#$&*
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 version 1
** **
Problem Number 1
The number of hours between dawn and dusk is modeled in a certain area by the function
• h(t) = 4.2 sin ( (2 `pi / 365) (t - 80) ) + 12,
where h is the number of hours and t the number of days since January 1.
Sketch a graph of this function and show on your graph the points where
• h is maximum (longest days)
• h is increasing fastest
• h is decreasing fastest
• h is minimum.
Determine the approximate values of t for each of these events.
.??I’m not very sure of how to set up this problem. Do I need to make up values of t to sketch the graph? But then how do I find the approximate values of t for each?
@&
This is a sine function of the form
y = A sin(omega * (t - b) ) + k,
which is constructed from the graph of y = sin(t) by a vertical stretch by factor A, a horizontal compression by factor omega, a horizontal translation b, and a vertical translation k.
If you don't understand this, and/or don't understand the circular definition of the sine function and how to graph it based on that definition, let me know and I'll refer you to some materials.
All this is standard in precalculus or first analysis courses, but not all courses are able to cover all the necessary material.
*@
.
.
.
.
.
.
.
.
.
Problem Number 2
The depth of water in a certain nonuniform container is y = .02 t4 + -2.6 t2 + 89, where depth y is in cm when clock time t is in seconds.
• At what average rate is the depth of water changing between clock times t = 17.2 and t = 17.3 seconds?
• At what average rate is the depth of water changing between clock times t = 17.2 and t = 17.21 seconds?
• At what average rate is the depth of water changing between clock times t = 17.2 and t = 17.201 seconds?
• What do you estimate is the rate at which water depth is changing at clock time t = 17.2 seconds?
????I know how to do this for a uniform container, what do I need to do differently in my calculations because this is a nonuniform container?
@&
You follow exactly the same procedure as for the uniform container. The only difference is that the function is not quadratic. The same concepts and procedures apply.
How do you find the average rate of depth change, given a time interval?
How does that procedure work out for this function?
*@
The rate at which water flows from a certain nonuniform cylinder is given by rate = .02 t3 + -2.6 t cm3 per minute, where t is in minutes. How much do water do you think will flow between clock times t = 17.3 minutes and t = 34.4 minutes?
???I think you would need to find the antiderivative here then subtract the two depths, but I am unsure of how to find the antiderivative of this function? Do I need to use the formula for the area of a cylinder somewhere in my calculation?
@&
You are correct that you need to find an antiderivative.
What, for example, is an antiderivative of t^3?
(The area of a cylinder is irrelevant to this question).
*@
.
.
.
.
.
.
.
.
.
.
Problem Number 3
If water depth is changing at the rate = 9 t / ( t + 9) at clock time t, with rate in cm/sec when t is in sec, then use two 2-interval approximations to estimate the change in depth between clock times t = 4 sec and t = 10 sec. One of your approximations should be an overestimate, the other an underestimate.
Dt = (10-4) / 2= 3 sec
T1= 4 sec, t2= 7 sec, t3= 10sec
Left: f(4) * 3sec + f(7) * 3 sec =
(9 * 4/ (4 + 9) * 3 + (9*7 / (7+9)) * 3 = 20.1205 cm
Right : f(7) * 3 sec + f(10)* 3 sec =
(9*7 /(7+9)) * 3 + (9* 10 /(10+ 9)) * 3 = 26.023 cm
20.1205 cm < depth change < 26.023 cm
????Is this the correct way to do this?
@&
Absolutely. Good job.
*@
.
.
.
.
.
.
Problem Number 4
Problem: Derive the expression for the derivative of the function y(t) = a t3 at clock time t.
Problem: If the rate of depth change is rate(t) = .042 t + -1.3, then what is the depth function if the depth at clock time t = 0 is 99? How long does it take for the depth to decrease from 82.87249 to 79.44964? What is the average rate which depth changes over this period?
.y(t) = a t^3
Y ‘ (t) = a * 2t^2
@&
"Derive" means "work out from the definition of".
The definition of the derivative is the limit as t -> 0 of the difference quotient (f(t+`dt) - f(t)) / `dt.
