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Mth 173
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** Question Form_labelMessages **
Theorem of Calculus
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These processes, differentiation and integration, are the two fundamental processes of calculus.
The Fundamental Theorem of Calculus, without which our technological civilization almost certainly would not exist in anything like its present form, essentially says that if we start with a rate function and integrate it we end up with a function whose derivative is the original rate function.
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Im having trouble understanding the meaning of:
if we start with a rate function and integrate it we end up with a function whose derivative is the original rate function.
To me this sounds essentially like the reflexive property. x=x
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It's much more like an inverse property, where a quantity combined with its inverse gives you the identity.
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If you start with a rate function, change it, and then take away your change, you have what you started with. It sounds like this too me as well. or
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If f(x) = x + 2, then g(x) = x - 2 is its inverse, since f(g(x)) = (x-2) + 2 = x.
Again, the operations of integration and differentiation are very nearly inverse operations (not inverse functions, but inverse operations).
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y=x+2
x=y-2
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Also, getting a derivative like this just sounds the same as getting the slope to me, or rather if I remember my applied calculus class I took at one point, the tangent line.
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The derivative of a function can be identified with the slope of its graph.
However you can't calculate the slope of, say, y = x^2 at the point x = 3 by finding a slope. To calculate a slope requires two points, and the x = 3 point is just one point.
The question of the slope of a function at a point is answered by the derivative function.
Having the derivative at, say, the x = 3 point of the graph of y = x^2 you can in fact calculate the tangent line. The derivative turns out to be 6, and the x = 3 point is (3, 9), so the tangent line is the line with slope 6 through the point (3, 9).
One thing you will learn within the next month or so is the rules for calculating derivatives of certain functions and combinations of those functions.
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#$&*
Mth 173
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
week2 quiz1 practice num1 redo
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RANDOMIZED PROBLEMS WEEK 2 QUIZ #1
If the function y = .023 t2 + -1.4 t + 67 represents depth y vs. clock time t, then what is the average rate of depth change
between clock time t = 8.3 and clock time t = 16.6? What is the rate of depth change at the clock time halfway between t =
8.3 and t = 16.6?
What function represents the rate r of depth change at clock time t? What is the clock time halfway between t = 8.3 and t =
16.6, and what is the rate of depth change at this instant?
????I believe the function meant to say t^2, i will assume it did, on my screen it says t2
plugging into the function for clock time t= 8.3(8.3,56.96447), for clock time t = 16.6(16.6,50.09788)
???the question asks what is the rate of depth change at the clock time halfway between 8.3 and 16.6, does that mean it
wants me to add them together and divide by 2, and find the rate of change of that clock time. or just find the slope between
8.3 and 16.6, and this would be its rate of depth change.
the rate of depth change at clock time halfway between 8.3 and 16.6 would be -6.86659/8.3
?????I dont understand what I am to do to find a function representing the rate r of depth change at clock time t if 'd is delta
or change in I think it might be
the function that represents r of depth change at clock time t r= (Depth-'dDepth)/(Time-'dTime)
The clock time halfway between 8.3 and 16.6 is 12.45.
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I believe you've missed the material in the Class Notes related to finding the rate function for a given depth function.
This occurs in Class Notes 3 and 4.
Check out that material, which I think you'll understand pretty easily, and submit another solution.
Your thinking on this questions is good, but you just don't yet have the information you need to solve the problem.
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REATTEMPT
What function represents the rate r of depth change at clock time t, I believe this wants the derivative of y = .023 t2 + -1.4 t
+ 67
r = .046t-1.4t = y'
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Good, but
r(t) = .046 t - 1.4, not .046 t - 1.4 t.
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what is the average rate of depth change between clock time t = 8.3 and clock time t = 16.6
well, if r stands for rate of dpth change at t, I should be able to plug these into my equation for r (???the derivative, correct?)
r=.046(8.3)-1.4(8.3)
r = 0.3818 -11.62
r = - 11.2382 for t=8.3
r=.046(16.6)-1.4(16.6)
r= .7636-23.24
r= -22.4764 for t=16.6
-11.2382 + -22.4764 = -33.7146
-33.7146/2= 16.8573
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You have calculated the average of the initial and final rates for the given interval. This is a good result, and it turns out that for quadratic functions, because of the linearity of their derivatives, the average of the initial and final rates is indeed the average rate.
However this will not in general calculate the average rate of depth change, which is (change in depth) / (change in clock time). It works only for quadratic functions.
To answer the question in general, you would calculate the initial and final depth, then divide by the 8.3 second time interval. You should take a minute and do so, so that you see in a convincing way that this result agrees with the result obtained by averaging initial and final rates.
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You also need to calculate these rates based on the correct r(t) function (see previous note).
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So the average rate between clock time t = 8.3 and t= 16.6 is 16.8573 (???what units would I put after this?)
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It's very good that you are thinking in terms of units.
Usually the depth and clock time will have specified units. However in this case they do not.
The rise between two points has units of depth, the run has units of clock time, so the given rate of change will have units of depth / units of clock time.
You could therefore specify your answer as
16.856 units of depth / units of clock time.
However since no units were specified, it's OK to just say that the average rate is 16.8573.
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What is the rate of depth change at the clock time halfway between t = 8.3 and t = 16.6?
I believe this question asks what is r at t = 12.45
r=.046(12.45)-1.4(12.45)
r=.5727-17.43
r= -16.8573
???? The average rate and the rate I got for doing this where the same
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Yes.
For a quadratic function the average rate of change on an interval, the average of the initial and final rates on that interval, and the instantaneous rate at the midpoint of the interval are identical.
This is a property unique to quadratic functions, and is the result of the linearity of the derivative.
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For the moment I'm accepting your rate function. If you calculate with the correct rate function, you will find that the two results still agree, and that they agree with your later result.
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##### I relooked at the notes and it said to find the average rate of two points do 'dy/'dt
'dy/'dt =(( .023(16.6^2)+-1.4(16.6) + 67) - (.023(8.3^2)-1.4(8.3)+67))/ 8.3
(50.09788-5696447)/8.3
=-.8273
????What is the difference between 'dy/'dt and r(2.5) in this case. I have confused myself
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If you calculate your earlier results using the correct rate function you will find that they agree with this result.
Due to the linearity of the derivative function, the average of the initial and final rates, the instantaneous rate at the midpoint, and the average rate over the interval (these are three distinct calculations) must agree.
This is the case for any quadratic function, and if it is the case for a given function over all possible intervals, that function must be quadratic.
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???If you do not mind, would you show me how to solve this problem, and let me use it as an example to attempt another problem.
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There are three distinct calculations here, all of which will agree if done correctly.
You have made an easily-correctable error in finding the derivative of the depth function.
I believe my notes will clarify the situation for you. If not, I know I can trust you to ask for additional clarification.
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