11.7#7 (to be completed)

Mr. Taylor,
I am having trouble with this problem, the values I'm finding
for the maximum and minimum values don't seem to be correct. How do you
find the absolute max and min on this region?

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I'd be happy to tell you.  BUT, if you want me to do that work I need you to do the work of telling me what *you* did first, and preferably also how what you did relates to what's in the textbook and my lecture notes--it's a lot better pedagogy to start with what you already do know than just to repeat the lecture I gave the other day that you didn't quite get in the first place.

webwork problem & the rules

Hello Prof. Taylor, I have worked out this webwork problem and am doing everything right, but there only two parts of the problem which are refusing.

I thought the two part just required finding the Magnitude of the vector above them, but I am getting it wrong. Some people got theirs right, so I don't understand why my answer is rejected. I therefore need your help!

Below is the problem;

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OK, 

1) first of all, lets talk about the RULES for getting my help on a webwork problem.  The rather than take a screenshot of the problem, click the "Email Instructor" button at the bottom of the problem--this will show me what you've done.  When you do this, you'll have a field to give me a message. In that field enter a brief description of what you have already done.  "Help, I don't know what to do" is NOT useful to you or to me. You have many resources, specifically the textbook, that covers the material, and each section of the webwork corresponds to a specific section of the book, which will tell you SPECIFICALLY what to do. 

2) The gradient 

∇f = <-160*2x/(x^2+y^2+2)^2, -160*2y/(x^2+y^2+2)^2>

   = -320/(x^2+y^2+2)^2 <x, y>

so ∇f(2,2) = -3.2 <2, 2> = <-6.4, -6.4>.  This means that the direction of maximal change, AS A UNIT VECTOR is u = <-1/√2, -1/√2> .  The answer to part c) that you have is NOT A UNIT VECTOR AND NOT THE GRADIENT, but it does point in the right direction. I don't know at all why you want that vector--but I guess you are trying to normalize incorrectly.  The norm of that vector is the answer you give. What you needed, though, is the directional derivative in that direction is
D_u = u . ∇f  = <-1/√2, -1/√2> . <-6.4, -6.4> = 6.4*2/√2 = √2*6.4
which is also equal to |∇f(2,2) |.  All in all I *guess* that the mistake you made was multiplying by
1/√2 while *not* taking the dot product 

Webwork Section 11.6 reopened

Due to the mass of requests, I've re-opened the section 11.6 webwork until Sunday night.

11.5#5

Hello Prof.Taylor,

This problem really needs your intervention:

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While non-interventionism has some attraction on a Friday night, OK.  This problem is really all about the chain rule (discussed in the book and in the lecture notes), for example 

∂W/∂s = ∂F/∂u ∂u/∂s + ∂F/∂v ∂v/∂s, 

and you are given all of the required components above for the values s=1 and t=0.

the most common mistake

The most common mistake people make when working a math problem is not that they don't know how to do the math, it's that they don't know what math to use and they don't know how to find out what math to use.  The reason that they don't know what math to use is because the don't know what the words mean (and so they stare at it and ponder and nothing gets done).  The reason that they don't know what the words mean is because they didn't read the textbook, or at least they didn't read the right part of the textbook.  Now anyone could be forgiven that last, because the textbook is so thick and heavy you could use it to kill poisonous snakes just by dropping it on them, so actually reading it might be fatal.

Except. It's so easy to find the right part of the textbook. Because. Each problem set refers to just one little section in the textbook, which is by itself quite light and slender. That means that the words that you don't understand are in that one little section.  And those words will have some equations that come right after that describe what those words mean, and those equations will be the math that you need to use to solve the problem.

problem

Hi Dr. Taylor,

 I need help answering this question. I dont know how to answer without an equation. Thank you.

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OK, first of all: at the bottom of every webwork page is a little button that says "Email Instructor".  If are having trouble with a problem you push that button and it will send me all of the info from the work you have been doing.
The equation you need here is the definition of directional derivative:
D_u f = ∇f . u = <∂f/∂x, ∂f/∂y>. u  for a unit vector u.  You are given a vector <4,6> which has unit vector <4/√52, 6/√52> and a vector <3,7> that has unit vector <3/√58,7/√58>.  This means that the directional derivatives given in the problem satisfy
6/√52 =  ∇f . <4/√52, 6/√52> or 6 =  ∇f . <4, 6> and
6/√58 =  ∇f . <3/√58, 7/√58> or 6 =  ∇f . <3, 7>.
These amount to the two equations
4 ∂f/∂x + 6 ∂f/∂y = 6
3 ∂f/∂x + 7 ∂f/∂y = 6
which you can solve for the two unknowns  ∂f/∂x, ∂f/∂y

11.5#1

Dear Professor Taylor,

I am struggling with this problem. For finding
dw/dt I am using the following for my partial derivative:
((y*(dx/dt)-x(dy/dt))/y^2)+((z*(dy/dt)-y*(dz/dt))/z^2). For each derivative
with respect to t I am taking the derivative of the expressions given for
each x,y,z.

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You wrote y/x^2 when you meant y/z^2.