course Mth 173
The depth of water in a certain uniform cylinder is given by the depth vs. clock time function y = .026 t2 + -2.9 t + 86.?
What is the average rate at which depth changes between clock times t = 14.8 and t = 29.6?
What is the clock time halfway between t = 14.8 and t = 29.6, and what is the rate of depth change at this instant?
What function represents the rate r of depth change at clock time t?
What is the value of this function at the clock time halfway between t = 14.8 and t = 29.6?
If the rate of depth change is given by dy/dt = .144 t + -1.6 then how much depth change will there be between clock times t = 14.8 and t = 29.6?
Give the function that represents the depth.
Give the specific function corresponding to depth 170 at clock time t = 0.
Response:
We find the average rate at which the depth changes through a couple of calculations. First, we plug 14.8 and 29.6 into the function. From here, we take the answers we get and find the average.
(22.9-48.8)/(29.6-14.8) = -1.75.
The clock time halfway between 14.8 and 29.6 is found by (29.6-14.8)/2 = 7.4. We then take 29.6 - 7.4 and get 22.2. The rate of depth change at this instant is -2.32.
this should be the same as the average rate of depth change over the interval; this is a characteristic of quadratic functions, and is unique to quadratic functions.
The function that represents the rate 'r' depth change at clock time 't' is y'(t) = 2(.026)t ?2.9. The value at 22.2, which is halfway between clock times 14.8 and 29.6, is y'(t) = 2(.026)(22.2) ?2.9. Which equals -1.7.
If the rate of depth change is given by dy/dt = .144 t + -1.6 then the depth change between clock times t = 14.8 and t = 29.6 would be 2.13. We find this by inserting 14.8 and 29.6 into our function. We then take the answers and find the difference in depth change by subtracting 2.6624 - .5312 = 2.13.
this is the difference between two rates of change, which is the change in the rate of change, not the change in the depth.
You could average the two rates of change; since the rate function is linear this would give you the average rate of change, which could then be used to find that change in depth.
In the rate of change function, dy/dt = .144 t + -1.6, we can convert this into the function that represents the depth simply by substituting the values into the quadratic depth function y(t) = at^2 + bt + c. When we do this, we come up with the function y(t) = .144t^2 ?1.6t + 170."
the rate function corresponding to this depth function would be rate = 2 * .144 t - 1.6 = .288 t - 1.6, which is different than the given rate function. The rate function that results from the depth function must match the given rate function.