course MTH 271
Feb. 23 - 5:40 P.M.I hope I did this right. If I messed up really bad, is there any way to redo and get credit??? thanks!
If the function y = .015 t^2 + -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 ?
Answer:
Y = .015t^2 + -1.7t + 93
Y = .015(13.9)^2 + -1.7(13.9) + 93
= 72.3
Y = .015(27.8)^2 + -1.7(27.8) + 93
= 57.3
Average Rate of Depth change:
‘dy/’dt = [ y(t + ‘dt) – y(t) ] / ‘dt
57.3 – 72.3 / 13.9
= -1.1
Rate of depth change halfway between:
‘dy/ ‘dt = [ y(t +’dt) –y(t) ] /’dt
57.3 – 20.85 / 13.9
= 2.6
You misinterpreted this part of the question; the answer would be identical to your later correct solution R(t) = 2(.015)(20.85) + -1.7 = -1.1.
What function represents the rate r of depth change at clock time t?
y’(t) = 2at + b
What is clock time halfway between t = 13.9 and t =27.8?:
Clock time = 20.85
The rate of depth change at this instant:
R(t) = 2at +b
R(t) = 2(.015)(20.85) + -1.7
R(t) = -1.1
Good.
If r(t) = .193t + -2.1, how much depth change will there be between clock times t =13.9 and t = 27.8?
R(t) = .193(13.9) + -2.1
R(t) = .58
R(t) = .193(27.8) +-2.1
R(t) = 3.3
Change: 3.3 -.58 = 2.72
3.3 and .58 are rates; so 2.72 is the change in the rate, not in the depth.
You could use your two rates to find the average rate, and then use this to find the change in the depth.
Depth function: t =0, depth function = 130
Y(t) = .5mt^2 +bt +c
Y(t) = .5mt^2 + bt +c
130 = .5m(0)^2 + b(0) +c
Good, but you know m and t from the given rate function. You should substitute those values for m and t.
You can solve the last equation for c. You would then substitute this value for c in your function y(t) = .5 m t^2 + b t + c.
You've done really well on most of these questions and you should be able to revise and modify the other questions without too much difficulty. 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). &#