course Mth 272
I really liked it when there were the correct answers after submission. What happened to that format?
??????????i??assignment #016016.
Applied Calculus II
10-19-2008
......!!!!!!!!...................................
00:14:59
Query problem 6.2.50 (was 6.2.48) solid of revolution y = x e^x x = 0 to 1 about x axis
......!!!!!!!!...................................
RESPONSE -->
I am unsure what this is asking. I could not find the match of this problem in the book, either. I looked through the whole section of the chapter and cannot find a problem like this one. Maybe I am overlooking something?
confidence assessment: 0
*&*& The volume over an interval `dx will be approximately equal to pi ( x e^x)^2 `dx, so we will be integrating pi x^2 e^(2x) with respect to x from x = 0 to x = 1.
Integrating just x^2 e^(2x) we let u = x^2 and dv = e^(2x) dx so that du = 2x dx and v = 1/2 e^(2x).
This gives us u v - int(v du) = 2x (1/2 e^(2x)) - int(1/2 * 2x e^(2x) ). The remaining integral is calculated by a similar method and we get a result that simplifies to
e^(2x) ( 2 x^2 - 2x + 1) / 4.
Our antiderivative is therefore pi e^(2x) (2 x^2 - 2x + 1) / 4. This antiderivative changes between x = 0 and x = 1 by pi ( e^2 / 4 - 1/4) = 5.02, approx.. *&*&
.................................................
......!!!!!!!!...................................
00:15:39
For your first step, what was u and what was dv, what were du and v, and what integral did you obtain?
......!!!!!!!!...................................
RESPONSE -->
I'm going to take a shot in the dark.
u= x
dv= e^x
du=dx
u= e^x
confidence assessment: 1
.................................................
......!!!!!!!!...................................
**** Problem 6.2.56 revenue function 410.5 t^2 e^(-t/30) + 25000
00:29:13
What expression did you evaluate to obtain the average daily receipts during the first quarter, and what was your result?
......!!!!!!!!...................................
RESPONSE -->
1st quarter= approx. 90 days
t= 90
410.5(90^2)e^(-90/30) + 25,000
revenue for 1st quarter= $190,544.49
confidence assessment: 3
The daily revenue function needs to be integrated to get the total revenue for a period.
*&*& If we integrate the revenue function from t = 0 to t = 90 we will have the first-quarter revenue. If we then divide by 90 we get the average daily revenue.
To get an antiderivative of t^2 e^(-t/30) we first substitute u = t^2 and dv = e^(-t/30), obtaining du = 2t dt and v = -30 e^(-t/30). We proceed through the rest of the steps, which are very similar to steps used in preceding problems, to get antiderivative
- 30•e^(- t/30)•(t^2 + 60•t + 1800).
Our antiderivative of 410.5 t^2 e^(-t/30) + 25000 is therefore 410.5 (- 30•e^(- t/30)•(t^2 + 60•t + 1800) ) + 25000 t.
The change in this antiderivative function between t = 0 and t = 90 is found by substitution to be about 15,000,000, representing $15,000,000 in 90 days for an average daily revenue of $15,000,000 / 90 = $167,000, approx.. *&*&
.................................................
......!!!!!!!!...................................
00:31:55
What expression did you evaluate to obtain average daily receipts during the fourth quarter and was your result?
......!!!!!!!!...................................
RESPONSE -->
How would this answer be different? If the year is split into quarters, then each value for ""t"" would be about the same, right? Wouldn't that lead to the same answer no matter what the quarter?
I'm sure that I have done it wrong. Please help with comments. There is not a worked-out example problem with a solution like this in section 6.2
confidence assessment: 0
.................................................
......!!!!!!!!...................................
00:34:52
What expression did you evaluate to obtain the year's total daily receipts and was your result?
......!!!!!!!!...................................
RESPONSE -->
t= 365
410.5(365^2)e^(-365/30) + 25,000
= 25,284
confidence assessment: 1
t starts at the beginning of the year, and does not represent a time interval but the running time, which in this case would be the day of the year. Thus t runs from 0 to 360.
You have to integrate the function to get the total revenue during the period. In this case you would integrate the function from t = 270 to t = 360.
.................................................
......!!!!!!!!...................................
**** Problem 6.2.62 c = 5000 + 25 t e^(-t/10), r=6%, t1=10 yr, find present value
00:50:50
What expression do you integrate to obtain the present value of the income function and what is your result?
......!!!!!!!!...................................
RESPONSE -->
5000 + 25te^(-.06*-.1t)dt
You tried to put too much into the exponent of your exponential. There are two different factors involving exponential functions, and they have very different meanings. See the given solution below:
u= 25t
dv= e^-.006t dt
v= -0.6e^-.006t
du= 25
25te^0.006t dt= 25t + 15e^-.006t
confidence assessment: 0
At 6% continuous interest the factor by which you multiply your principal to find its value after t years is e^(.06 t). So to find the principal you need now to end up with a certain amount after t years, you divide that amount by e^(.06 t), which is the same as multiplying it by e^(-.06 t).
*&*& The income during a time interval `dt is ( 5000 + 25 t e^(-t/10) ) `dt. To get this amount after t years we would have to invest ( 5000 + 25 t e^(-t/10) ) `dt * e^(-.06 t).
Integrating this expression from t = 0 to t = 10 we obtain
int(( 5000 + 25 t e^(-t/10) ) * e^(-.06 t) dt , t from 0 to 10).
Our result is $38,063.
Note that the entire income stream gives us int(( 5000 + 25 t e^(-t/10) ) dt , t from 0 to 10) = $50,660 over the 10-year period.
The meaning of our solution is that an investment of $38,063 for the 10-year period at 6% continuously compounded interest would yield $50,660 at the end of the period. So $38,063 is the present value of the income stream. *&*&
.................................................
......!!!!!!!!...................................
00:51:23
Explain the meaning of the expression you integrated--why does this function give the present value?
......!!!!!!!!...................................
RESPONSE -->
I kind of became lost in my problem, so I wanted to check with you if I was on the right track before continuing further.
confidence assessment: 0
.................................................
......!!!!!!!!...................................
00:54:01
Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE -->
I had a lot of problems with this assignment. Actually, with the past two assignments, really. But I was wondering if you could comment and make suggestions on my work on Assignment 16 and then I could re-work the problems and submit it again. Are these assignments graded for accuracy? If so I may be making a big mistake by submitting a barely-completed assignment. But since I am unsure I would like to give you what I have and get feedback before continuing. I was just very confused as these problems did not seem like the ones solved within the chapter, so it made it hard for me to follow. Please help. Thanks
confidence assessment: 3
.................................................
Feedback is the main purpose of the assignments; don't worry about the grading. You're making an excellent and mostly successful effort, and your homework grade will be fine.
I've been debating for some time about whether to use the question-answer format for the last two chapters. Students tend to prefer it, and it would certainly save me a lot of work in the long run. On the other hand, once a student has been successful to this point, there is some indication that they tend to be more successful if they submit the problems and self-critique the posted response. As a result I haven't made the change.
See my notes, which should answer many of your questions. You are required to do so, but you are encouraged to submit a copy of this document along with self-critiques, questions, comments etc. Insertions should be marked with &&&& before and after so I can easily distinguish them from your previous work and the comments I have inserted here.
Of courses works better when I'm able to respond promptly. Having been out of town for three days, the timing wasn't particularly good for you.