A point moves along the curve y= x^2+1 in such way that when x=4 the x-coordinate is increasing at the rate of 5ft/sec at what rate is the y coordinate changing at that time?

1 Answer
Mar 6, 2018

#40ft#/s#ec#

Explanation:

#y=x^2+1#, Differentiating implicitly with respect to t.....

#dy/dt=2xdx/dt+0#. When #x=4, dy/dt=2[4]dx/dt#

So #dy/dt=8dx/dt#. we are told #dx/dt=#5ft#/sec# .........and so #dy/dt=40ft#/sec.