#$&*
course Mth 173
I had some trouble with this problem. 9-15-10 6:34 p.m.
If the function y = .022 t2 + -1.6 t + 72 represents depth y vs. clock time t, then what is the average rate of depth change between clock time t = 9.1 and clock time t = 18.2? What is the rate of depth change at the clock time halfway between t = 9.1 and t = 18.2?What function represents the rate r of depth change at clock time t? What is the clock time halfway between t = 9.1 and t = 18.2, and what is the rate of depth change at this instant?
If the function r(t) = .176 t + -1.9 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 9.1 and t = 18.2?
• What function represents the depth?
• What would this function be if it was known that at clock time t = 0 the depth is 140?
You are answering here for the function y = .022t^2- 1.6t + 72; this part of the solution should have followed that question, not the (different) question where you are given the r(t) function. No problem for me, but this might be the source of some of you confusion on the r(t) question.
The rate of depth change of t=9.1 for the function y = .022t^2- 1.6t + 72 y’=2(.022)(9.1) – 1.6 = -1.1996
The rate of depth change of t=18.2 for the function y = .022t^2- 1.6t + 72 y’=2(.022)18.2)-1.6= -.7992
The average rate of depth change between the two is = -.9994
The average rate of change between clock time t=9.1 and clock time t=18.2 is -.9994.
the average rate of depth change with respect to clock time, between two clock times is (change in depth) / (change in clock time)
the average rate of depth change with respect to clock time is not generally equal to the average of the initial and final rates of change( which is what you found), though for one specific class of functions this will be the case
So is the average rate in this case equal to the average of the two rates?
-1.1996 + -.7992 = -1.9988/2 = -.9994
The rate of depth change at the clock time halfway between t = 9.1 and t= 18.2 is -.9994
• 2(.022)(13.65)-1.6=-.9994
Everything you have calculated is useful and related to the overall question, but you haven't yet found the average rate, at least not in a way that would give the correct result for any function.
After this point I am uncertain as to How to work the remainder of this problem. The following questions I am unsure how to answer, if you could guide me throw this process I would appreciate it greatly.
What function represents the rate r of depth change at clock time t? What is the clock time halfway between t = 9.1 and t = 18.2, and what is the rate of depth change at this instant?
For the y = .022t^2- 1.6t + 72 function you have calculated the rate function, which is r(t) = y ' (t), and applied it.
You now have a new r(t) function, unrelated to the preceding. What must be the y(t) function for this rate function? Find that function and use it to answer the questions:
If the function r(t) = .176 t + -1.9 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 9.1 and t = 18.2?
• What function represents the depth?
• What would this function be if it was known that at clock time t = 0 the depth is 140?
?R(t) =Ro(-.944)^t???
This is an exponential function. The given rate function was a linear function. Unsure of where the exponential function came from.
You're definitely on the right track. See my notes.
&#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).
Be sure to include the entire document, including my notes. &#
#$&*