*@
82.87249 = .021t^2 - 1.3t + 99
A = .021 b= -1.3 c= 16.123
After quadratic equation: X= 17.157
79.44964 = .021t^2 - 1.3t + 99
A= .021, b= -1.3, c= 19.55036
After quadratic equation: x = 25.75
25.75 - 17.157 = 8.593 seconds
Average rate = 79.44964 - 82.87249depth / 8.592 seconds = -.3984 depth/ sec
@&
@&
It isn't required, but it's good to know that since the rate function is linear, its average value will occur at the midpoint of the interval. If you were to find the midpoint and plug it into the expression for rate(t), you should get the same result.
This would be a useful way to check your solution.
Doing the arithmetic in my head, I get something close to -.4 units of depth / units of time.
*@
*@
.
.
.
.
.
.
.
.
Problem Number 5
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 2916 its rate of change is known to be 300. If this is the t=0 state of the population, then approximately what will be the population at t = 1.7? What then will be the population at t = 3.4?
.rate(t) = kP
Rate (300) = k(2916)
K= .103
????I’m still confused on how to find the population at the different times. I had trouble with this before the major quiz, can you possibly walk me through the calculation for the first time t= 1.7?
@&
The rate at which population changes is
dP/dt,
so your equation will be
dP/dt = .103 * P.
Now, at the initial rate how much will the population change during the first 1.7 second interval?
What therefore will be the new population?
Using dP/dt = .103 * P you can find the new rate.
At this rate, how much will the population change during the next 1.7 second interval?
What therefore will be your new population?
*@
.
.
.
.
.
.
.
.
.
Problem Number 6
Determine the functions f(z) and g(x) for which each of the following is a composite of the form f(g(x)):
3 e ^ ( 5 x^2)
F(z) = 3 e^ (z)
G(x) = 5 x^2
5 sin( 5 t^2 - 7 t + 8)
F(z) = 5 sin(z)
G(x) = (5 t^2 - 7t + 8)
.
.
.
.
.
.
.
.
.
.
Problem Number 7
The velocity of a car coasting downhill increases at a rate that is pretty much proportional to the slope. A car coasts down a hill whose downward slope is increasing. Answer the following by sketching and describing graphs:
• What would a graph of the car's velocity v(t) vs. clock time t look like?
Would this graph be linear starting in the upper left hand corner and crossing through the origin, then going down the other side of the graph. Because even though the velocity is increasing it will eventually reach zero. Or would this graph be a curve or parabola?
• Would the graph of v(t) vs. t be increasing or decreasing?
Would y(t) vs. t be the distance vs time?
If so the graph would be increasing
• Would it be doing so at an increasing or a decreasing rate?
increasing
• What would a graph of v ' (t) vs. t look like?
The graph would be concave up because the slope is increasing at an increasing rate
• Sketch a graph of v '' (t) vs. t corresponding to your graph of v ' (t) vs. t.
The graph would also be concave up
@&
In any case:
If v(t) is increasing at an increasing rate, what can you say about the trend of its slopes?
Whatever can be said about the trend of those slopes can be said about the graph of v ' (t), since the slope of the v(t) graph at any point is v ' (t).
Now what can be said about the slopes of the v ' (t) graph? Be careful you don't say more than can be inferred from the shape of the v(t) graph.
*@
@&
In terms of physics, what is the meaning of v ' (t) and what is the meaning of v '' ( t)?
*@
I’m still having a little troubling visualizing these graphs
** **
I have went through a version of test #1 and filled in what i do and do not understand. I was hoping you could help answer my questions and see if I'm going in the right direction on the problems I completed. thank you
@&
Just a friendly caution:
I don't ordinarily correct grammer, but you've made an error here that could limit you. The use of "have went" is not correct, and is never correct. It's "have gone".
If you go out of this area and use "have went", it will be likely to lower the way others estimate you.
*@
@&
In any case, see my notes above.
*@
@&
I've answered most of your questions with questions, which tends to drive people nuts, but it's for your own benefit (the question part, not the part about 'nuts').
I'll be glad to follow up. If so, treat it as a revision, according to the usual:
&#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. &#
*@