course MTH 173
7/6/10 9:30 p.m.
If the function y = .015 t2 + -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 ?
Avg rate of depth change between clock times 13.9 and 27.8
y(13.9)=.015(13.9)^2 - 1.7(13.9) + 93
y(13.9)= 72.2682
y(27.8)=.015(27.8)^2 - 1.7(27.8) + 93
y(27.8)= 80.9626
Which gives us these pair of points on the graph (13.9,72.2682) and (27.8,57.3326)
y2-y1= -14.9356
x2-x1= 13.9
dividing y/x= -1.0745 cm/s
The rate of change half way between these points are (27.9 - 13.9)*1/2= 20.9
y(20.9) = .015(20.9)^2 - 1.7(20.9) + 93
y(20.9) = 64.0222 giving us the point on the graph (20.9,64.0222)
To find the instantaneous rate of change we have to find the limit as delta(t) approaches 0
y(20.9 + dt)= .015(20.9 + dt)^2 - 1.7(20.9 + dt) + 93
=.015(dt^2 + 41.8dt + 436.81) -35.53 - 1.7dt + 93
=.015dt^2 -1.073dt + 64.0222
dividing this by dt we have .015dt - 1.073
limit dt > 0 = -1.073 the rate of change at t=20.9 the time half way between the inital pair of points.
the function y = .015t^2 - 1.7t + 93 represents the rate at which depth changes with respect to clock time
the clock time half way between t=13.9 and t=27.8 = 20.9
r(t)=.193t + -2.1
r(13.9) = .5827
r(27.8)= 3.2654
depth change = .5827-3.2654 = -2.6827 change in depth
r(t) represents the depth
if clock time t=o with this function and the depth was 130 the function would have to read r(0)=.193(0) + 130
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