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course Mth 173
2/12 8:35 PM
week 3 quiz 1
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course Mh 173
2/11 4:19 AM
Version 2Determine the average rate of change of the function y(t) = .5 t 2 + 80 t + -10 between x and x + Dx.
What are the growth rate, growth factor and principal function P(t) for an initial investment of $ 990 which is compounded
annually at 10% interest?
that is a quadratic formula, and the derivative would be the rate of change.
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for terminology sake....I believe I would then call this an expression
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Terminology:
That's a quadratic function.
A quadratic expression is an expression of the form a t^2 + b t + c.
A quadratic equation sets a quadratic expression equal to zero.
The quadratic formula is used to solve a quadratic equation.
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That's just terminology, but it helps to keep it straight.
You are right about the derivative, and that's what's really important here.
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the derivative of a quadratic is
y'=2at+b
this specific function would be
y'=1t+80
y'=t+80
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Very good.
This is related to the correct answer to the question, but it isn't the correct answer. The derivative gives the instantaneous rate of change. The question asked for an average rate of change.
The average rate of change between t = x and t = x + `dx would be based on the change in y and the change in x.
The change in y would be the second value minus the first.
The second value would be y(x + `dx), and the first would be y(x).
What therefore are the expressions for y(x + `dx), y(x), and the change in y?
What is the change in x?
What therefore is the average rate of change?
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you said What therefore are the expressions for y(x + `dx), y(x), and the change in y?
What is the change in x?
this sounds like the slope to me. which as you've seen me state previously, the derivative seems to be a type of slope, and I'd use( y(x + `dx) - y(x) ) /( change in x) which sounds like the derivative.
but if I remember correctly now I always crafted my segments of lines, finding the average rate of change, or so I called it.
This has got me so confused now.... but I'll try again . the only thing I can think of currently is if i take two instantaneous rates, and get the average...that is an average of rates so
(((x+'dx) +80)+(x+80)/2 would be the average rate of change between x and x + 'dx
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You can't reliably find the average rate of change on an interval by averaging the initial and final rates. That will actually work if the derivative of the function is linear, which is the case here, but it's not a valid general approach to finding the average rate of change. It does work to find an approximate average rate of change, but that's not what was requested here.
The question can be, and needs to be, answered without reference to the derivative.
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The calculation is straightforward.
The average rate of change of y with respect to t is
average rate = (change in y) / (change in t).
As t changes from x to x + `dx, what are the two corresponding values of y, and what therefore is the change in y?
What is the change in t?
What therefore is the average rate of change?
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Having found the average rate of change, you can then ask what happens as `dx approaches zero. That will lead you to the formula for the derivative.
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so does that mean the average is before I use the limit as x approaches 0?
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growth rate is the &&&&proportional&&&& increase, &&&&????or the percent increase as a decimal&&&& .1, the growth factor is rate+ 1, so 1.1.
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Terminology again:
.1 is not the percent increase. The percent increase would be 10%, since .1 is the same as 10%.
.1 would be called the proportional increase.
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the function is P=P0(1+r)^t
for this specific instance
p=990(1.1)^t
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Good work and good thinking throughout.
I've inserted a couple of notes regarding terminology.
You did not, however, answer the question about the average rate of change correctly, though did a good job of showing how to find the instantaneous rate using the derivative (that wasn't requested but it was well done).
I would like you to submit a revision showing the average rate of change, per my notes.
&#Please see my notes and submit a copy of this document with revisions, comments 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.
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Self-critique (if necessary):
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Self-critique rating:
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You are making very good use of the concept of the derivative, but the average rate of change is not calculated using the derivative. It is based on values of the original function itself.
The calculation is very simple, and in fact you do understand it. You just need to associate that calculation with the question.
The slope of the graph of y vs. t does represent the average rate of change of y with respect to t. You don't need a graph or a slope to find a rate of change, so the rate of change and the slope are in fact two different things. In other words, the two are not identical, but one does represent the other in a way that makes them seem as if they are.
In any case, try one more revisions, and this time use #### to indicate your insertions.
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