12.3#10

I've tried almost two times to figure out what exactly this problem is
asking for in the incorrect box. I've tried a ton of different expressions,
and the generic f(r,theta), f(r,t), etc. I am just very confused as to what
exactly it wants as an answer.

I can handle picky problems, but this is
one silly demand here.

Any idea what is wrong with it?


Thanks!

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Sometimes webwork does picky, silly things, but this isn't one of those times. You  are given a function of x&y, f(x,y) You just decide to replace x by r and y by t, but that's not the way it works. Remember : x=r cos(t) and y = r sin(t), and you need to substitue these in for x&y.

Things not to do. (updated 10/27/16)

OK. I'm grading the exam and not pulling my hair out, but only because I don't have enough hair to do that. So, purely for your edification, I thought I would make a rogue's gallery of things not to do.

1) DO NOT REVERSE THE ORDER OF AN ITERATED INTEGRAL BY SIMPLY REWRITING THE INTEGRALS IN THE REVERSE ORDER.  (Limits of integration that are functions can never, ever,wind up in the outside integral.  

2) Do not simply change replace the variables in the limits of integration for the inside integral without changing the functional form.  This will never work either.

3) The antiderivative of a function is NOT the same as the derivative.  For instance 

∫ 1/(1+x^5)dx≠5x^4(1+x^5)^-2

4) 1/(a+b)≠1/a + 1/b.  NEVER, EVER.

5) Do not write a negative answer for an integral when the integrand is strictly non-negative on the domain of integration.

6) When you are computing the value of an iterated integral, be sure you compute the antiderivative of the inside integral with respect to the inside variable,  for instance
∫_a^b∫_f1(x)^f2(x) x dy dx = ∫_a^b x y |_y=f1(x)^y=f2(x) dx = ∫_a^b x f2(x)-x f1(x) dx

but
∫_a^b∫_f1(x)^f2(x) x dy dx ≠ ∫_a^b 1/2 x^2 |_y=f1(x)^y=f2(x) dx = ∫_a^b 1/2(f2(x)^2-f1(x)^2) dx

Section 12.3 HW not due until next friday

Currently the homework 12.3 is due this Saturday and I was wondering if that was a mistake or not considering we have not gone over it yet and it's not on the test? Thank you for the clarification.

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Thanks for bringing this to my attention. I've moved the due date for section 12.3 back to Friday the 28th.

MAT Test 2 review materials

Excuse me Professor Taylor, but do you have any practice materials up for the second midterm?

Thanks!

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Same place (as noted on the syllabus): <https://math.asu.edu/resources/math-courses/mat267>

Estimated Grade as of 10-17

If I had to give you a grade today, these are the grades I would give you.  Here's how I figured it:  Since your final grade depends 50% on your three midterms, 25% on the final exam, 15% on the homework and 10% on quizzes and other activities, I gave a 75% weight to the one midterm you've taken 15% weight to your homework percentage and 10% to your quiz percentage.  Then I assigned grades on the principle <90%=A, <80%=B, <70%=C, <60%=D.  I did not bother to put +/- grades at this time.

11.6 #12

Hi Dr. Taylor!

I am not sure on how to take the partial derivative of
this function with x, y, and z variables.

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This question is designed to fool you.   :-)

Yes! If you try to treat this equation as the graph of a function of two variables, you'll wind up with a z on both sides of the equation, which will make you crazy.  Instead, just treat the surface as the 2-level surface {(x,y,z): f(x,y,z)=2} where f(x,y,z)=z(1-xe^y cos(z)).  Then you take partials with respect to z the same way you take any partial derivative. The equations for the tangent plane of a level surface of a function of three variables is covered n the textbook (remember that?) and in the lecture notes for 10/3/16.

11.6 #14

I have a question about the first part of this problem.

I have tried getting the gradient vector, composed of the partial derivative of Z with respect to x and Z with respect to y. I then made it a unit vector by dividing the vector evaluated at (1,1) by the magnitude. I then got this answer, which is incorrect.

Could I know what I am doing wrong?
Thank
you very much!

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The only thing you've done wrong is that you're giving the vector going up the mountain, while you need to go down.

11.6 #18

Dr. Taylor,

Hello,
I have a few problems I am stuck on. And they are always one of three things.
The first is when it asks me to find a normal line I struggle. I can find the tangent plane equation easily. I understand the normal line is the given point dot product t times grad f. But, the point has an x, y, and a z even though the equation is only in two variables(like x=... in this case). I have tried putting in everything for the partial derivative for x in the normal line equation and yet I invariably get it wrong. What key concept am I missing?
Second, how do I do the problems that say find the angle above horizontal? I understand I would have to use <0,1> for a 2d plane and <0,0,1> for a 3d plane to start the problem, but that is all I know.
Third, what is the value of maximal increase? Like on one I found the vector of direction of maximal increase (easy) and then it asked, what is the value of maximal increase and I took the magnitude of the vector I had and that got me the wrong answer. I think my issue with that is just more conceptual.
If you could clear up any of those, that would be greatly appreciated! Thank you in advance.

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That's a lot of questions.
First, there are some different x,y,z's that can get confused here.  The tangent plane is the tangent plane at a point.  That means you have a specific gradient vector at the point, and a specific tangent plane that is perpendicular to that specific gradient vector. So before you can get those, the x,y,z that are in the definition of the graph z=f(x,y) of the function have disappeared and been replaced by some specific numbers (a,b,c) that satisfy c=f(a,b).  This means that the point (a,b,c) is on the graph of the function. It's also on the tangent plane of the graph z=f(x,y), which passes through that point. In addition a normal vector to that tangent plane is -del(f)(a,b)+k; this means that the equation of the tangent plane is -∂f/∂x(a,b)(x-a)-∂f/∂y(a,b)(y-b)+(z-c)=0.  There are a lot of normal lines to that plane; one of them passes through the point (a,b,c). A parametric equation to this normal line is <a,b,c>+t<-∂f/∂x(a,b),-∂f/∂y(a,b), 1>. Hope this clarifies.

Second, the tangent of an angle in a triangle is the opposite divided by the adjacent, which you could rephrase as the rise divided by the run. But the rise divided by the run has another name: it's called the slope.  When you have a motion along a curve in the plane this means that the tangent of the angle above the horizontal is dz/ds, where s is arclength, or equivalently (dz/dt)/(ds/dt). The numerator you can get from the chain rule, and the denominator is the speed.

Third when you find the direction of maximal increase--that's given by a unit vector--it just tells you direction to go in, not how fast you go up when you go in that direction.  How fast you go up is given by the directional derivative. The directional derivative in the direction of the greatest increase has a special form, which is discussed in the lecture notes on directional derivatives, *and* in the text book.

Attendance though 10/12/16

Lecture Notes 10/3/16-10/7/16

Lecture Notes 10/3/16

Lecture Notes 10/5/16

Lecture Notes 10/7/16

Lecture Notes 9/26/16-9/30/16

Lecture Notes 9/26/16

Lecture Notes 9/28/16

Lecture Notes 9/30/16