Assignment 17

course Mth 174

A couple of my answer were not inserted correctly. I am not sure why. I may have hit the spacebar or something without meaning to. But I do feel like I have grasped section 11.1 really well. And as usual, I need to practice reading and interpreting graphs, but 11.2 was not very difficult to grasp.

xΐ–Λτ‚ΆΏ»ιvκƝ†{½]ω}assignment #017

017. `query 17

Cal 2

08-02-2008

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18:40:07

Query problem 11.1.10 Find a value of omega such that y '' + 9 y = 0 is satisfied by y = cos(omega * t).{}{}Give your solution.

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My first step in this was to find y''.

y'' = omega*cos(omega*t)

therefore

omega * cos(omega * t) + 9 * cos(omega * t) = 0

My next step was to factor out cos(omega * t) to get cos(omega * t) * (omega + 9) = 0

I set each term equal to zero: cos(omega * t) = 0 and omega + 9 = 0

This means that when omega = -9 the function = 0.

I also concluded that when omega * t = pi/2 it equals zero also, but only because cos(pi/2) is equal to zero.

Good attempt, but you didn't take your derivatives correctly.

y = cos(omega*t) so

y' = -omega*sin(omega*t) and

y"" = -omega^2*cos(omega*t)

Now substituting in y"" + 9y = 0 we obtain

-omega^2*cos(omega*t) + 9cos(omega*t) = 0 , which can be easily solved for omega:

-omega^2*cos(omega*t) = -9cos(omega*t)

omega^2 = 9

omega = +3, -3

Both solutions check in the original equation.

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19:34:48

Query problem 11.1.14 (3d edition 11.1.13) (was 10.6.1) P = 1 / (1 + e^-t) satisfies dP/dt = P(1-P)

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I worked on this problem for longer than usual. My first mistake was that I was unsure about the derivative of 1 / (1+e^-t).

This is the same as (1 + e^-t) ^ -1. So I found the derivative to be -1 * (1 + e^-t) ^ -2. Also, by the quotient rule, the derivative is e^-t / (1 + e^-t) ^ 2.

The derivative of e^-t is - e^-t, by the chain rule. So the derivative would be -e^-t / (1 + e^-t)^2. Using this derivative your approach would work. See my subsequent note for another alternative.

With my first derivative, which is equal to (-1)(P)^2, this would mean that (-1)P^2 = P(1 - P). Dividing by P we get (-1)P = 1 - P. Then divding by P again we get, -1 = 1 / P - 1, adding 1 to both sides we get 0 = 1/P. Which is 0 = 1 / (1 / 1 + e^-t) = 1 + e ^ -t. Which means that e ^ -t = -1. I could not find a value for t which would result in this, so I found out this is wrong.

Using the quotient rule derivative I used this equation: (e ^ -t) (P^2) = P(1 - P) Using the same steps as above I equated that e ^ -t = 1/P - 1 = e ^ -t = 1 + e^-t - 1. Which then means that e^-t = e^-t. This hopefully satifies the problem.

P= 1/(1+e^-t) ; multiplying numerator and denominator by e^t we have

P = e^t/(e^t+1) , which is a form that makes the algebra a little easier.

dP/dt = [ (e^t)’ ( e^t + 1) – (e^t + 1) ‘ e^t ] / (e^t + 1)^2, which simplifies to

dP/dt = [ e^t ( e^t + 1) – e^t * e^t ] / (e^t+1)^2 = e^t / (e^t + 1)^2.

Substituting:

P(1-P) = e^t/(e^t+1) * [1 - e^t/(e^t+1)] . We simplify this in order to see if it is the same as our expression for dP/dt:

e^t/(e^t+1) * [1 - e^t/(e^t+1)] = e^t / (e^t + 1) – (e^t)^2 / (e^t + 1)^2. Putting the first term into a form with common denominator

e^t ( e^t + 1) / [ (e^t + 1) ( e^t + 1) ] – (e^t)^2 / (e^t + 1 ) ^ 2

= (e^(2 t) + e^t) / (e^t + 1)^2 – e^(2 t) / (e^t + 1) ^ 2

= (e^(2 t) + e^t – e^(2 t) ) / (e^t + 1)^2

= e^t / (e^t + 1)^2.

This agrees with our expression for dP/dt and we see that dP/dt = P ( 1 – P).

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19:34:51

how did you show that the given function satisfies the given equation?

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19:35:16

What is the derivative dP/dt?

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(e ^ -t) / (1 + e ^ -t) ^2

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19:43:20

Query problem 11.1.16 (3d edition 11.1.15) (was 10.1.1) match equation with solution (cos x, cos(-x), x^2, e^x+e^-x, `sqrt(2x) )

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20:36:42

which solution(s) correspond to the equation y'' = y and how can you tell?

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y = x^2 corresponds when sqrt(2) = x. y'' of x^2 = 2. So for 2 = x^2, x must equal sqrt(2).

Also, e^x + e^-x corresponds. The second derivative of this is itself, so y'' = y.

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20:39:38

which solution(s) correspond to the equation y' = -y and how can you tell

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The only one I found that corresponds was y = x^2. y' = 2x. 2x = -(x^2) Therfore when x = -2, this works. 2(-2) = -4 and -(-2*-2) = -4.

