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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 **
test 1 practice 1
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Just getting some practice in before the test.
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Problem Number 1
Explain how to get a tangent-line approximation to the function y(t) at the t = t0 point. Use a labeled sketch in your explanation.
???I am unsure how to acquire this with no explicit function to evaluate.
You would first need to evaluate y(t0) to get the (t0, y(t0)) coordinate, and find the derivative to get the rate. you could then use the point slope equation to find a line tangent to this
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That's it. Good.
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????could the derivative not also be a possible tangent line
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The derivative gives you just the slope of the tangent line. You need more than just the slope to get the equation of a straight line.
In this case you have a point on the line, which gives you the rest of the information you need to specify the line.
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Problem Number 2
Sketch the graph of a smooth and continuous function which is zero at x = -3, x = -.4999999 and x = 2.2, which is not zero at any other point, and whose derivative at x = -3 is double the derivative at x = 2.2.
Sketch a graph of the derivative of this function.
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so at x= -3, x -.4999999999, and x=2.2, y= 0
the derivative beind double would mean the slope is twice as much, so at -3, if the slope is one, (-3,0) would change to (-2, 1),
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For a smooth function the graph would be continually curving, so the slope would be continually changing, so it wouldn't go through (-2, 1), but it might well go close to (-2, 1).
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and that would mean at (2.2, 0) the slope would have to be .5, and the next point would be (3.2, .5),
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Again, probably somewhat close to (3.2, .5).
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but im guessing these are measured at instanteous rates, so it would have to already be doing this right? so does that mean for me to represent this id have to aproach it positively, after from -.49999 i was going in a negative direction?
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Using -.5 instead of -4.999999 abd constructing the simplest possible smooth graph:
The graph would have to start turning toward (-.5, 0) as it passes through x = -3, so its slope would be decreasing, and it would pass somewhat below (-2, 1).
Then after passing through (-.5, 0), where its slope would be negative, it would start turning toward (2.2, 0), requiring its slope to begin increasing. Having passed through (2.2, 0) with slope 1/2 and increasing slope, it would then pass somewhat above (3.2, .5).
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(it was hard for me to do this with my mouspad, but i tried to get some picture up)
Problem Number 3
Explain in terms of rates and amounts why and how, given the values of a derivative function y ' (t) at two nearby t values, we can estimate the change in its antiderivative y(t) between these t values by averaging two values of y ' (t) and multiplying by the corresponding time interval.
say we have y'(t)== 2x and t values of 1 and 1.1
y'(t)=2(1) = 2
y'(t)=2(1.1) 2.2
the derivative of a function is its rate, which tells you HOW MUCH y changes at a point.
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The derivative doesn't tell you how much y changes, it tells you the rate, with respect to t, at which y changes. Thus the average derivative gives you the avearge rate, which is equal to (change in y) / (change in t).
Of course you know this, as the rest of your statement shows.
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so knowing how much it changes at one point, and another, allows you to know a sort of min/max values of change, which when you find the average, gives you the, well average. which gives you a sort of slope value aka your rate.. then multiplying this by the current elapsed time, give you the proper line with that rate.
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Right.
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thinking about the differential may help with these thoughts, i am not sure
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The differential is definitely related. However the differential uses just the slope at the original point, which makes it simpler but in fact tends to make it less accurate as a tool of approximation.
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A slightly more succint statement equivalent to yours:
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y ' (t) is the rate at which the amount represented by y(t) is changing.
If you average two nearby rates, you get a very good approximation of the average rate.
Multiplying the average rate by the change in t gives you the change in the quantity for that interval, and the change in the quantity is the change in y(t).
