#$&*
course MTH 271
5:01 pm 27 oct 2011
If the function y = .018 t2 + -1.7 t + 88 represents depth y vs. clock time t, then what is the average rate of depth change between clock times t = 12.8 and t = 25.6?the derivative of this function is y'=.036t-1.7 which will give us the rate of depth change.
Y'(12.8)=0.4608-1.7=-1.2392cm/sec
Y'(25.6)=0.9216-1.7=-0.7784cm/sec
Now we take these two rates and average them: (-0.7784+-1.2392)/2=-2.0176/2=-1.0088cm/sec
What is the rate of depth change at the clock time halfway between t = 12.8 and t = 25.6?
We must plug into the derivative function the time t=19.2sec:
Y'=.036(19.2)-1.7=0.6912-1.7=-1.0088cm/sec
What function represents the rate r of depth change at clock time t?
r=.036t-1.7
What is the value of this function at the clock time halfway between t = 12.8 and t = 25.6?
it's r=-1.0088cm/sec
If the rate of depth change is given by dy/dt = .057 t + -2.5 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 12.8 and t = 25.6?
f(12.8)=0.7296-2.5=-1.7704cm/sec
f(25.6)=1.4592-2.5=-1.0408cm/sec
Now we multiply the average of these two by the timeinterval:
(-1.0408+-1.7704)/2=-2.8112/2=-1.4056cm/sec
25.6-12.8=12.8seconds
12.8*-1.4056=-17.99168 cm over the interval, this means 17.99168 cm were lost during this time
Give the function that represents the depth. What would this specific function be if at clock time t = 0 the depth is 180?
y=.114t^2-2.5t+180"
Self-critique (if necessary):
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#$&*
course MTH 271
5:01 pm 27 oct 2011
If the function y = .018 t2 + -1.7 t + 88 represents depth y vs. clock time t, then what is the average rate of depth change between clock times t = 12.8 and t = 25.6?the derivative of this function is y'=.036t-1.7 which will give us the rate of depth change.
Y'(12.8)=0.4608-1.7=-1.2392cm/sec
Y'(25.6)=0.9216-1.7=-0.7784cm/sec
Now we take these two rates and average them: (-0.7784+-1.2392)/2=-2.0176/2=-1.0088cm/sec
What is the rate of depth change at the clock time halfway between t = 12.8 and t = 25.6?
We must plug into the derivative function the time t=19.2sec:
Y'=.036(19.2)-1.7=0.6912-1.7=-1.0088cm/sec
@& Good thinking, but this is not how you calculate the average rate of change.
The average rate of change is (change in y) / (change in x).
For quadratic functions, which is what you happen to have here, the average of the initial and final rates of change will be the average rate of change. This is, however, not so for any other class of functions.*@
What function represents the rate r of depth change at clock time t?
r=.036t-1.7
What is the value of this function at the clock time halfway between t = 12.8 and t = 25.6?
it's r=-1.0088cm/sec
If the rate of depth change is given by dy/dt = .057 t + -2.5 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 12.8 and t = 25.6?
f(12.8)=0.7296-2.5=-1.7704cm/sec
f(25.6)=1.4592-2.5=-1.0408cm/sec
Now we multiply the average of these two by the timeinterval:
(-1.0408+-1.7704)/2=-2.8112/2=-1.4056cm/sec
25.6-12.8=12.8seconds
12.8*-1.4056=-17.99168 cm over the interval, this means 17.99168 cm were lost during this time
@& This works because the rate function is linear. If it wasn't, then this wouldn't give you the correct change in the quantity.
You've expressed a good insight nevertheless.
What does work is to find an antiderivative function (as you have done below). The change in an antiderivative is the change in the depth.*@
Give the function that represents the depth. What would this specific function be if at clock time t = 0 the depth is 180?
y=.114t^2-2.5t+180"
&#This looks good. See my notes. Let me know if you have any questions. &#