assn 8 query

course Mth173

???????k?????assignment #008

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

?????D????????Calculus I

10-06-2007

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16:36:02

Query class notes #09

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RESPONSE -->

okay

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16:38:44

What functions f(z) and g(t) express the function 2^(3t-5) as a composite f(g(t))?

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RESPONSE -->

Breaking apart the functions we get:

f(g(t)) = 2^(3(t)-5)

f(z) = 2^(z)

g(t) = 3(t)-5

z= g(t)

f(g(t)) = 2^(3(t)-5)

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16:39:07

** g(t) = 3t - 5, f(z) = 2^z.

The z is a 'dummy' variable; when we find f(g(t)), the g(t) is substituted for z and we get f(g(t)) = 2^(g(t)) = 2^(3t-5). **

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RESPONSE -->

okay

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16:45:05

describe in some detail how we can numerically solve a differential equation dy /dt = f(x), given a point (x0, y0) on its solution curve, an interval (x0, xf) and an increment `dx

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RESPONSE -->

the x axis is the same as the t axis so when given a point on the curve we can plot the point, and given an intreval of (x0, xf) we get 'dx, using the slope formula 'dy/'dx = f(x) we can find another point.

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16:47:30

** You start with a point (x0, y0) on the y vs. x graph.

You evaluate the function y' for x=x0 and y=x0, which gives you a slope for your y vs. x graph.

Using the chosen increment `dx you then multiply `dx by the slope to determine how much change there will be in y, and you use this information to obtain a new approximate point on your y vs. x graph by adding the change in y to y0 and the change in x to x0.

You then repeat the process starting with the new point. **

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RESPONSE -->

My explaination was a little confusing to me and you too I'm sure but I was trying to say what is said in the answer window, I think that I get the concept, although I find it a little hard to explain without some concrete terms.

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16:50:25

explain why a numerical solution to differential equation is only an approximate solution in most cases

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RESPONSE -->

Because as 'dt gets larger and the rate changes the differential equation doesn't account for it, the answer will be thrown off, as long as 'dt is small then the answer for y won't be off much, and so the differential is only an aproximation.

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16:51:15

** You assume the slope at the initial point, but that slope generally changes at least a bit by the time you get to the second point. So you are assuming a constant slope when the slope actually changes.

If your interval is small enough the change in slope will have a small effect. **

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RESPONSE -->

the rate change I was talking about would equate to the slope of the line/function that is stated in the answer window.

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16:59:06

query Problem 1.4.8 Solve 2 * 5^x = 11 * 7^x

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RESPONSE -->

2*5^x = 11*7^x

2= 11 *(7/5)^x

2/11 = 7/5 ^x

x = log(2/11) / log(7/5)

x= -5.067

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17:00:46

** Taking logs of both sides and applying the laws of logarithms we get

log 2 + x log 5 = log 11 + x log 7. Rearranging we obtain

x log 5 - x log 7 = log 11 - log 2 so that

x ( log 5 - log 7) = log 11 - log 2 and

x = (log 11 - log 2) / (log 5 - log 7).

This can be approximated as -5.07. ** DER

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RESPONSE -->

I rearanged before taking my logs at the end, is this okay? I still ended up with the same answer this time, or is it just easier to take the log before rearanging?

You should be able to get this expression into the form

x = (log 11 - log 2) / (log 5 - log 7),

which has the precision advantage of expressing the answer in terms of logs of whole numbers.

If the equation was a * b^x = c * d^x, you would want to be able to get the equation into this form.

However I believe you could easily do this, and for the current problem your solution is fine.

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17:09:49

Problem 1.4.28 simplify 2 ln(e^A) + 3 ln(B^e)

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RESPONSE -->

2ln(e^A) + 3 ln(B^e)

2A +3ln(B^e)

because ln(e^A) is comprised of inverse functions it cancells out to A

I can't figure out what ln(B^e) would simplify too.

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17:12:31

** Starting with 2 ln (e^A) + 3 ln (e^B) we use the fact that the natural log and exponential functions are inverses, expressed by the law of logarithms ln(e^x) = x, to get

2 * A + 3 * B or just

2A + 3B. **

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RESPONSE -->

typo of ...3ln(B^e) instead of ...3ln(e^B). But as stated in the answer window and my answer they are inverse functions and would cancell out to just B.

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17:24:29

query Problem 1.4.31 (was 1.7.26) P=174 * .9^twhat is the function when converted to exponential form P = P0 e^(kt)?

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RESPONSE -->

P(t) = P0 b^t

P = 174 * .9^t

converting to e form

b^t = e^(kt)

t ln(b) = kt

ln(b) = k

ln(.9) = -.1054 = k

P = P0 e^((-.1054)t)

or

P= P0 e^((ln(.9))(t))

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17:24:56

** 174 * .9^t = 174 * e^(kt) if

e^(kt) = .9^t, which is the case if

e^k = .9. Taking the natural log of both sides we get

ln(e^k) = ln(.9) so that

k = ln(.9) = -.105 approx.

So the function is

P = 174 e^(-.105 t). **

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RESPONSE -->

okay

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17:34:56

Query problem 1.4.44 (was 1.6.24) population function for exponential growth if 40 meg in 1980 and 56 meg in 1990; doubling timegive a population function and the doubling time

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RESPONSE -->

y = y0b^t

56 = (40) b^10

1.4 = b^10

10th root (1.4) =b

b= 1.03

y=y0(1.03)^t

doubling time

(40 *2) = (40)(1.03^t)

80 = 40 (1.03^t)

2=(1.03^t)

log2/log1.03 = t

t=20.6 doubling time

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17:37:55

** The population function is exponential and has form P = P0 * e^(kt).

Let t be the time since 1980 and population be in millions. Then we have

40 = P0 e^(k * 0) and

56 = P0 e^(k * 10).

