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Old 12-02-2004, 10:46 AM   #1
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Calc problems, need help please

Here are some problems im having trouble with. If anyone could help, it would be greatly appreciated.

1)For t years since 1980, the population, P, of mexico (in millions), is given by:
P = 67.38(1.026)^t

- What was the average population of mexico between 1980 and 1990?

2)Calculate the present value of a continuous revenue stream of $1000 per year for 5 years at an intrest rate of 9% per year compunded continuously.

3)figure a gives the relative growth rate of popuation.
a)on what interval is the population increasing? by what percentage does the population increase during this interval?
b)on what interval is the population decreasing? By what percentage does the opoulatin decrease during this interval?
c)by what percentage does the population change during the 15 year period shown?

4)figure b is the graph of the annual yeild, y(t), in bushels per year, for an orchard t years after planting. THe trees take about 10 years to get established, but for the next 20 years they give a substantial yield. after about 30 years, however, age and disease start to take their toll, and the annual yield falls off.
a)represent on a sketch of the figure the total yield, f(m), up to M years, with o<= m <= 60. write an expression for f(m) in terms of y(t).
b)sketch a graph of f(m) against m for 0 <= M <= 60
c)write an expression for the average annual yield, a(m), up to m years.
d)when should the orchard be cud down and replanted? assume that we want to maximize avverage revenue per year and that fruit prices remain constant, so that this is achieved by maximizing average annual yield. use the graph of y(t) to estimate th etime at which the average annual yield is a maximum. explain your answer geometrically and symbolically.
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Old 12-02-2004, 10:56 AM   #2
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Old 12-02-2004, 11:12 AM   #3
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Hm...I'm in cal...and don't understand a bit of that. Thank God I'm done with advanced math after monday night.
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Old 12-02-2004, 01:32 PM   #4
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for average population, I got 76.8.... I'm assuming that's in millions
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Old 12-02-2004, 01:40 PM   #5
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dude, just so you know, this is more of a differential equations thing than a Calc. thing... but yeah, it does start in Calc.
for problem 2, I just assumed that you had $1000 initially, and then after every year passed, you added $1000 to it, instead of just averaging adding $1000 added each year (like about $900 a month or something). And I get exactly $6603
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Old 12-02-2004, 01:47 PM   #6
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C is kind of just a guess... but I still like my answers
0-10 increasing and by 10%
10-15 decreasing and by 2.5%

you can do this by taking the equation y=(-.02/10)x+.02 and taking the integral from 0 to 10 with respect to x, that gives you .1, which is 10%
for the rest of it, you can just change your limits of integration to 10 to 15, and you get -.025, which is 2.5%
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Old 12-02-2004, 01:49 PM   #7
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Old 12-02-2004, 02:11 PM   #8
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that's an idea for the last one.... I got bored. I have a suggestion for you. Get a graphing calculator, and plug " -(.75*(x-33))^2+500" in and see what you get, it's your graph you put on here... take the integral of that, you get,
-.1875x^3+18.5625x^2-112.563x.... that should help you some, there's a +C on the end of that, but yeah, I don't care. Lastly, in the pic I attached, that's what the sketch of what your yield would be looks like. The flattened A cup that's facing up, that's your original graph, and the two red lines are the two lines on your original graph that are on each side of the maximum value, just to give you an idea.... but now I'm bored, and going to do something else, enjoy
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Old 12-02-2004, 05:52 PM   #9
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i hate math
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Old 12-02-2004, 05:57 PM   #10
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I don't.... I'm in Calc. III got an A in it too
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Old 12-02-2004, 07:51 PM   #11
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1)For t years since 1980, the population, P, of mexico (in millions), is given by:
P = 67.38(1.026)^t

- What was the average population of mexico between 1980 and 1990?

t = 10 (number of years)

P = 67.38(1.206)^10

AVG = P/10 or population per year growth.

2)Calculate the present value of a continuous revenue stream of $1000 per year for 5 years at an intrest rate of 9% per year compunded continuously.

Formula

S Final value of investment
P Initial value of investment
r Annual percentage rate (APR)
t Number of years

Value of investment after t years:

S = Pert

Where e is the transcendental number 2.7182818285...

Notice that the output, S, is an exponential function of t. That is, if we consider the final value of the investment as a function of the length of time for the investment, then t, the length of time for the investment, is in the exponent position, and this makes S an exponential function

so you need to do the following:
each year the amount varies so calculate the first year, then do the same for the amount it has for each additional year.

S = 1000e(.09)(1) + 2000(.09)(1) + 3000(.09)(1) + 4000(.09)(1) + 5000(.09)(1)

Im at work fixing to get off ill mess with the others later, make sure all the graphs are attatched.
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Old 12-02-2004, 07:58 PM   #12
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Quote:
Originally Posted by jimmy_beaner
I don't.... I'm in Calc. III got an A in it too
I used to hate math but after having: 100, 110, 112, 113, 115, 116, 125, 126, 245, 345, 431, and im sure others.... ive learned to at least somewhat enjoy it. I did love cal II I pulled a 98 average in there and it was NOT a cake walk class.... I had to do tons of hw and crap
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Old 12-02-2004, 08:16 PM   #13
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Quote:
Originally Posted by SpectorV
1)For t years since 1980, the population, P, of mexico (in millions), is given by:
P = 67.38(1.026)^t

- What was the average population of mexico between 1980 and 1990?

t = 10 (number of years)

P = 67.38(1.206)^10

AVG = P/10 or population per year growth..
nice try, but not that easy. You'd be right if the equation were a linear equation. Just like if you have the average value of a sin wave over the first 180 degrees.... it's not sin(180)/180.... that's not how it works. You have to take the integral of the function given with the limits of integration from 0 to 10, and then divide by 10.

Quote:
Originally Posted by SpectorV
2)Calculate the present value of a continuous revenue stream of $1000 per year for 5 years at an intrest rate of 9% per year compunded continuously.

Formula

S Final value of investment
P Initial value of investment
r Annual percentage rate (APR)
t Number of years

Value of investment after t years:

S = Pert

Where e is the transcendental number 2.7182818285...

Notice that the output, S, is an exponential function of t. That is, if we consider the final value of the investment as a function of the length of time for the investment, then t, the length of time for the investment, is in the exponent position, and this makes S an exponential function

so you need to do the following:
each year the amount varies so calculate the first year, then do the same for the amount it has for each additional year.

S = 1000e(.09)(1) + 2000(.09)(1) + 3000(.09)(1) + 4000(.09)(1) + 5000(.09)(1)

Im at work fixing to get off ill mess with the others later, make sure all the graphs are attatched.
ummm, sorry, but I don't know that your last equation was proved?... I took what you started with, 1000e^(.09*1)... and then added 1000 to that answer, and took it times e^.09... and so on.
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Old 12-02-2004, 08:18 PM   #14
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I started out at Calc. I and Calc. based Honors Physics.... I loved them... then, I transferred, and my current college laughed in my face when I asked if there was something there I could take.... "Calc. III.... but no one hardly ever takes that"
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Old 12-02-2004, 08:29 PM   #15
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im not going to read thru this whole thread, but PM me or email me if you still need help

redrum86gt@gmail.com
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Old 12-02-2004, 09:04 PM   #16
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how 'bout you just post up some answers redrum.... we've got two listings on here with answers already.... humor us
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Old 12-03-2004, 06:39 AM   #17
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thanks for the help guys, i worked through and got around the same answers you did for the first three, the last one im still working on, im glad this is my last math class
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