course Mth 271
The depth vs. clock time function y = .024 t2 + -1.7 t + 74 indicates the depth y of water in a certain uniform cylinder at clock time t. At what average rate does depth changes between clock times t = 9.6 and t = 19.2?
y = .024t^2 1.7t + 74
t = 9.6
y = .024(9.6)^2 1.7(9.6) + 74 = 59.89
t = 19.2
y = .024(19.2)^2 1.7(19.2) + 74 + 50.20
dy/dt = 50.20 59.89/19.2 9.6 = -9.69/9.6 = -1.009
What clock time lies midway between t = 9.6 and t = 19.2, at what rate is depth changing at this instant?
y = .024t^t 1.7t + 74
y = .048t 1.7
t = 9.6 + 19.2 = 28.8/2 = 14.4
y = .048(14.4) 1.7 = -1.0088
Note that, up to roundoff error, this is identical to the average rate over the entire interval.
For quandratic functions (and only for quadratic functions) the average rate on every interval is the same as the rate at the midpoint of the interval.
What is the function that represents the rate r of depth change at clock time t?
r = change depth/change time = dy/dt = 2at + b
Evaluate this function at the clock time halfway between t = 9.6 and t = 19.2.
r = change depth/change time = dy/dt = 2at + b
y = 2at + b
y = .048t 1.7
t = 9.6 + 19.2 = 28.8/2 = 14.4
y = .048(14.4) 1.7 = -1.0088
If the rate of depth change is given by dy/dt = .176 t + -1.5 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 9.6 and t = 19.2?
dy/dt = .176t 1.5
t = 9.6 + 19.2 = 28.8/2 = 14.4
dy/dt = .176(14.4) 1.5 = 1.0344
Give the function that represents the depth. What would this specific function be if at clock time t = 0 the depth is 150?
To find the function that represents depth, I took the integral of the given function.
dy/dt = .176t 1.5
y = .176t^2/2 1.5t + c
dy/dt = .088t^2 1.5t + c
t = 0
150 = .088(0)^2 1.5(0) + c
150 = c
y = .088t^2 1.5t + 150
&#This looks good. See my notes. Let me know if you have any questions. &#