query 8

query 8

Not bad. I've given you lots of notes, since you asked a lot of good questions. Hopefully they'll clarify the point of a lot of this. Exponential functions and natural logs are natural for engineering and for calculus applications, as I explain a little further in my notes.

â{a\úlXipx~ assignment #008 008. `query 8 Calculus I 05-18-2005

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16:54:08 What functions f(z) and g(t) express the function 2^(3t-5) as a composite f(g(t))?

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RESPONSE --> did it.

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16:54:16 ** 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 --> yup

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16:56:00 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 --> know it.

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16:56:29 ** 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 --> yup

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16:58:16 explain why a numerical solution to differential equation is only an approximate solution in most cases

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RESPONSE --> it is only approximate if the point picked is close to the first one. how close depends on the equation.

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16:58:42 ** 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 --> ok.

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17:26:23 query Problem 1.4.8 Solve 2 * 5^x = 11 * 7^x

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RESPONSE --> 2 * 5^x = 11 * 7^x 2 * log{5}x = 11 * log{7}x i understand that log is automatically base or root number 10 unless otherwise specified(like you did with the {} stuff). thats how chemistry calculations in scientific notation work. ok. as on the notes where you say that log{base2}x = 2^x

If I said that let me know where.

The correct statement would be

log{base2}(2^x) = x.

ok. so: how the hell can: log 2 + x log 5 = log 11 + x log 7 be the answer? a)x is on the wrong side of the log

Gotta know and apply the laws of logarithms:

log(2 * 5^x) = log(2) + log(5^x) because log(ab) = log(a) + log(b). log(5^x) = x log(5) because log(a^b) = b log(a).

So if you take the log of both sides the left-hand side becomes

log 2 + x log 5.

The log of the right-hand side follows by a very similar process.

b)since when does the base become the power c) right now the inverse of (not being confused)=me.

Takes a little getting used to; the laws of logarithms and the laws of exponents are inverse and you need to be sure you know both. Once you've got them straight this type of problem isn't bad.

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17:26:35 ** 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 --> uuuhhh...

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17:42:55 Problem 1.4.28 simplify 2 ln(e^A) + 3 ln(B^e)

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RESPONSE --> 2 ln(e^A) + 3 ln(B^e) 2* log of {base natural #} A + 3* log of {base B }natural # so 2A + 3* log of {base B }natural #. in your anwser, it says e^b but here the problem says b^e -??? i have to be missing something...

Looks like you've just about got it. See below.

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17:44:06 ** 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 --> h-e-l-p

The solution is correct for the problem it solves, but it did get the second expression backwards. So here is the correct solution:

Starting with 2 ln(e^A) + 3 ln(B^e) we use the fact that ln(a^b) = b ln a to get

2 A ln(e) + 3 e ln(B). Since ln(e) = 1 we have 2 A + 3 e ln(B).

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17:54:30 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 --> ok where is e in the first equation? don't understand what you want me to do?

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18:23:47 ** 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 --> ok. i see what i was supposed to do.

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18:51:12 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 --> p = 40(***^t) 40% rise in 10 yrs. 4% rise every year.

This doesn't follow. A 4% rise compounded every year gives you significantly more than a 40% rise in 10 years--this is because 1.04^10 > 1.4.

p = 40(1.04^t) 80 = 40(1.04^t) 2 = 1.04^t (log 2) / (log {base 1.04} ) = t

Not bad except for the error I pointed out above.

That denominator would just be log(1.04). You can take your calculator and do log(2) / log(1.04). Just be careful you don't accidentally do log(2/1.04), which is very different.

???

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19:02:55 ** 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 --> and population be in millions. where did the millions part come from? if you said that instead of"meg" i would know.

Don't try to read the problems by the abbreviated statements I use here. These statements are just guidelines for you, abbreviated references for me. Use the statements in the text.

why is e used (of course i mean instead of reg. logs & bases?)

This could have been done in base 10, or any other base you might want to choose. You would get the same answer.

However the most convenient choices are base e and base 10, because calculators have buttons for both of these bases and the corresponding logs.

e is the natural base to use for exponential growth; as shown in the modeling assignment, exponential growth continuously compounded at rate r gives you the function e^(rt). Most exponential growth in nature and in engineering is continuous in nature, so this is the function we want to use.

Also as you'll see soon enough, calculus with base e is much, much easier than with any other base.

stuff like this is making it more difficult for me.

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19:16:31 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 --> i'll try this: i do it using my logic and you translate into log stuff. 10% or .1 of 1 so H.L.1=.5 H.L.2=.25 H.L.3=.125 H.L.4=.0625 so almost 4 half lives are required. 29x4=116 so almost 116 years. mabye i can understand better like this, and relate it.

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19:17:44 ** 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 --> so e is for-?-?-?

As with my previous note, e is the natural base to use for continuous exponential growth, and e is the easiest base to use for calculus.

The latter is because the derivative function for e^t is just good ol' e^t. The natural exponential function has a growth rate equal to itself.

For comparison the derivative of 10^t = ln(10) * 10^t. Not bad, but e^t is a whole lot simpler and easier to understand.

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19:19:43 Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> i know i can get this... guess i'll do it over again....

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19:21:26 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 --> -

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