dx = 2x^2*dx

using eq.1

2dx = 2x^2*dx
1=x^2
x=1 or -1

using the value of x and equation given, it'll give 2 points
(1,5/3) and (-1,1/3)

Hope it helped. ^^