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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). **......!!!!!!!!...................................
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. **......!!!!!!!!...................................
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. **......!!!!!!!!...................................
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
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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......!!!!!!!!...................................
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...
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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. **......!!!!!!!!...................................
RESPONSE --> h-e-l-p
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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). **......!!!!!!!!...................................
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.
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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. **......!!!!!!!!...................................
RESPONSE --> and population be in millions. where did the millions part come from? if you said that instead of"meg" i would know.
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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. **......!!!!!!!!...................................
RESPONSE --> so e is for-?-?-?
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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)?......!!!!!!!!...................................
RESPONSE --> -
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