Callahan

9.2: 2, 3, 4, 15, 16, 17, 34, 35, 66, 67

9.3: 3, 4, 7, 9, 11, 13, 15

2) Obtain a contour plot of on the domain centered at the point , and magnify the plot until the contours look straight, parallel, and equally spaced. Compare your results with the plots on page 475.

3) The purpose of this exercise is to estimate the rate of change at , using the most highly magnified contour plot you constructed in the last exercise:

a) What is the horizontal spacing between the two contours closest to the point ?

From my graph it looks like:

b) Find the z-levels and of those contours, and then compute :

So the change is between contour lines at the point is about .

c) Compare the values of you now obtain with the slope in the x-direction that you estimated in exercise 1:

4) Repeat all the work of the work of the last exercise, this time for :

a) What is the vertical spacing between the two contours closest to the point ?

From my graph it looks like:

b) Find the z-levels and of those contours, and then compute :

So the change is between contour lines at the point is about . This is close to what we found in the last problem, they should be the same.

c) Compare the values of you now obtain with the slope in the x-direction that you estimated in exercise 1:

15) The figure on the left, above, is the countour plot of a function :

a) What are the values of ?

b) What are the partial rates and ?

16) What are the values of ?

Using the equation (p477):

a) Find so that . Find so that .

17) Write the intercept form of the formula for .

We have been using it the whole time!

34) Find the microscope equation for the function at the point .

The microscope equation is of the form: (p485)

So we need the partial derivatives of evaluated at the point:

So, our microscope equation is:

35)

a) Use the microscope equation to estimate the values of and .

These values are close to so we can calculate from this point:

For :

For :

b) Calculate the exact values of the quantities in part (a), and compare those values with the estimates. In particular, indicate how many digits of accuracy the estimates have.

66,67)

9.3: 3, 4, 7, 9, 11, 13, 15

3) Inspect a contour plot of to find the maximum value of subject to the constraints:

So it looks like the maximum is near the point . Lets find it exactly, first we need the equation of the curve created by the intersection of the constraint plane and the surface. We find this by solving the constraint for a variable (y or x) and substituting it into the original surface equation to eleminate one variable:

Now we just take the derivative of the this "projection onto the y-z plane" and set it equal to zero to find the maximum y-value:

Substituting this y-value back into the constraint equation we find :

So the maximum is at the point .

To find the maximum z-value we substitute back into the original surface equation:

4) Add the constraint to the preceding three, where is a parameter that takes values between 0 and 5. Find the maximum value of subject to all four constraints. Describe how the position of the constraint depends on the value of the parameter .

As the value of the parameter decreases the maximum will still be at until . After this, the maximum will fall on the boundary line of the constraint until . Finally, the maximum will be on the line defined by until reaches zero. This can bee seen in the countour plot above.

To find the line defined by :

To find the point where the line crosses the constraint boundary we solve the constraint for and substitute it into the equation:

So it crosses at the point .

7) Use a graph to locate the maximum value of the function:

So it looks like the maximum is near the point . Lets find it exactly by takeing the gradient and setting it equal to the zero vector:

We can graphicly see the maximum point by plotting the functions and and seeing where they cross:

Algebraicly we can find exactly where they cross by solving one for a given variable and substituting it into the other solved for the other variable:

So the maximum occurs at .

9)

a) Locate the position of the minimum of as a function of the parameter .

We use the gradient and set each component equal to zero to find critical points:

So we can see that any value of gives us at least one critical point (apoint on the α vs x curve). To find the value of where it changes from haveing 1 critical point, to 2 or 3 critial points, we take the derivative of the function and set it equal to zero to find its critial points:

So when we get 2 or 3 critial points, when we get only 1 critical point.

So, to get an idea of what the function looks like, lets let . We should get 3 critical points:

So we can see the 3 critical points in the graph above, they occure at where the lines cross. From the gradiant field we can see that there are two minimums at , and the maximum at .

At the function is plotted below. We get 2 critical points, one is a maximum, the other is anothe strange type!

b) The position of the minimum jumps catastrophically when passes through a certain value. At what value of does this happen, and what jump occurs in the minimum?

At .

By looking at the α vs. x graph and the contour plots for the surface at varius values of (above), we can see that as comes down from positive 2 there is only one minimum with a positive x-value (on the right). Suddenly at we get anothe strange critical point! Again, suddenly at the strange critical point devides into a local maximum and a local minimum. But while the initial minimum is still the global minimum. At we have two equely minimum points and no global global minimum. With the minimum with a negative x-value (on the left) becomes the global minimum. As we continue down with more negative values of the trends that we saw for positive values occure in the reversed order until we have only one global minimum with a negative x-value (on the left).

11) Find the maximum value of in the first quadrant, subject to the constraint :

So it looks like the maximum is near the point . Lets find it exactly, first we need the equation of the curve created by the intersection of the constraint plane and the surface. We find this by solving the constraint for a variable (y or x) and substituting it into the original surface equation to eleminate one variable:

Now we just take the derivative of the this "projection onto the x-z plane" and set it equal to zero to find the maximum x-value:

Substituting this x-value back into the constraint equation we find :

So the maximum is at the point .

To find the maximum z-value we substitute back into the original surface equation:

13)

a)

So it looks like the maximum is near the point . Lets find it exactly by takeing the gradient and setting it equal to the zero vector:

We can graphicly see the maximum point by plotting the functions and and seeing where they cross:

f)

So it looks like the maximum is near the point , but it looks like there are other critical points. Lets find them exactly by takeing the gradient and setting it equal to the zero vector:

Substituting into the equation solved in terms of , we have:

So we get the first two critical points:

Substituting into the equation , we have:

It looks like we need the quadratic formula to find our values:

And we get the second two critical points:

Below is the combined graph of a gradient field and a level curves plot, an implicit function plot for the partial derivatives set equal to zero, plus a parametric plot for the vertical lines (because they don't show up in the implicit function plot).

15)

Converted by