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Problem Number 4
If an ice cream cone 5 inches high contains 211.25 cal, then how many cal would we expect a geometrically similar ice cream cone 17 inches high to contain? Using the differential estimate the additional calories in a cone of height 5.01 inches as compared to a cone of height 5 inches.
this is that proportionality I believe, because of the words geometrically similar it seems to give a clue.
basic proportionality is y=kx. a cone is volume though, so do i use basic, or something like y=kx^3?i will stick with the basic y=kx
211.25=k * 5
k=42.25
y=42.25x
y= 42.25 * 17
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The key is to use the correct proportionality. Ice cream cones are not linear entities. They occupy volumen in three dimensions. So y = k x^3 would be appropriate to this question.
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y=7 18.25 for an ice cream cone 17 inches high.
dy= rate * dx
dy= .01 * 42.25
dy= .4225
y= 211.6725 at 5.01
Problem Number 5
The cost in dollars of producing n steel bars is C(n).
What are the units of C’(n)?
What would it mean for C’( 4300) to equal 6149?
What could you conclude if you knew that C’( 5000) = 10670 and C( 5000) = 5500, and what would be the practical meaning of your conclusion?
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if C(n) is dollars for n number of bards
then C'(n) is the rate, or change in cost at a given amount of bars, so cost/bars
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Right units.
C ' (n) is rate at which the cost of producing bars is changing.
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In terms of average rates, C ' (n) is the limiting value of (change in cost) / (change in number of bars) when producing n bars.
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so C'(4300)=6149 is saying it the rate is increasing 6149 dollars, at the 4300 bars instant.
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The rate is not increasing 6149 dollars. It's the cost that's increasing.
The rate at which the cost is increasing is 6149 dollars / bar, when the number of bars produced passes 4300.
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if the rate is climbing at 10670 at 5000 bars, and the cost of bars at 5000 is 5500, then the cost is climbing, because the derivative is positive.
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More than just not linear. The data given here are absurd. But we still need to interpret it to see that it is so.
This would mean that the next bar would cost about 10670 more than the current bar.
That doesn't fit well with the information given about the cost of the bars (for example C ' (4300) = 6149 and C ' (5000) = 10 670 would imply an average rate around 8000 dollars / bar, so that the change in cost for the 700 bars between 4300 and 5000 bars would be about 8000 dollars / bar * 700 bars = 5 600 000 dollars, when the change in cost is in fact -649 dollars.
So the given information is inconsistent, but the interpretation would be as stated.
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???I also want to say that the function doesnt seem linear, but I don't know for sure anything else
Problem Number 6
Sketch a smoothly curving continuous graph with three points A, B and C, each to the right of the preceding, such that the following quantities occur in the given order, from least to greatest:
The slope at C
The average slope between B and C
The slope at A
The average slope between A and B
The slope at B.
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!!!I am fine with this problem, so no worries here.
Problem Number 7
The rate at which a certain quantity grows is rate = 7 * log( .1 t + 1), where the rate is in units / hour and t is clock time in hours. Use two 2-interval approximations to estimate the change in velocity between clock times t = 5.1 hours and t = 10.8 hours. One of your approximations should be an overestimate, the other an underestimate.
????this sounds like the....I believe it was called Riemann sums? the left and right sums? but im not sure by what it means 2-interval approximations. does this mean it wants me to only use 3 data points, say at 0, 5.1 and 10.8?
x= , 5.1,7.95 10.8
y= 1.252838631,1.77845117, 2.226443345
interval 2.85
2.85 * 1.252838631 = 3.570590098, 2.85 * 1.77845117 = 5.068585835. = 8.639175933
2.85 * 1.77845117 = 5.068585835, 2.85 * 2.226443345= 6.345363533. = 11.41394937.
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Very good.
That dangling = sign after the period in each line is grammatically incorrect.
You would want to write this as
2.85 * 1.252838631 = 3.570590098, 2.85 * 1.77845117 = 5.068585835. Total = 8.639175933
2.85 * 1.77845117 = 5.068585835, 2.85 * 2.226443345= 6.345363533. Total = 11.41394937.
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average is about 10.03
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You appear to be in good shape here. It's clear that you understand these concepts very well.
Check my notes.
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