course mth271
I am having a lot of trouble with the corse it seems that things are changing as how to get to and the set up of the material. The amount of time to do them has been much more that I had thought. I did find a tutor to help some as of 20 July 2009. My biggest concern is that I will not have time to complet this cource. With this quiz the instructions that I printed off this am when I got to the tutor at lunch time today I could not find the same instructions on line.
You don't specify which quiz this is. However some quizzes do randomize, changing every 5 minutes.
You can always copy the quiz you are working on and email it to yourself, so you will still have access to the original.
Regarding the question of finding things, have you completed the Orientation for the course? If not, you're doing a very good job of finding things, since you have submitted a number of assignments. However you should run through the Orientation in order to know how to use the materials.
The depth of water in a certain uniform cylinder is given by the depth vs. clock time function y = .02 t2 + -1.5 t + 63. What is the average rate at which depth changes between clock times t = 9.9 and t = 19.8?
We substitute 9.9 for t y=.02(9.9to the2nd) – 1.5(9.9)+63 y= 50.1102
Then the same with 19.8 and y= .02(19.8 to the 2nd ) –1.5(19.8)+63 y=41.1408
To get the average dy/dt (19.9-9.9)/(41.1408-50.1102) = -.906cm/sec if the times are in sec and the depths are in cm
· What is the clock time halfway between t = 9.9 and t = 19.8, and what is the rate of depth change at this instant? 14.85 and the instant rate of change is - .906
· What function represents the rate r of depth change at clock time t?
· R(t)=.04t – 1.5
· What is the value of this function at the clock time halfway between t = 9.9 and t = 19.8?
.04(14.85) – 1.5 = - .0906
should be -0.906, and will agree with the average rate you found earlier.
If the rate of depth change is given by dy/dt = .256 t + -1.1 then how much depth change will there be between clock times t = 9.9 and t = 19.8?
(9.9 , 1.4344 ) ( 19.8 , 3.9688) the total change in depth is 1.4344 – 3.9688 = - 2.5344 and the average drop for this time is - 2.5344/ ( 9.9 –19-8 ) = .256
The values of this function are rates of depth change, not depths.
To get the change in depth you need to find the average of the rates of depth change, and multiply by the time interval.
· Give the function that represents the depth.
Y= .128 t (to the 2nd ) – 1.1t + c
· Give the specific function corresponding to depth 250 at clock time t = 0.
Y = .128( to the 2nd ) –1.1 t+ 250
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Your work is very good, see my notes for one minor arithmetic mistake, and more importantly for one fundamental error.