Tomorrow's test! (a request)

Hello Dr. Taylor,

I had a question about tomorrows test. I was wondering if I could bring some blank graph paper in, to use as scratch paper. It will help me draw pictures to scale (for when switching limits of integration), and sometimes I like to write out the solution as it helps me figure out step by step what I need to do to solve the problem. Do you think this is a possibility? I will be okay with or without it. It just helps me draw my pictures which can be key to the answer. Let me know what you think. Thanks!

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I'd prefer to supply the scratch paper; the functions in the limits of integration will be very simple and any switching of limits of integration will depend on an understanding of these functions and their inverse functions and not on any precise rendering. 

Partial homework extension

I have extended the due date for section 12.5 to next Monday at midnight.

No cheat sheets allowed on the exam

Would you be so kind as to let us have a HAND WRITTEN cheat sheet
with general formulas for ourselves? I think if you allowed us to have
example problems it would literally be cheating. But, perhaps if we could
have formulas like for linear approximation, and what information the
determinate tells us about our critical points. We could turn it in with
our tests so you know what was on there and that students don't share one
copy. I hope you take this into consideration as something allowed for this
test.

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Sorry, I asked the course coordinator and the rest of the instructors for this course, the consensus is that there we cannot allow cheat sheets.

Lecture Notes 10-14-15 through 10-19-15

Lecture Notes 10/14/15

Lecture Notes 10/16/15

Lecture Notes 10/19/15

12.2 no 3

I'm having difficulty understanding how to solve this problem.  Where
should I start?

Well, the first step is to read section 12.2 in the textbook and the notes. The basic notion is that you are trying to write the double integral as an iterated integral.  For part (a), this means that you are integrating first with respect to y and then with respect to x, this means that you need to integrate from the horizontal line at the bottom of the triangle--which gives you the lower limit of integration of the inside integral, to the slanting line at the top of the triangle.  In order to do this, you *must* find the equation of the line relating from the graph of the line; this means that you write a linear equation. for y as a function of x. This function of x becomes your upper limit of integration.
For part (b) you must turn this around--the vertical line on the right gives you the lower limit of the inside integral, and the upper limit becomes the equation for x in terms of y that you get by solving the equation of the slanting line for x.

12.1 no 1

Can you tell me what I'm doing wrong here?

I  think so.  When dividing the square [0,2]x[0,2] into four equal squares you get squares [0,1]x[0,1], [0,1]x[1,2], [1,2]x[0,1], [1,2]x[1,2].  It looks like you misinterpreted the questions--for example in part (A) you have supplied the value of the integral

But you should instead give the value of the Riemann sum approximation for the integral

with that partition of the square with values at the points (0,0),(1,0),(0,1),(1,1)

(25-0^2-0^2)ΔA_{11}+(25-1^2-0^2)ΔA_{21}+(25-0^2-1^2)ΔΑ_{12}+(25-1^2-1^2)ΔΑ_{22}

and since all of the ΔA's take the value 1, the correct answer for part (A) should be 

25+24+24+23=96

Review for Test 2!!!

The review for test 2 is in WEXLER 116 at 8:15pm on Oct 19th, 2015.

11.6 problem 20 (updated)

Dear Professor Taylor,

I'm not really sure how to do this. I'm fairly
certain the grad f vector point towards the upper right of the graph at
every point. But I think I just happened to guess correctly to get the
first two answers. Can you explain what how to do this?

Thank you,


At the point P=(1,4) in the direction v=(i+j)/√2, the directional derivative is approximately ______

The first step is to know where you are; in this case at the intersection of the vertical line x=1 and the horizontal line y=4. If you go up from the intersection of the two red lines, the contour changes from about 4.0 to half way between 4.0 and 6.0--i.e. 5.0. while the y-value changes from 4 to 5--hence the rise over the run is 1/1=1. Likewise the x-value changes from 1 to 2.5 while the contour changes from 4.0 and 10.0--the intersection of the red horizontal line with the yellow vertical line--which gives you another rise and a run to compute a difference quotient. For the direction vector v you add the two difference quotients and divide by √2.

Btw, grad(f) points to the upper right at a few points on the graph, but not at most points: toward the top of the graph near the y-axis it points mostly to the right, on the x-axis except near x=0 it points mostly up, on the diagonal line y=x it points to the upper right corner.

