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course MTH 173
I completed a practice test for your review. I did not fill in all of the problems because I feel comfortable with several of them. If you can please review the ones that I did fill in and answer several questions that are on the form it will be much appreciated.
Calculus I Test 1Problem Number 1
Sketch a graph of f(t) vs. t with the following characteristics:
At some point A, f ' (t) > 0 and f '' (t) < 0.
At some point B, f ' (t) and f (t) have the same value.
At some point C, f '' (t), f(t) and f ' (t) all have different signs.
At some point C, f '' (t)* f(t) * f ' (t) is positive, but f'', f' and f do not all have the same sign.
This problem was fairly easy to draw and label so I wont ask for too much review here. I would like to ask about the last two items having to do with Point C. If all I have to work with is positive and negative, how can they all have different signs? Does one have to be positive, one negative and the other zero?
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The don't all have different signs, they just don't all have the same sign.
In otherwords, either two are positive and one negative, or two are negative and one positive.
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Problem Number 2
Problem: The quadratic depth vs. clock time model corresponding to depths of 40.81827 cm, 27.50627 cm and 20.06398 cm at clock times t = 11.0583, 22.11659 and 33.17488 seconds is depth(t) = .024 t2 + -2 t + 60.
Show the system of equations we would solve to get this model.
Then use the model to determine whether the depth will ever reach zero.
Problem: The quadratic depth vs. clock time model corresponding to depths of 40.81827 cm, 27.50627 cm and 20.06398 cm at clock times t = 11.0583, 22.11659 and 33.17488 seconds is depth(t) = .024 t2 + -2 t + 60.
Show the system of equations we would solve to get this model.
According to the depth function what is the average rate at which depth changes between clock times t = 25.2 and t = 1025.2 seconds?
What is the instantaneous rate of depth change at each of these clock times?
How does the average of the instantaneous rates compare with the average rate, and why should the comparison be as it is?
#### I feel very comfortable with this problem so I will not include it here for review.####
Problem Number 3
Determine the average rate of change of the function y(t) = .1428571 t 4 between t and t + `dt.
What are the growth rate, growth factor and population function P(t) for an initial population of 1430 which has an annual growth rate of 10%?
#### For the average rate of change of the function y(t) = .1428571 t 4 between t and t + `dt, are you looking for me to use the difference quotient and simplify or rather get the derivative and simplify as much as possible? ####
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The derivative doesn't give you the average rate of change on an interval, it gives you the instantaneous rate of change at a point.
Of course you could find the expression for the derivative and calculate the average value of that expression over the interval, which among other things would require integrating it over the interval, but that would just bring you back to the original function.
So the short answer to your question is that you're stuck with the difference quotient, or an equivalent calculation.
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For population,
P(t) = 1430(1.10)^t
where the growth rate is 10 percent and the growth factor is 1.10.
Problem Number 4
.Draw the graph of a continuous function f(x) satisfying all the following conditions:
f ' (x) > 0 for x > 4
f ' (x) > 0 for x < 7
f ' (x) < 0 for 7 < x < 4
#### I feel comfortable with this problem. ####
Problem Number 5
Use both 2-interval and a 5-interval approximations to find the left and right Riemann sums of the function y = 4 t / ( t + 6) from t = 8 to t = 13.
Based on these approximations make your best estimate of the integral.
What would be the difference between left- and right-hand sums if we used 100 intervals for the approximation?
For the two interval I have:
2.286(3) + 2.588(2) = 6.858 as the left hand sum.
2.588(3) + 2.736(2) = 13.236 as the right hand sum.
This gives the Riemann sum of (6.858 + 13.236)/2 = 10.047
For the five interval I have:
2.286 + 2.4 + 2.5 + 2.2588 + 2.667 = 12.442 for the left hand sum
2.4 + 2.5 + 2.2588 + 2.667 +2.736 = 12.892 for the right hand sum
(12.442 + 12.892) / 2 = 12.6670 as the Reimann sum
For the actual integral I came up with 12.67084. As we increase the number of divisions, or rectangles, we can increase the precision of the approximation under the curve.
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Good.
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Problem Number 6
Write the differential equation expressing the hypothesis that the rate of change of a population is proportional to the population P. Evaluate the proportionality constant if it is known that the when the population is 2174 its rate of change is known to be 200. If this is the t=0 state of the population, then approximately what will be the population at t = 1.6? What then will be the population at t = 3.2?
`dP/`dt = P = 200
Does this problem say that the slope or tangent line at 2174 is 200? I looked through the notes to find an example and didnt see much. Where can I look for help or can you tell me where to go with this problem?
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y is proportional to x if there exists a constant k such that y = k x.
Thus to state that the rate of change of the population is proportional to P is to say that
dP/dt = k * P for some constant k.
The statement that the rate of change is 200 when the population is 2174 gives you information to plug into this relationship and determine k.
For this and closely related topics you should review Class Notes 8 - 10, either in the posted documents accessible from your Assignments Page or on your DVD's.
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Problem Number 7
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 B
The slope at A
The average slope between A and B.
For the drawing, my graph looks like an exponential function; say sqrt(x).
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An exponential function is a function of the form y = b^x.
Power functions are of the form y = x^p for some fixed power p.
sqrt(x) = x^(1/2) is not exponential, it's a power function.
I see the term 'exponential' misapplied very frequently; in fact I think I caught an article in Scientific American or the Smithsonian magazine that made this error.
It's a significant error, since the implications of exponential growth are very different than those of power-function growth.
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I have point a at around x = 1, and it has the greatest slope. Point b a bit further to the right and point c further to the right still. This should meet all of the requirements above.
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Almost.
From your description, thought, the average slope between A and B, though, won't be greater than the slope a A.
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You appear to be in pretty good shape.
Check my notes and let me know if you have additional questions.
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