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20:41:44

which solution(s) correspond to the equation y' = 1/y and how can you tell

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I only found one that corresponds. It is y = x^2 when x = (1/2) ^ (1/3)

2(1/2) ^ (1/3) = 1 / (1/2) ^ (1/3) ^ 2

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20:47:01

which solution(s) correspond to the equation y''=-y and how can you tell

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Both y = cos(x) and y = cos(-x) correspond with this equation. y'' of the first is -cos(x) and y'' of the second is -cos(-x).

For the first: -cos(x) = cos(x), it is this simple.

For the second: -cos(-x) = -cos(-x)

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20:48:03

which solution(s) correspond to the equation x^2 y'' - 2y = 0 and how can you tell

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I found y = x^2 to correspond with this for all x. The sceond derivative is 2. Therefore x^2(y'') - 2y = 0 is equal to 2x^2 - 2x^2 = 0

You got many of these. Check your solutions against the following:

The function y = cos(x) has derivative y ‘ = -sin(x), which in turn has derivative y ‘’ = - cos(x). If we plug these functions into the equation y ‘’ = -y we get –cos(x) = - ( cos(x)), which is true, so this function satisfies the differential equation y ‘’ = - y.

The function y = cos(-x) has derivative y ‘ = sin(-x), which in turn has derivative y ‘’ = - cos(-x). If we plug these functions into the equation y ‘’ = -y we get –cos(-x) = - (cos(x)), which is true, so this function satisfies the differential equation y ‘’ = - y.

The function y = x^2 has derivative y ‘ = 2 x , which in turn has derivative y ‘’ = 2. Plugging into the equation x^2 y ‘’ – 2 y = 0 we get x^2 * 2 – 2 ( x^2) = 0, or 2 x^2 – 2 x^2 = 0, which is true.

The function y = e^x + e^(-x) has derivative y ‘ = e^x – e^(-x), which in turn has derivative y ‘’ = e^x + e^(-x). It is clear that y ‘’ and y are the same, so this function satisfies the differential equation y ‘’ = y.

The function y = sqrt(2x) has derivative y ‘ = 2 * 1 / (2 sqrt(2x)) = 1 / sqrt(2x) and second derivative y ‘’ = -1 / (2x)^(3/2). It should be clear that y ‘ = 1 / sqrt(2x) is the reciprocal of y = sqrt(2x), so this function satisfies the differential equation y ‘ = 1 / y.

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20:59:11

Query problem 11.2.5 (3d edition 11.2.4) The slope field is shown. Plot solutions through (0, 0) and (1, 4).{}{}Describe your graphs. Include in your description any asymptotes or inflection points, and also intervals where the graph is concave up or concave down.

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Obviously I have two points: (0, 0) and (1, 4)

The solution at (0, 0) is a horizontal line.

the solution at (1, 4) is a line that starts at y = 0 and is concave up to (1, 4) then changes to concave down when y = 10

Good, but I believe that to the right of (1, 4) the solution through (1, 4) is concave down (as you say) and that it approaches a horizontal asymptote around y = 8.

To the left of (1, 4) the graph will approach the negative x axis as an asymptote.

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21:07:00

Query problem 11.2.10 (was 10.2.6) slope field

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21:14:27

describe the slope field corresponding to y' = x e^-x

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Graph III. It's slopes to the left of the y-axis start out as zero and then slowly change into undefined as it gets closer to the y-axis.

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21:15:24

describe the slope field corresponding to y' = sin x

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This graph actually looks like the sin function itself.

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21:15:52

describe the slope field corresponding to y' = cos x

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Similar to the behavior of the sin graph, this slope field acts simlarily to its actual graph.

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21:17:42

describe the slope field corresponding to y' = x^2 e^-x

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Graph V. From left to right it has slopes of zero, then as it gets closer to the y-axis it rises, then levels off backto zero on the opposite side as it began to rise.

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Good thinking. Compare your reasoning with the following:

Field I alternates positive and negative slopes periodically and is symmetric with respect to the y axis, so slope at –x is negative of slope at x, implying an odd function. The periodic odd function is sin(x). So the field is for y ‘ = sin(x).

Field II similar but slope at –x is equal to slope at x, implying an even function. The periodic even function is cos(x). So the field is for y ‘ = cos(x).

Field III is negative for negative x, approaching vertical as we move to the left, then is positive for positive x, slope increasing as x becomes positive then decreasing for larger positive x. The equation is therefore y ‘ = x e^(-x).

Field IV has all non-negative slopes, with near-vertical slopes for large negative x, slope 0 on the y axis, slope increasing as x becomes positive then decreasing for larger positive x. These slopes represent the equation y ‘ = x^2 e^(-x).

Field V has all non-negative slopes, about slope 1 on the x axis, slopes approaching 0 as we move away from the x axis. This is consistent with the equation y ‘ = e^(-x^2).

Field VI has all positive slopes, with slope approaching zero as x approaches infinity, slope approaching infinity (vertical segments) for large negative x. The field therefore represents y ‘ = e^(-x).

You're doing well on these questions; your overall approach is consistently good. See my notes to identify a errors on a few details.