hw asst  26

course Mth 272

g؀Uܿa裛Ζ{񋝭assignment #026

026.

Applied Calculus II

07-28-2008

......!!!!!!!!...................................

15:24:36

Query problem 7.3.38 level curves of z = e^(xy), c = 1, 2, 3, 4, 1/2, 1/3, 1/4.

......!!!!!!!!...................................

RESPONSE -->

im not sure here,

The contour curves appear to be very steep around the x axis and then level out as they move away from the x axis

confidence assessment: 1

.................................................

......!!!!!!!!...................................

15:28:13

What is the level curve z = c for the given function?

......!!!!!!!!...................................

RESPONSE -->

it's the plane for the intersection of the surface at z = e^xy.

for the c values of 1,2,3,4,1/2,1/3,1/4 the level curve is very steep around the x axis and levels out as you move away from the x axis

confidence assessment: 1

.................................................

......!!!!!!!!...................................

15:48:44

Describe how the level curves look for the given values of c, and how they change from one value of c to another.

......!!!!!!!!...................................

RESPONSE -->

from c = 1 to c = 2, the level curve is not very steep,

from c= 2 to c = 3, the level curve is not very steep at all,

from c = 3 to c = 4, the level curve is not steep,

from c = 4 to c =1/2, the level curve is steep and is right next to curve of c = 1

from c = 1/2 to c = 1/3, the level curve is even steeper

from c = 1/3 to c =1/4, the level curve is even steeper

confidence assessment: 2

** The z = c level curve of e^(xy) occurs when e^(xy) = c.

We solve e^(xy) = c for y in terms of x. We first take the natural log of both sides:

ln(e^(xy)) = ln(c), or

xy = ln(c). We then divide both sides by x:

y = ln(c) / x.

For c = 1 we get y = ln(1) / x = 0 / x = 0. Thus the c = 1 level curve is the x axis y = 0.

For c = 2 we get y = ln(2) / x = .7 / x, approximately. This curve passes through the points (1,.7) and (-1, -.7), and is asymptotic to both the x and y axes.

For c = 3 we get y = ln(3) / x = 1.1 / x, approximately. This curve passes through the points (1,1.1) and (-1, -1.1), and is asymptotic to both the x and y axes.

For c = 4 we get y = ln(4) / x = 1.39 / x, approximately. This curve passes through the points (1,1.39) and (-1, -1.39), and is asymptotic to both the x and y axes.

For c = 1/2 we get y = ln(1/2) / x = -.7 / x, approximately. This curve passes through the points (-1,.7) and -1, -.7), and is asymptotic to both the x and y axes.

For c = 1/3 we get y = ln(1/3) / x = -1.1 / x, approximately. This curve passes through the points (-1,1.1) and -1, -1.1), and is asymptotic to both the x and y axes.

For c = 1/4 we get y = ln(1/4) / x = -1.39 / x, approximately. This curve passes through the points (-1,1.39) and -1, -1.39), and is asymptotic to both the x and y axes.

The c = 2, 3, 4 level curves form similar hyperbolas in the first and third quadrant which progressively 'bunch up' closer and closer together. Similar behavior is observed for the c = 1/2, 1/3, 1/4 hyperbolas, which occur in the second and fourth quadrants.

**

.................................................

......!!!!!!!!...................................

15:58:55

Query problem 7.3.46 queuing model W(x,y) = 1 / (x-y), y < x (y = ave arrival rate, x = aver service rate).

......!!!!!!!!...................................

RESPONSE -->

ok, so y is average arrival rate and x is the average service rate.

at (15, 10)

W(15,10) = 1 / (15-10) = .2

at (12,9)

W (12, 9) = 1 / (12-9) = .33

at (12, 6)

W (12, 6) = 1 / (12-6) = .50

at (4, 2)

W (4,2) = 1 / (4-2) = .50

interpreting this I think you would say that there's no difference in the avg. amount of time that a customer waits in line for service from point c to do, but when there is a lesser percentage distance between x and y like in points a and b, then customer, wait in line is going to be shorter

confidence assessment: 2

.................................................

......!!!!!!!!...................................

15:59:08

What are the values of W at (15, 10), (12, 9), (12, 6) and (4,2)?

......!!!!!!!!...................................

RESPONSE -->

previously answered

confidence assessment: 2

.................................................

......!!!!!!!!...................................

16:03:54

You may take extra time with the following: What is the nature of the worst combination of x and y, and why is this bad--both in terms of the behavior of the function and in terms of the real-world situation?

......!!!!!!!!...................................

RESPONSE -->

the nature of the worst combination of x and y I'd say would be when y is as close as it can be to x.

for example if you have a point (15,14), then putting them in the equation 1 / (15-14) gives you 1/1 = 1,

this is not good for the function and in a real world like the last problem this would mean your longest wait

confidence assessment: 2

Good. More generally:

The wait is bad when W(x,y) is a large positive number. This occurs when the denominator is smalli.e., when x-y is small. This means that x would be close to y.

Interpretation: x close to y means that the service rate is only a little greater than the arrival rate.

.................................................

&#Let me know if you have questions. &#