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course Mth 271
Submittted 01-24-11 @6:45pm
If the function y = .015 t2 + -1.7 t + 93 represents depth y vs. clock time t, then what is the average rate of depth change between clock time t = 13.9 and clock time t = 27.8? What is the rate of depth change at the clock time halfway between t = 13.9 and t = 27.8?What function represents the rate r of depth change at clock time t? What is the clock time halfway between t = 13.9 and t = 27.8, and what is the rate of depth change at this instant?
If the function r(t) = .193 t + -2.1 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 13.9 and t = 27.8?
• What function represents the depth?
• What would this function be if it was known that at clock time t = 0 the depth is 130 ?
Substituting the clock time of 13.9 into the function y = .015t^2 + (-1.7t) = 93 I found the rate of depth change to be 72.3. Substituting the clock time of 27.8 into the function I found the rate of depth change to 57.3 Calculating the clock time halfway between 13.9 and 27.8 is equal to 6.95 then add to 13.9 = 20.8. Substituting this clock time into the function I found the rate of depth change to be 64.1
To find the average rate of depth change using the function y(t+dt) – y(t) / dt =
57.3-72.3 / 27.8-13.9 = -15 / 13.9 = -1.079
The clock time halfway between t=13.9 and 27.8 = 20.8.
(dy/dt – derivative of depth function with respect to time) giving derivative of original function of y=.015t^2 – 1.7t +93
The function r(t) = .03t – 1.7. Substituting the clock time of 20.8 into this function would give you the instant rate of change of -1.0745.
The function that represents depth is y(t) = .015t^2 – 1.7t +93.
The function would be y(0) = .015(t)^2 – 1.7(0) + 130
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&#Very good work. Let me know if you have questions. &#