course Mth271
5 July 2009, 6:56p.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?
depth@ t=13.9 = .015(13.9)^2+ -1.7(13.9)+91= 70.268
t=27.8= .015(27.8)^2 = -1.7(27.8) + 91= 55.333
dy/dt=(55.333-70.268)/(27.8-13.9)= -14.935/13.9 =-28.835 cm/sec
Right, up to the last step where you apparently subtracted instead of dividing.
-14.935 - 13.9 =-28.835
-14.935/13.9 is not -28.835
What is the rate of depth change at the clock time halfway between t = 13.9 and t = 27.8?
clocktime halfway bewteen= 20.85sec
(.015)(20.85^2)+(-1.7)(20.85)+91= 62.076cm/sec
This would be the depth at the halfway point, not the average rate of change of depth with respect to clock time.
What function represents the rate r of depth change at clock time t?
Since the average rate is represented by the slope, dt progressing toward zero, we can obtain the actual rate. I think the answer is the general rate of
change function where the basic quadratic function is generalized: y'(t)=d'y/d't= at + b
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?
t halfway= 20.85seconds, at this instant means as d't approaches zero.y'(t)=2at +b= 2(.015)(20.85)+(-1.7)= -1.075cm/sec
Right. And your previous calculation -14.935/13.9 is equal to -1.075. Look back and be sure you see the connection.
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?
rate function=.193t - 2.1
.193t = 2a, a=.193/2= .097, b=-2.1 y=at^2 + bt + c = (.097)(t^2)+ (-2.1)(t) +c 2 clocktimes: t=13.9, t=27.8
y(13.9)= (18.741 + -29.19)= -10.499, y(27.8)= (74.965+ -58.38)= 16.585 Change in depth, y= y(27.8)- y(13.9)= 27.034 cm
Good. Alternatively your could have evaluated the rate function at the two points. Since the rate function is linear, the average rate will be the average of the two rates. Multiplying the average rate by the time interval would give you the change in depth. The result should be the same (up to roundoff; your.097 would actually be .0965).
What function represents the depth?
quadratic
The function would be your previous
(.097)(t^2)+ (-2.1)(t) +c.
What would this function be if it was known that at clock time t = 0 the depth is 130 ?
y=130?
If you plug t = 0 into your function (.097)(t^2)+ (-2.1)(t) +c the result should be equal to the depth 130.
This allows you to evaluate c, which you then substitute into the form of your function.
Very good work. Only one arithmetic error, and one problem not quite completed.
&#Let me know if you have questions. &#