course Mth 271
Please let me know if there are things I did wrong. I'm going to ask my high school teacher about my work, but I feel very confident about what I'm turning in.
The depth of water in a certain uniform cylinder is given by the depth vs. clock time function y = .019 t2 + -1.2 t + 50. What is the average rate at which depth changes between clock times t = 7.4 and t = 14.8?
The average rate of change is represented by change in depth/change in time (dy/dt). This function is equal to the derivative of the original function: y = 2at + b.
This function would have been y ' = 2 a t + b, not y = 2 a t + b.
y = .038t + -1.2
t = 7.4, 14.8
y = .038(7.4) + -1.2 = -.9188
y = .038(14.8) + -1.2 = -.6376
dy = -.9188 – -.6376 = -.2812
dt = 14.8 – 7.4
dy/dt = -.2812/7.4 = -.038
This would be the change in y ' divided by the change in t, not the change in y divided by the change in t.
Using `d for 'change in' (`d is written as the capital Greek Delta, the triangular symbol for 'change in'), what you found was `d y ' / `dt. This would correspond to the approximate value of the second derivative of y with respect to t, which is indeed equal to the constant value -.038.
You should find the change in y and divide it by the change in t, then compare your result with subsequent results to gain a valuable insight.
• What is the clock time halfway between t = 7.4 and t = 14.8, and what is the rate of depth change at this instant?
t = 7.4 + 14.8/2 = 11.1
y = .019t^t + -1.2t + 50
y’ = .038t + -1.2
y’ = .038(11.1) – 1.2
y’ = -.7782
• What function represents the rate r of depth change at clock time t?
r = change depth/change time = dy/dt = 2at + b
• What is the value of this function at the clock time halfway between t = 7.4 and t = 14.8?
t = 7.4 + 14.8/2 = 11.1
y = 2at + b
y = 2a(11.1) + b
y = 2(.019)(11.1) + -1.2
= -.7782
If the rate of depth change is given by dy/dt = .25 t + -1.5 then how much depth change will there be between clock times t = 7.4 and t = 14.8?
• Give the function that represents the depth.
To find the function that represents depth, I took the integral of the above function.
dy/dt = .25t + -1.5
dy/dt = .25(t^2/2) + -1.5t + c
y = .125t^2 + -1.5t + c
• Give the specific function corresponding to depth 190 at clock time t = 0.
y = .125t^2 + -1.5t + c
190 = .125t^2 + -1.5t + c
190 = .125(0)^2 + -1.5(0) + c
190 = c
y = .125t^2 + -1.5t + 190
Excellent work, but you went a little too far on the first question.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end). &#