query8

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course Mth 173

217 9:41 PM

008. `query 8

Question `q Query class notes #09

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

Your solution

g(t)= 3t-5

f(z)=2^z

Confidence Assessment 3

Given Solution

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).

Self-critique (if necessary) OK

Self-critique Rating OK

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

Solution 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.

STUDENT QUESTION

Could you give me, when you critique this assignment, a problem that I could work out to see how this is actually done

It seems a bit confusing to me still.

INSTRUCTOR RESPONSE The Class Notes give additional examples.

If dydt = t^2 + y, and we know that y(2) = 6, then we can approximate y(2.1) as follows

y(2) = 6 so when t = 2 we have

dydt = 2^2 + 6 = 10.

Therefore if `dt = .1, we obtain the approximation `dy = (dydt) `dt = 10 .1 = 1.

This approximation assumes that the rate of change dydt remains constant between y = 2 and y = 2.1. This isn't completely accurate, but since the interval is small the error is also small.

We conclude that y(2.1) = y(2) + `dy = 6 + 1 = 7.

Now we approximate y(2.2).

We know that y(2.1) is about 7, so when t = 2.1 we have

dydt = 2.1^2 + 7 = 11.41.

So if `dt = .1, we have the approximation `dy = 11.41 .1 = 1.141.

Once again this is an approximation which assumes an unchanging value of dydt for the entire interval. Again the error is small, but of course it is added to (and in part based on) the error in the preceding step.

Our approximation is thus y(2.2) = y(2.1) + `dy = 7 + 1.141 = 8.141.

The process could continue. For example, we could do 7 more steps and obtain an approximation to y(3). This approximation would accumulate errors at every step, so the accuracy would decrease with every step.

We could improve our accuracy by using a smaller interval. For example we could assume intervals of .01 rather than .1.

This would require10 times as many steps, and would accumulate 10 times as many errors; however the errors would tend to be much smaller, and the total error in approximating, say, y(3) using intervals of .01 would be smaller than the total error that would result from intervals of .1.

Self-critique (if necessary) OK

Self-critique Rating OK

Question

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

Your solution

it assumes a same rate of change, for a different point, when generally the rate of change will be a little different.

Confidence Assessment 3

Given Solution

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.

Self-critique (if necessary) OK

Self-critique RatingOK

Question `q query Problem 1.4.10 Solve 4 * 3^x = 7 * 5^x

Your solution

I am not sure how to do this, I know we use logarithms for when x is in the exponent, . The only way I know to do a logarithm is in the form A^N=X is logA(x)=N. So it just seems weird.

I guess we could consider it just equal to take the logarithm of both sides so I guess im using the log(xy)= log(x) + log(y)

log(4 * 3^x) = log(7 * 5^x)

log(4) + log(3^x) = log(7) + log(5^x)

now for log(x^n) it is equal to nlog(x) so

log(4) + xlog(3) = log(7) + xlog(5)

we have to normal logs and two xlogs, so put common relationship together i guess, don't know the proper words to say.

xlog(3)-xlog(5)=log(7)-log(4)

take out the x

x(log(3)-log(5))=log(7)-log(4)

I guess now I have to take the log of those numbers

????on a major quiz or test how will I be able to do that, my normal calculator cannot perform these operations, or the other ""basic"" calculator I have performs them wrong.

x * -0.2218487496 = 0.2430380487

x =-1.095512367

Confidence Assessment 1

Given Solution

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

log 4 + x log 3 = log 7 + x log 5. Rearranging we obtain

x log 5 - x log 3 = log 4 - log 7 so that

x ( log 5 - log 3) = log 4 - log 7 and

x = (log 4 - log 7) (log 5 - log 3).

This can be approximated as -1.095. DER

Self-critique (if necessary) OK

Self-critique RatingOK

Question `q Problem 1.4.6 simplify 2 ln(e^A) + 3 ln(B^e)

Your solution

I am unsure how to do this problem.

Confidence Assessment

Given Solution

Starting with 2 ln (e^A) + 3 ln (B^e) we first 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 + e ln(B^e).

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This should be 2 A + 3 e ln(B)

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We then use the fact that ln(x^y) = y ln(x), applied to the expression ln(B^e), to get

2 A + 3 e ln(B).

Self-critique (if necessary)

What happened to the 3 in 3ln(b^e) for the first step?

i thought ln(e^x) would be xln(e). OHH okay that the same as LOGA(x)=1 as long as x=a

so 2ln(e^x) is same as 2*x * 1 or 2x

so 2A for this equation

so 2A + 3 ln(B^e)

doing the same process we get

2A+ 3eln(B)

Self-critique Rating 3

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There was an error in one step of the given solution. Check my note.

