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course mth 173
Week 4 quiz 2#$&*
course mth 173
The velocity of an automobile coasting down a hill is given as a function of clock time by v(t) = .001 t^2 + .14 t + 1.8, with v in meters/sec when t is in seconds. Determine the velocity of the vehicle for clock times t = 0, 15 and 30 sec and make a table of rate vs. clock time.Sketch and label the trapezoidal approximation graph corresponding to this table and interpret each of the slopes and areas in terms of the situation.
Evaluate the derivative of the velocity function for t = 22.5 sec and compare with the approximation given by the graph.
By how much does the antiderivative function change between t = 0 and t = 30 seconds, what is the meaning of this change, and what is the graph's approximation to this change?
v(t) = .001 t^2 + .14 t + 1.8
v(0) = .001 (0)^2 + .14 (0) + 1.8
v(0)= 1.8
(0, 1.8)
v(15) = .001 (15)^2 + .14 (15) + 1.8
v(15) = .225 + 2.1 + 1.8
v(15) = 4.125
(15, 4.125)
v(30) = .001 (30)^2 + .14 (30) + 1.8
v(30) = .9 + 4.2 + 1.8
v(30) =6.9
(30, 6.9)
X y
0 1.8
15 4.125
30 6.9
Area under the curve during first 15 sec. interval:
Ave ht * width = area
(4.125 + 1.8)/2 = 2.963
2.963 * 15 = 44.445 m/sec^2
Slope of curve during first 15 sec. interval:
y’ = (4.125 - 1.8)/(15)
y’ = .155 m/s
Area under curve during the second 15 sec. interval:
(6.9 + 4.125)/(2) = 5.513
5.513 * 15 = 83.695 m/sec^2
y’= (6.9 - 4.125)/ (15)
y’ = .185 m/sec
v(t) = .001 t^2 + .14 t + 1.8
v’(t) = 2at + b
v’(t) = .002t + .14
v’(22.5) = .002(22.5) + .14
v’(22.5) = .185 m/s
When t=22.5, it is during the second interval of 15 sec. and is approximately .185 which was determined using the trapezoid approximation graph above.
If v’(t) = .002t + .14 is the derivative, then the anti-derivative is v(t) = .001 t^2 + .14 t + 1.8
v(0) = .001 (0)^2 + .14 (0) + 1.8
v(0)= 1.8 m
(0, 1.8)
v(30) = .001 (30)^2 + .14 (30) + 1.8
v(30) = .9 + 4.2 + 1.8
v(30) =6.9 m
(30, 6.9)
6.9 - 1.8 = 5.1m
This means that in 30 seconds the graph moved from 1.8 in 0secs to 6.9 in 30 seconds, which is a difference of 5.1 meters.
I’m not too sure about the graph’s approximation of the difference of this change.
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Good answers on the graph and the derivative function v ' (t).
The question about the antiderivative referred to the antiderivative of the original function. You used the antiderivative of the derivative of the original function, which is just the original function.
What is an antiderivative of the original (quadratic) velocity function? By how much does it change during each interval? How are these changes related to your trapezoidal graph?
&&&&&This is a little confusing. The differentiation is the process of taking the derivative to get the rate of change function and to integrate is to take the antiderivative to get the original function, which in this case is v(t) = .001 t^2 + .14 t + 1.8. I’m not sure what the antiderivative of the original function is. I have looked over the notes, what am I not seeing? &&&&&
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You know that the derivative of x^3 is 3 x^2, the derivative of x^2 = 2 x and the derivative of x is 1.
From this you should be able to figure out what you would have to take the derivative of to get x^2, and what you would have to take the derivative of to get x (you've already applied the latter in finding an antiderivative of a linear function).
In other words, what is an antiderivative of x^2, and what is an antiderivative of x?
The problem is to then find an antiderivative of
v(t) = .001 t^2 + .14 t + 1.8.
Don't spend more than 10 minutes. If you can't get it by then, just ask and I'll tell you. But better you spend a few minutes and figure it out yourself.
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&#Please see my notes and, unless my notes indicate that revision is optional, 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).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you.
Spend a reasonable amount of time on your revision, but don't let yourself get too bogged down. After a reasonable amount of time, if you don't have at least a reasonable attempt at a solution, insert the best questions you can showing me what you do and do not understand, and I'll attempt to clarify further.
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end document
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&#Please see my notes and, unless my notes indicate that revision is optional, 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).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you.
Spend a reasonable amount of time on your revision, but don't let yourself get too bogged down. After a reasonable amount of time, if you don't have at least a reasonable attempt at a solution, insert the best questions you can showing me what you do and do not understand, and I'll attempt to clarify further. &#
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