From the first equation we get

40 = P0 so the second equation becomes

56 = 40 * e^(10 k) or

e^(10 k) = 56 / 40 = 1.4. Taking logs we get

10 k = ln(1.4) so that

k = ln(1.4) / 10 = .0336, approx.

Thus our equation is

P = 40 e^(.0336 t).

This doubles when

e^(.0336 t) = 2. Taking the ln of both sides we have

.0336 t = ln(2) so that

t = ln(2) / .0336 = 20.6, approx.

Doubling time is about 20.6 years. **

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RESPONSE -->

i used the exponential function we learned earlier for compound intrest and got the same answer for both, just a different method, would you rather me use the one you have in the answer box instead?

The only difference is that you used the A b^t form of an exponential and the given solution used A e^(kt). You understand the relationship between these forms and could have used either form had it been specified. So your solution is fine.

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17:45:20

query Problem 1.4.50 (was 1.7.42 but changed) time to get to 10% of strontium 90 if half-life 29 yrs

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RESPONSE -->

50=100b^29

.5=b^29

29th root of .5= b

b=.9764

%left = 100(.9764^t)

10 = 100(.9764 ^t)

.1 = .9764 ^t

t = log.1/log.9764

t= 96.411 year to 10%

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17:45:41

** If the model is Q = Q0 e^(kt) then since half-life is 29 years we know that

e^(k * 29) = 1/2. Taking ln of both sides

29 k = ln(1/2) so that

k = ln(1/2) / 29 = -.0239.

So the model is

Q = Q0 * e^(-.0239 t).

Now if we want to get 10% of the original quantity we have Q = Q0 / 10 so that

Q0 / 10 = Q0 e^(-.0239 t) and

e^(-.0239 t) = 1/10.

Taking logs of both sides and solving for t we get

t = ln(1/10) / -.0239 = 96.3 approx. **

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RESPONSE -->

okay

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17:48:15

Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

is there an advantage to using the p = po e^(kt) vs the p=po b^t method other than being able to use the ln instead of b'root?

It's easiest to do calculus with the e^(kt) form so you should know that.

It's easiest to understand the e^(b t) form so you should know that.

When dealing with half-lives and doubling times there are advantages to the 2^(kt) form so you should know that.

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17:48:40

Problem 1.4.31 P=174 * .9^t

What is the function when converted to exponential form P = P0 e^(kt)?

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RESPONSE -->

P = P0 b^t

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17:49:29

If P=174(.9)^t is put in the form P = P0 e^(kt) then P0 = 174 and e^(k t) = .9.t so that e^k = .9.

It follows that

e^k = .9 so that

ln(e^k) = ln(.9) or

k = ln(.9) = .105.

The function is therefore

P=174 e^-(.105 t).

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RESPONSE -->

okay

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17:50:35

problem 1.4.44 population function for exponential growth.

If 40 meg in 1980 and 56 meg in 1990 give the equation and find the doubling time

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RESPONSE -->

answered a few questions ago

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17:58:50

P=Po b^t is the form of the function. Initial quantity is 40 * 10^6 so Po = 40 * 10^6. Substituting Po = 40 * 10^6:

P=40*10^6 b^t. At t = 10 we have P = 56 * 10^6 so we substitute for P and t:

56*10^6=40*10^6 b^10. We solve for b:

1.4=b^10

b=1.03

P=40*10^6(1.03)^t is our function.

doubling time occurs when the 40^10^6 grows to 80*10^6:

80*10^6=40*10^6(1.03)^t

2=1.03^t

log2=tlog1.03

t=23.4498

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RESPONSE -->

I see you, want me to use the other way now, I have a way of reversing instructions. For the population doubling.

function: P0e^(.023t)

doubling time:

80 = 40 e^(.023t)

2=e^(.023t)

.023t =ln2

t=23.45

(i used the saved k from my calculator instead of .023 as not using it would have thrown of my answer .023 is just a representation)

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17:58:55

10:32:42

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RESPONSE -->

okay

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18:05:30

Problem 1.7.42 percent of original strontium -- 90 after century; 2.47% annual decay rate.

What percent of the original strontium -- 90 would remain after a century?

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RESPONSE -->

2.47% decay then b= -.0247

100(-.0247^(100)) = 0% left

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18:05:38

10:34:19

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RESPONSE -->

okay

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18:05:44

I did not understand this problem, but this is what I have:

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RESPONSE -->

okay

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18:05:52

Q=Qoe^(-kt)

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RESPONSE -->

okay

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18:06:01

Q=Qoe^-.0247t

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RESPONSE -->

okay

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18:06:06

That`s all that I can do with that problem at this point

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RESPONSE -->

okay

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18:07:22

** The model is Q(t) = Qo * e^(kt).

You know that you lost .0247 of the quantity in a year. Thus

Q(1) = Qo e^(k* 1) = (1 - .0247) Qo.

So Qo e^(k* 1) = (1 - .0247) Qo.

This equation is easily solved for k.

Then you substitute t = 100 back into the function, using your newly found k. **

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RESPONSE -->

I think I forgot to sub in 1 to find my b and thus threw my answer off

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&#

This looks good. See my notes. Let me know if you have any questions. &#