Hello Professor Taylor,

I did what you said to do for this problem.
I get
that
df/dx= (10-4)/(3-2)=4i
and
df/dy=(5-4)/(5-4)=1j

so the answer along
the vector v I get
(4i+1j)/sqrt(2) I kept getting 5/sqrt(2) but I changed
it to six/sqrt(2) and that happened to be the right. What am I doing wrong?

you've got a little problem using the dot product correctly, since you kept the i and j after you did the dot product, and they're supposed go away. basically you did this correctly though, the answer 5/√2 is correct and 6/√2 is incorrect.  The problem was in the software and I fixed it, it should work now

Section 11.7 Problem 8

Hey Professor Taylor,
I am very much stuck on problem 8 from section 11.7. I was hoping you would be able to help me out. Thank you very much. The problem states the following:

Find the absolute maximum and absolute minimum of the function f(x,y)=2x^3+y^4 on the region {(x,y)|x^2+y^2≤81}.  It also asks for the x and y coordinates at which the maximum and minimum values are obtained at.

Sincerely,

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(Yep, this one is a bit of a detail-challenge)

OK, the domain is the closed disk of radius 9 centered at the origin. The first thing you need to do is to compute the gradient of f:  

It follows that the critical points inside the disk are when 6x^2=0 and 4y^3=0; i.e. only at the point (0,0). Since f(0,0)=0 and f can take both positive and negative values on the x-axis inside the disk, for example f(1,0)=2 and f(-1,0)=-2,  it follows that the absolute min and absolute max are on the boundary of the domain,  i.e. must be on the circle of radius 9 centered at (0,0).  

We find the absolute max and min on the unit circle by parameterizing the circle of radius 9 as 

x(t)=9 Cos(t), y(t)=9 Sin(t).  Then  f(x(t),y(t))=9^3(6 Cos^3(t)+9 Sin^4(t)), which is a differentiable function of t, and so the absolute max and min will be at critical points of this function of one variable, i.e. satisfy the equation 

so the critical points are where Sin(t)=0 AND where Cos(t)=0 AND where - Cos(t)+6 Sin^2(t)=0.  Thus we get t=0,π/2,π, 3π/2 and solutions of the equation Cos(t) - 6 + 6Cos^2(t)=0 (where we used the identity Sin^2(t)=1-Cos^2(t)).  BUT notice that the equation is quadratic in Cos(t).  Using the quadratic formula we get that Cos(t)=(-1±√(1^2-4(1)(-6))/12=(-1±√25)/12=1/3, -1/2.  Then, since Sin(t)=±√(1-Cos^2(t)), we need to check the points (-1/2,±√3/2) and (1/3,±2√2/3), hence we have eight points to check: plugging in the numbers we get the values (9,0,1458), (0,9,6561), (-9,0,-1458),(0,-9, 6561), (-4.5, 9√3/2,14580), (-4.5, -9√3/2,14580), (3, 6√2,5238), (3, -6√2, 5238). 

Thus the absolute max value is 14580 which takes place at (-4.5, ±9√3/2) and the absolute min value is -1458 which takes place at (-9,0). 

Lecture Notes This Week 10/5/15 to 10/9/15 (Updated)

Lecture Notes 10/5/15

Lecture Notes 10/7/15

Lecture Notes 10/9/15

Section 11.3 problem 11

I'm stumped on how exactly to get this answer.  I just can't see the
relation between Fy and the contour lines.

The notion is that the partial derivative is the limit of the difference quotient 

this means that when h is small, the partial derivative is approximately equal to the finite difference quotient:

This is is an important and useful fact for an engineer, because a lot of times the only thing you know about the function is some contours or some data points.  For this problem you can start at the point (1,3), where you're right on the contour for 10, so you can get f(x,y)=10, and if you increment y by about h=2.6 you get the  f(x,y+h)=8

--that's all the information you need to get the approximation

Attendance through 9/25/15

Lecture Notes Week of 9/28/15-10/2/15

Lecture Notes 9/28/15

Lecture Notes 9/30/15

Lecture Notes 10/2/15

Lectures Notes Week of 9/21/15-9/25/15

Lecture Notes 9/21/15

Lecture Notes 9/23/15

Lecture Notes 9/25/15