Question

Let C be the curve of intersection of the parabolic cylinder x^2 = 2y and the surface 3z = xy. Find the exact length of C from the origin to the point (6,18,36)

Answer 1

well, wolfram bites

http://www.wolframalpha.com/input/?i=plot+0%3Dxy%2F3+-+z+and+x^2-2y%3D0

Answer 2

i cant seem to recall parametric to well today

Answer 3

argh yea i have trouble reparametrizing

Answer 4

I just dont understand how or what the rationale is for just choosing a random x=t or some other parameter

Answer 5

\(\frac{x^2}{2} = y\) and the surface 3z = xy

\(\frac{x^2}{2} = y\) and the surface z=\(\frac{x^3}{6}\)

x y z
0 0 0
1 1/2 1/6
2 2 4/3
3 9/2 9/2
4 8 32/3
5 25/2 125/6
6 18 36


x = t
y = t^2/2
z= t^3/6

might define the curve parametrically

Answer 6

by restating everything in terms of "t"; we can relate it all into one variable

Answer 7

\[\int f(x,y,z)\sqrt{x'^2+y'^2+z'^2}dt\]
\[\int f(x,y,z)\sqrt{1+t^2+\frac{t^4}{4}}dt\]
\[\frac{1}{2}\int f(x,y,z)\sqrt{4+4t^2+t^4}dt\]
i just cant seem to recall what to do with the f(x,y,z)

Answer 8

if i take my lead from this: http://tutorial.math.lamar.edu/Classes/CalcIII/VectorArcLength.aspx

it says we just need to integrate the tangent vectors mangnitude

Answer 9

\[\frac{1}{2}\int_{0}^{6}\sqrt{4+4t^2+t^4}dt\]
i really do need to brush up on my line integrals i spose