d= √500t^2 -240t -29
(b) d=d(t). For what value of t is d smallest?
t≈ hours
To find the value of t for which d is smallest, we need to find the derivative of d(t) and set it equal to zero to find any critical points. Then we can use the second derivative test to determine if the critical point is a minimum.
Given d = √(500t^2 - 240t - 29)
d' = (1/2)*(500t^2 - 240t - 29)^(-1/2) * (1000t - 240)
d' = (250t - 60) / √(500t^2 - 240t - 29)
Setting d' = 0:
250t - 60 = 0
t = 60/250
t = 0.24 hours
To determine if this is a minimum, we need to find the second derivative:
d'' = (250) / √(500t^2 - 240t - 29) - (500t - 240)*(250t - 60) / (2*√(500t^2 - 240t - 29))^3
Plugging in t = 0.24:
d'' = (250) / √(500*0.24^2 - 240*0.24 - 29) - (500*0.24 - 240)*(250*0.24 - 60) / (2*√(500*0.24^2 - 240*0.24 - 29))^3
Calculating d'' at t = 0.24 will tell us whether the function has a minimum at that point.