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Question `q query Problem 1.4.27 5th; 1.4.26 4th; 1.4.31 (was 1.7.26) P=174 .9^t

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

Your solution

wel if P=174 then

174=P0 e^kt

that is confusing, the functino is P= but our value says P=174, yet we put it in P0, that tricked me for a moment

P=174e^kt

not sure what is next

the rest of the equation i have to fill is

e^kt and the rest of the information I hae left is .9^t. i think .9 is relative to my constant k, so I should have both variables to check and make them equal

so

e^kt = .9^t

taking the square root of t on both sides

e^k=.9

this has taken me a while to think of what to do, I am not used to dealing with e, in fact half the time I think of it as an unknown constant, but i keep forgetting it does have a value. the thing i can think of is ln is the inverse so

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

at first I didnt see a point in doing this, but then remembered the logarithm rule

kln(e)=ln(.9)

k * 1 = ln(.9)

k=-0.1053605157

going back to the equation

P=174e^(-0.1053605157t)

Confidence Assessment 1

Given Solution

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).

Self-critique (if necessary) OK

I understand how to get to the other, but I don't understand the importane of the different ways the information was set up, and the importance of being able to switch between the two equations..I also don't understand the importance of e. I remember my teacher in how school talking about it ALL the time, but i just never caught on to the importance of e. Also I don't know the difference, but I took Applied Calculus in high school.

Self-critique Rating OK

@&

No criticism of your teacher, but 2/3 of the applied calculus courses in high school usually lack the combination of background, time, study skills and aptitude to learn the subject.

The courses are overenrolled, so the teacher is stuck having to pass a bunch of students who shouldn't pass the course.

As a result nobody learns much.

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Modeling Project 3 shows how e is the limiting value of (1 + 1/n)^n, approached there as the limiting interest rate on 100% interest, as the number of compoundings approaches infinity.

So e arises naturally from an easily understood situation.

If interest rate r is compounded n times, the equivalent interest rate is (1 + r / n)^n, which can also be written (1 + 1/n)^(r * n). This expression approaches e^r, rather than e.

There is more, but this is the heart of the relationship.

Also the derivative of e^x is just e^x, whereas the derivative of a^x is ln(a) * a^x. It should be clear than the rule for e^x is simpler. It is particularly easy to do calculus with exponential functions using base e.

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Though the e representation is the easiest with which to do calculus, the form b^x is often easier to associate with a given application. The 2^x form also arises naturally. The bottom line is that to do calculus with these functions, it is often easiest to convert them to 'e' notation.

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Question `q Problem 1.4.27 was 1.4.26 P=174 .9^t

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

Your solution

this is a repeat of the previous question

Confidence Assessment

Given Solution

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).

Self-critique (if necessary)

Self-critique Rating

Question `q problem 1.4.40 was 1.4.32 population function for exponential growth.

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

Your solution

my problem for chater 1 section 3 problem #32 in the 6th edition calculus is this:

The population of a region is growing exponentially. There were 40,000,000 people in 2000 (t = 0) and 48,000,000 in 2010. Fina an expression for the population at any time t, in years. what population would you predict for the year 2020? What is the doubling time?

P=P0(1+r)^t

not sure, but in 10 years that was an 8 million increase. if i figure that out per year though...thats an average...does that work for exponentials that are compound? wait, i could probably just plug it in for the amount in 2010

48,000,000=40,000,000(1+r)^10

mathematically not sure how to solve that, but I can plug in numbers,

r=.01 = a little under 45 million

r=.02 is over, its over 48.7million

r=.015 is 46.4 million

r=.017 is 47.3 million

r=.018 is 47.8 million

r=.019 is 48.2

so between .018 and .019

r=.0187 is 48.1 million

r=.0186 is 48.09 million

r=.0184 is 48.00029 million

r=.0183 is 47.95 million

so between .0183 and .0184

r=.01835 is 47.97 million

r=.01838 is 47.9908 million

r=.01839 us 47.99558 million

so between .01839 and .0184

r=.018395 is 47.9979 million

r=.018398 is 47.99935 million

r= .018399 is 47.999822 million

so between .018399 and .0184

r= .0183993 is 47.999964 million

r=.0183994 is 48.0000112425 million

so between .0183883 and .0183994

r=.01838837 is 47.9999971 million

r=.01838838 is 48.0000018 million

so between .01838837 and .01838838

r=.018388376 is 47.999999930 million

r=.018388377 is 48.0000004 million

so between .018388376 and .018388377

r=.0183883761 is 47.9999999778 million

r=.0183883762 is 48.0000000249 million

so between .0183883761 and .0183883762

r=.01838837614 is 47.999999996689 million

r=.01838837615 is 48.0000000014 million

so between .01838837614 and .01838837615

r=.018388376147 is 47.9999999999885

r=.018388376148 is ...what a minute what does it matter this far for me to test after knowing over 8 digit places

