The discussion revolves around a vector-valued function defined as r(u,v)=2ucosv i + 2usinv j + u^4, with specified ranges for u and v. Participants are exploring the relationship between the parameters and the resulting surface, particularly focusing on the equation z=(x^2+y^2)^2/16.
The conversation is active, with participants providing insights and questioning each other's reasoning. Some have offered guidance on manipulating the equations, while others are clarifying their interpretations of the relationships between the variables.
There is a noted concern regarding the complexity of the equations and the potential for confusion in variable substitution. Participants are also grappling with the implications of the original function's parameters on the derived equations.
stevecallaway said: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)?
stevecallaway said:I was in the middle of editing when you two responded, will you please re-look at my attempt at a solution?
stevecallaway said: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