Adv Multi V Calc questions

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  • #1
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Homework Statement



.pdf file of assignment is here: http://www.math.washington.edu/~sullivan/3264_sp10.pdf

So, I have a few questions about the assignment.

Problem #1 I need to know how to find the 14 points.

Problem #2 When it tells me to computer ∂f/∂x amd ∂f/dy at (0,0), does that mean I just I take the limit as (x,y)-->(0,0), since obviously the two are undefined at (0,0)?

Problem #3 The two conditions for differentiability at a point are that f(x,y) is continuous at that point and that lim (h,k)-->(0,0) |f(x+h,y+k)-f(x,y)-(Ah+Bh)| / √(x2+y2) = 0, where A= [f(x+h,y)-f(x,y)]/h, B=[f(x,y+k)-f(x,y)]/k, according to my textbook. Right? How am I supposed to show that such approaches 0 is h,k approach zero?


Homework Equations



They're scattered around the assignment

The Attempt at a Solution



The only one worth showing my work is Problem 1. I used the theorem on the front.

grad(F)=<1, 3y2, z3>
grad(G)=<x2, y2, z2>

grad(F), then, is never the zero vector.
grad(G) is at the point (0,0,0).

If I make grad(F) a scalar multiple of grad(G), my system of equations looks like

1=ßx2
3y2=ßy2
z3=ßz2
x2+y2+z2=1.

------> 1/ß + y2 + z3/ß = 1 -----> ß2 + 3y2 + ßz2=1.

Now, how do I solve for the 13 other points with that equation????

Thanks in advance.
 

Answers and Replies

  • #2
lanedance
Homework Helper
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so the way i read the question, the function & constraint are:
[tex] F(x,y,z) = x + y^3 + z^3[/tex]
[tex] G(x,y,z) = x^1 + y^2 + z^2 = 1[/tex]

The only one worth showing my work is Problem 1. I used the theorem on the front.

grad(F)=<1, 3y2, z3>
grad(G)=<x2, y2, z2>

these gradients don't look right
 
  • #3
lanedance
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[tex] \nabla F = <\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}> [/tex]
 
  • #4
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[tex] \nabla F = <\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}> [/tex]

I have no idea what I was thinking!

Let me redo that part.

grad(F)=<1,3y2,3z2>
grad(G)=<2x,2y,2z>

1=2ßx;
3y2=2ßy;
3z2=2ßz;
x2+y2+z2=1.

Now what?
 
  • #5
lanedance
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ok so you have 4 equations, try & solve them...
 
  • #6
986
9
ok so you have 4 equations, try & solve them...

Hmmmm.....

To simplify,

ß=1/2x;
ß=(3/2)y;
ß=(3/2)z;
x2+y2+z2=1;

So,

(2ß)2 + (2ß/3)2 + (2ß/3)2 = 1.

----> 4ß2 + 4ß2/9 + 4ß2/9 = 1

----> 36ß2 + 8ß2 = 1

----> 44ß2 = 1

----> ß= 1/√44

How am I supposed to get 13 points? That would only give me one.
 
  • #7
lanedance
Homework Helper
3,304
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Hmmmm.....

To simplify,

ß=1/2x;
shouldnt this be
[tex] \beta = \frac{1}{2x} [/tex]
ß=(3/2)y;
ß=(3/2)z;
to get these you have divided through by y & z respectively, which is ok if tehy are noth non-negative, but what about the cases when one or both are zero?

x2+y2+z2=1;

So,

(2ß)2 + (2ß/3)2 + (2ß/3)2 = 1.

----> 4ß2 + 4ß2/9 + 4ß2/9 = 1

----> 36ß2 + 8ß2 = 1

----> 44ß2 = 1

----> ß= 1/√44

How am I supposed to get 13 points? That would only give me one.

if last step was correct, which i don't think it is based on the first comment, when you take a squareroot you get a positive & negative solution
 
  • #8
986
9
shouldnt this be
[tex] \beta = \frac{1}{2x} [/tex]

to get these you have divided through by y & z respectively, which is ok if tehy are noth non-negative, but what about the cases when one or both are zero?



if last step was correct, which i don't think it is based on the first comment, when you take a squareroot you get a positive & negative solution

Attempt 3:

x=1/(2ß),
y=(2ß/3),
z=(2ß/3),

into the 4th equation gives me

1/(4ß2) + 4ß2/9 + 4ß2/9=1.

Multiply everything by 36ß2 and I get

9 + 16ß4 + 16ß4 = 36ß2.

----> I don't know how to solve that, but I assume there would be 4 solutions, (By the way, how would I solve that?) which would then given me 4 different points (x,y,z).

I wouldn't matter is y or z were negative, because they'd just become positive when I plug them into the equation of a sphere, right?

As for working with zeros, that would give me 6 more solutions......

Am I getting closer?

Explain, please.
 
  • #9
lanedance
Homework Helper
3,304
2
Attempt 3:

x=1/(2ß),
y=(2ß/3),
z=(2ß/3),

into the 4th equation gives me

1/(4ß2) + 4ß2/9 + 4ß2/9=1.

Multiply everything by 36ß2 and I get

9 + 16ß4 + 16ß4 = 36ß2.

----> I don't know how to solve that, but I assume there would be 4 solutions, (By the way, how would I solve that?) which would then given me 4 different points (x,y,z).
solve for another variable [itex] \alpha = \beta^2 [/itex] then solve for beta. also not that beta cannot be zero
I wouldn't matter is y or z were negative, because they'd just become positive when I plug them into the equation of a sphere, right?
right, so if it doesn't matter if y or z are negtive, they are both separate solutions to the problem
As for working with zeros, that would give me 6 more solutions......

Am I getting closer?

Explain, please.
maybe, you'll have to try and see, try the cases:
[tex] y = 0, z \neq 0 [/tex]
[tex] y \neq 0, z = 0 [/tex]
[tex] y = 0, z = 0 [/tex]

14 solutions is a lot, so you'll have to try & be systematic in working through
 
  • #10
986
9
solve for another variable [itex] \alpha = \beta^2 [/itex] then solve for beta. also not that beta cannot be zero

right, so if it doesn't matter if y or z are negtive, they are both separate solutions to the problem

maybe, you'll have to try and see, try the cases:
[tex] y = 0, z \neq 0 [/tex]
[tex] y \neq 0, z = 0 [/tex]
[tex] y = 0, z = 0 [/tex]

14 solutions is a lot, so you'll have to try & be systematic in working through

So, I got ß= +/- ( (9 +/- √153) / 16). That's four different values which can be plugged in for four points (x,y,z).

If I set y=0, z=z, that's the same as z=0, y=y, ..... Okay, I get it.
 
  • #11
lanedance
Homework Helper
3,304
2
so i think you'll get 4 more for y=0, 4 more for z=0, then 2 for y=z=0
 

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