im just going to g with r=.018388376148

so

P=40,000,000(1.018388376148)^20

about 57,587,558 in 2020

doubling time would be

80,000,000

so 80,000,000=40,000,000(1.018388376148)^t

2=(1.018388376148)^t

logarithms are used to get variables from exponents

log(2)= log((1.018388376148)^t)

log(2)=tlog(1.018388376148)

log(2)=t * 0.00791343365

0.30102999566= t * 0.00791343365

t=38.04037652606

Confidence Assessment 1

Given Solution

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=4010^6 b^t.

At t = 10 we have P = 56 10^6 so we substitute for P and t

5610^6=4010^6 b^10.

We solve for b

1.4=b^10

b=1.03

Substituting P0 and b into the original form

P=4010^6(1.03)^t is our function.

doubling time occurs when the 40^10^6 grows to 8010^6

8010^6=4010^6(1.03)^t

2=1.03^t

log2=tlog1.03

t=23.4498

103242

STUDENT QUESTION

How should I set up the initial problem

INSTRUCTOR RESPONSE

The form of the function is P = P0 b^t, where P0 and b are regarded as unknowns.

To solve for two unknowns you need two equations.

You have the values of P and t at two different points, which allows you to write down two equations, which you then solve simultaneously to get P0 and b.

The two equations would be

40 = P0 b^0 and

56 = P0 b^10.

b^0 = 1 so the first equation gives us P0 = 40.

This makes the second equation

56 = 40 b^10,

which is solved for b in the manner shown above.

Self-critique (if necessary)

i left mine as (1+r), but at ^0 would still be 1, which I didn't catch for the inital amount.

Self-critique Rating 2

Question `q Problem 1.4.50 was 1.4.45 percent of original strontium -- 90 after century; 2.47% annual decay rate.

I dont see this problem, I have 6th edition, and my chapter 1 section 4 problem 50 is size of exponentially growing bacteria colony. and my chapter 1 section 4 problem 45 is the populations of china and india.

P=P0 * .9753^t

thinking a century is a year

90=P0 * .9753^100

90=P0 * 0.082001618

P0=1097.539319

and half would be

548.7696595

548.7696595=1097.539319 * .9753^t

1/2= .9753^t

log(.5) = log(.9753^t)

log(.5)=tlog(.9753)

-0.301029996 = t * -0.010861775

t = 27.71462270

Your solution

Confidence Assessment 2

Given Solution

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

103419

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

Q=Qoe^(-kt)

Q=Qoe^-.0247t

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

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.

STUDENT QUESTION

How would Qo e^(k 1) = (1 - .0247) Qo

INSTRUCTOR RESPONSE

Q(1) = Q0 e^(k 1) is the amount after one year.

The sample loses 2.47% in a year, so the amount after a year is Q0 - .0247 Q0 = Q0 ( 1 - .0247).

Setting equal the two expressions for the amount after one year we get

Q0 e^(k 1) = Q0 (1 - .0247), so that

e^(k 1) = (1 - .0247).

STUDENT QUESTION

Like above I am having difficulty setting the problem up to solve, I get how to solve the actual problems in the book

assignment but I am having trouble with the word problems. How should I get the initial equation, I can solve the equation

its just getting it to start that is causing me difficulty

INSTRUCTOR RESPONSE

You need to know that this situation involves a constant decay rate, and that a constant growth or decay rate gives us the form

Q(t) = Q0 e^(kt).

Since we are looking for the percent change rather than the actual amount, it doesn't matter what the initial amount Q0 is. All that matters is Q(t) Q0.

So the only important unknown quantity is k.

The given information tells us that when t = 1, 2.47% has been lost, so that Q(1) = (1 - .0247) Q0 = .9753 Q0.

Q(1) = Q0 e^(- k 1 ), so we have two expressions for Q(1). Setting them equal we have

.9753 Q0 = Q0 e^(-k).

Self-critique (if necessary)

I don't know the difference of when to use P=P0(1+r)^t and the P=P0e^(kt), so did I just do it completely wrong?

Self-critique Rating 1"

@&

Your solution works.

However the 'e' representation is easier to do calculus with.

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Self-critique (if necessary):

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#*&!

&#Good responses. See my notes and let me know if you have questions. &#