query 8

query 8

You're close on the worksheet/class notes questions.

Need to dig into that section of the text a little more. Be sure you know the laws of logs and the laws of exponents. You'll be using them.

I'll be glad to answer questions.

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15:55:58 What functions f(z) and g(t) express the function 2^(3t-5) as a composite f(g(t))?

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RESPONSE --> if 2^(3t-5) = f(g(t)) then f(z) =2^x and g(t) =3t-5

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15:56:30 ** g(t) = 3t - 5, f(z) = 2^z.

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RESPONSE --> ok

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16:03:24 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 --> not sure what you mean, but i think i take point 1, and add the interval, ot distance to it to get point 2.

Right idea througout. In your terms, if the slope is (y2-y1)/(x2-x1):

Using your notation the known point would be (x1, y1) instead of (x0, y0). That's OK. We can think in terms of (x1, y1) instead. x2 would be x1 + the increment, i.e., x2 = x1 + `dx. The goal is to find y2, which you aren't assumed to know. What you are assumed to know is the function f(x), which you evaluate at x = x1 to get dy/dx (that's what it means to say that dy/dt = f(x) ).

So dy/dx = f(x1) is the slope at x = x1. Again, if you have the function f you can plug in x1 and get this quantity.

Now, (y2-y1)/(x2-x1) is the slope between the x=x1 point and the x = x2 point of the graph.

If the interval `dx = x2 - x1 is small and f(x) is reasonably well-behaved, f(x) isn't gonna change much over that interval, so the slope won't change much.

As a result dy/dx and (y2-y1)/(x2-x1) will be pretty nearly the same.

So where are we? We know dy/dx, we know x1 and x2, and we assume we know y1. And we know that dy/dx = (y2-y1)/(x2-x1), approximately. So if we solve this equation for y2, we get a pretty good approximation.

This process is described in some detail in the class notes.

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16:09:34 ** You start with a point (x0, y0) on the y vs. x graph.

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RESPONSE --> i confused. increment isn't the difference between 2 points? ???if i had a better description like a graph, i would know what you want...

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16:24:57 explain why a numerical solution to differential equation is only an approximate solution in most cases

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RESPONSE --> because the # at that point is usually a decimal figure, and gets rounded.

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16:26:50 ** 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.

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RESPONSE --> oh. i knew that but i thought we were talking about the same point. not a diffrent one in the same equation.

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16:57:03 query Problem 1.4.8 Solve 2 * 5^x = 11 * 7^x

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RESPONSE --> 2 * 5^x = 11 * 7^x ??? i could graph it, but i give.

Gotta use logs to get the variable out of the exponent so you can work on it.

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17:02:50 ** Taking logs of both sides and applying the laws of logarithms we get

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RESPONSE --> ok.

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17:11:37 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 ln(e^A)+b^3 i give up.

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17:13:40 ** 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

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RESPONSE --> ok

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17:16:42 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 --> confused again going to study that more. and try it again tomorrow.

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