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The depth of water in a certain uniform cylinder is given by the depth vs. clock time function y = .019 t^2 + -1.2 t x y 7.4 42.16 11.1 39.02099 14.8 36.40176

These y values do not appear to correspond to the given function. Can you clarify?

What is the average rate at which depth changes between clock times t = 7.4 and t = 14.8? (36.40176 - 42.16) / (14.8 - 7.4) = (-5.7583)/7.4 = -.778 The average rate is -.778 * 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? The clocktime halfway between 7.4 and 14.8 is 11.1, and the depth is 39.02. The rate of depth change at this instant is -.778 lim x-> 0 = -.778 * What function represents the rate r of depth change at clock time t? lim t->0 r=0.19dt-.778

This would be lim{t->0}[(.019 (t + `dt)^2 - 1.2 ( t + `dt) - ( .019 t^2 - 1.2 t) ) / `dt ].
The limiting value of this expression for y = a t^2 + b t + c is y ' = 2 a t + b.
So the rate function corresponding to a = .019 and b = -1.2, with c = 0, would be r = y ' = 2 ( .019) t - 1.2.

* What is the value of this function at the clock time halfway between t = 7.4 and t = 14.8? Its value is -.778

At t = 0, .019 t - .778 would be -.778. Not so at t = 11.1.

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? I'm really not sure how to use this equation to derive an answer that you're looking for, I know that the change will be some kind of rate between the change in time, but how to derive that I am a little shady on. I thought that by doing .25*(7.4) - 1.5 = dy/(7.4) would give a change in y that would be the correct answer, but the answer I got for that ( .35) seemed a bit small for such a large gap in time.

Your instinct here isn't far off, but .25(7.4)-1.5 would be the rate at t = 7.4.
Between t = 7.4 and t = 14.8 the rate changes, and the rate at t = 14.8 would be as relevant as the rate at t = 7.4.
So you could figure .25 ( 14.8) - 1.5 and get the rate at t = 14.8.
Then if you average the two rates, you get a reasonable approximation of the average rate. In fact since the rate function is linear, you in fact get the average rate. Specifically this is the average rate of depth change with respect to clock time.
Having the average rate you multiply by the time interval to get the change in depth.

* Give the function that represents the depth. * Give the specific function corresponding to depth 190 at clock time t = 0. "

See if you can modify your work on this assignment in accordance with my notes, and submit a revision.
You have a pretty good grasp of the main ideas, just need to put them together a little differently.
Let me know if you have questions.

These y values do not appear to correspond to the given function. Can you clarify?

This would be lim{t->0}[(.019 (t + `dt)^2 - 1.2 ( t + `dt) - ( .019 t^2 - 1.2 t) ) / `dt ]. The limiting value of this expression for y = a t^2 + b t + c is y ' = 2 a t + b. So the rate function corresponding to a = .019 and b = -1.2, with c = 0, would be r = y ' = 2 ( .019) t - 1.2.

At t = 0, .019 t - .778 would be -.778. Not so at t = 11.1.

See if you can modify your work on this assignment in accordance with my notes, and submit a revision. You have a pretty good grasp of the main ideas, just need to put them together a little differently. Let me know if you have questions.

This would be lim{t->0}[(.019 (t + `dt)^2 - 1.2 ( t + `dt) - ( .019 t^2 - 1.2 t) ) / `dt ].
The limiting value of this expression for y = a t^2 + b t + c is y ' = 2 a t + b.
So the rate function corresponding to a = .019 and b = -1.2, with c = 0, would be r = y ' = 2 ( .019) t - 1.2.

See if you can modify your work on this assignment in accordance with my notes, and submit a revision.
You have a pretty good grasp of the main ideas, just need to put them together a little differently.
Let me know if you have questions.