Multi-Variable Calculus Assignment Questions

[Jamin2112]

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.

 

[lanedance]

so the way i read the question, the function & constraint are:
$$F(x,y,z) = x + y^3 + z^3$$
$$G(x,y,z) = x^1 + y^2 + z^2 = 1$$

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

 

[lanedance]

$$\nabla F = <\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}>$$

 

[Jamin2112]

$$\nabla F = <\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}>$$

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?

 

[lanedance]

ok so you have 4 equations, try & solve them...

 

[Jamin2112]

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.

 

[lanedance]

Hmmmm...

To simplify,

ß=1/2x;

shouldnt this be
$$\beta = \frac{1}{2x}$$

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

 

[Jamin2112]

shouldnt this be
$$\beta = \frac{1}{2x}$$

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.

 

[lanedance]

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 $\alpha = \beta^2$ 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:
$$y = 0, z \neq 0$$
$$y \neq 0, z = 0$$
$$y = 0, z = 0$$

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

 

[Jamin2112]

solve for another variable $\alpha = \beta^2$ 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:
$$y = 0, z \neq 0$$
$$y \neq 0, z = 0$$
$$y = 0, z = 0$$

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.

 

[lanedance]

so i think you'll get 4 more for y=0, 4 more for z=0, then 2 for y=z=0