scores and estimated grade

Please refer to the scoring described in the course syllabus.  The score listed is computed from that description with the difference that the estimated score assigns 75% value to the two midterms instead of 50% to three midterms and 25% to the final exam--because you haven't taken the third midterm or final yet.
Note that you have two estimated grades, the first referring to the linear grade scale, the second referring to the curved grade scale, your estimated grade is the better of these two. (please click on the image to enlarge)

linearization and equation of tangent plane

Dr, Taylor

What is the difference between linearization and finding the eq of the tangent plane at a specific point. I find them to be practically the same.

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They are almost the same thing.  The linearization is just a rearrangement of the equation of the tangent plane; from the equation of the tangent plane 

∂f/∂x(x_0,y_0)(x-x_0) + ∂f/∂y(x_0,y_0)(y-y_0) - (z-z_0) = 0 

you can rearrange to get the linearization

                              z = ∂f/∂x(x_0,y_0)(x-x_0) + ∂f/∂y(x_0,y_0)(y-y_0) + z_0

Lecture Notes 10/16/17 and 10/18/17 (to be completed)

Lecture Notes 10/16/17


Lecture Notes 10/18/17 (to be completed)

Review Session and Practice for Friday's Exam

1) On Wednesday we'll have a review session for the exam on Friday. Please come prepared with questions.

2)  The generic mat267 practice exams are available here.

3) An old practice exam written by me can be accessed here.

Lecture Notes 10/11/17 through 10/13/17

Lecture Notes 10/11/17

Lecture Notes 10/13/17

11.7#10

Dr. Taylor,Could you help me with this problem. My Fy is slightly different than the one from the 

solution. But I think I got it right. It is underlined on Red.

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Congratulations, you found an error in the webwork computation of f_y; your computation is the correct one.  It looks like they were thinking about the answer you got, because they got the correct critical point, as I guess you did too.

11.7#13

Hey Dr. Taylor,

How am I suppose to get this? I thought when looking at contour plots for local extrema that we look for concentric circles that come around a point. This diagram only shows parts of it, and I've been
guessing the points and level curves because the problem's picture only shows part of the curves, and not even the parts that would show the extrema (concentric circles around a point, the point representing a max or min).
I know we aren't supposed to guess on answers, but I don't know how else to solve this?? I literally guessed the first coordinate by estimating where the circle's center would be by looking at the section of the circle that was given in this section of the contour diagram... But I can't estimate the second one's coordinates.
Thank you!

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you did just fine. as you point out, guessing is the only thing you can do here. that happens a lot when you have incomplete information as you do here.

11.7#6

Hello. I found the derivative with respect to x which is [ y-4xy-5y^2] and the derivative with respect to y which is [ x-10xy-2x^2]. I then used those equations to find the critical points. I got that x=0,x=1/2 and y=0,y=1/5. So using a combination of these I got three out of the four critical points, but I can't get the fourth from what I have.

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Well, the derivative with respect to x is (1 - 4 x - 5 y) y and the derivative with respect to y is
x(1-2x-10y) so you have the equatoins (1 - 4 x - 5 y) y=0 and x(1-2x-10y)=0. The first equation can be solved if either y=0 or (1 - 4 x - 5 y)=0 and the second equation if  either x=0 or (1-2x-10y) =0. One of each pair must be true at the same point (x,y), so that the options are (x=0,y=0)-->(0,0), or
(x=0, 1-4x-5y=0)-->(0,1/5), or (1-2x-10y=0, y=0)-->(1/2,0) or
(1-2x-10y=0, 1-4x-5y=0)-->(1/6, 1/15).

11.6#9

I'm not sure what I'm doing wrong with this, so I'm probably doing it in
the wrong way altogether

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OK, first of all, you didn't tell me anything that you did to get to that point like you are supposed to.  Second, notice that you are engaging in a little psychological warfare against yourself? Stop that, it's harmful to yourself.  Webwork even told you got the first answer right, so why say that you're altogether wrong? Given that how far wrong could you be about the second question?

Using the arctan is the correct thing to do.  The argument of the arctan is not the ratio of the y and x directions though, since what this is talking about is the angle above the horizontal.  The horizontal in this case is any direction in the xy-plane.  You have to get the slope of z in the direction of the unit vector you use in the first part