IntegrateMe Messages 214 Reaction score 1 Thread starter Apr 13, 2010 #1 When x = 16,the rate at which [tex]\sqrt x[/tex] is increasing is [tex]\frac {1}{k}[/tex] times the rate at which x is increasing. What is the value of k?
When x = 16,the rate at which [tex]\sqrt x[/tex] is increasing is [tex]\frac {1}{k}[/tex] times the rate at which x is increasing. What is the value of k?
IntegrateMe Messages 214 Reaction score 1 Apr 13, 2010 #2 I thought it would be 4 but the answer is 8.
jrosen13 Messages 31 Reaction score 0 Apr 13, 2010 #3 Its the ratio of the derivatives evaluated at x=16, (2 sqrt(x))^-1