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,

**********

(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

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Lecture Notes 9/23/15

Lecture Notes 9/25/15