Graph the surface represented by the vector valued function

[stevecallaway]

Homework Statement

r(u,v)=2ucosv i + 2usinv j + u^4
0<=u<=1
0<=v<=2pi

Homework Equations

don't care about the graph as much as how the book got to the answer equation
z=(x^2+y^2)^2/16


3. The Attempt at a Solution [/b
cosu=x/2v sinu=y/2v cosu^2+sinu2=1 (x/2u)^2+(y/2u)^2=1 x^2+y^2=(4)^2
(x^2+y^2)=(4z)^2 (x^2+y^2)/16=z^2 z=sqrt((x^2+y^2)/16)
Why isn't the final answer z=sqrt...

 

[lanedance]

Homework Statement

r(u,v)=2ucosv i + 2usinv j + u^4
0<=u<=1
0<=v<=2pi

Homework Equations

don't care about the graph as much as how the book got to the answer equation
z=(x^2+y^2)^2/16

The Attempt at a Solution

x=2ucosv y=2usinv z=u^4
since u is in all three different parts of the equation, we can't get rid of a variable, then that is why the answer equation is "z="?
Does x=the differentiation of y? Or can we factor out the 2u leaving 2u(cosv i + sinv j)?

x=2ucosv
y=2usinv
z=u^4

try substituting x & y into the equation
z=(x^2+y^2)^2/16
and see what you get

 

[Dick]

What is x^2+y^2 in terms of u?

 

[stevecallaway]

I was in the middle of editing when you two responded, will you please re-look at my attempt at a solution?

 

[Dick]

I was in the middle of editing when you two responded, will you please re-look at my attempt at a solution?

It's pretty garbled. But I'm with you through x^2+y^2=16*u^2, which is what I hope you meant to write. Then did you just put u=z? What did you do?

 

[stevecallaway]

Sorry about the garbled-ness. I thought I put more spaces in between the equations. Yes, I did substitute u=z and then I dividev both sides by 16 to get (x^2+y^2)/16=z^2. Then I took the sqrt of both sides which would give z=sqrt((x^2+y^2)/16) but the book gives an answer of z=(x^2+y^2)^2/16

 

[Dick]

Sorry about the garbled-ness. I thought I put more spaces in between the equations. Yes, I did substitute u=z and then I dividev both sides by 16 to get (x^2+y^2)/16=z^2. Then I took the sqrt of both sides which would give z=sqrt((x^2+y^2)/16) but the book gives an answer of z=(x^2+y^2)^2/16

But you can't put u=z! z=u^4. What does that make u^2?

 

[stevecallaway]

u^2=sqrt(z)...got it. Brilliant. Thanks.