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

1. find the equation of line of intersection.

Answer 2

6z=x^3

Answer 3

a line integral is similar to arclength, except for a small modification .... right?

Answer 4

pfft, line integral is something else ... :)

Answer 5

is it easier to parametrise this, or .... other 3d coord it?

Answer 6

yes

Answer 7

parametrization will reduce the number of integrals involved

Answer 8

@amistre64 what substitution do you think?

Answer 9

and the surfaces can be thought of as:

z = x^2-2y
z = xy/3

Answer 10

x seems the highest order.. x = t?

Answer 11

or, are you thinking of a y = x^2/2 sub :)

Answer 12

i did that the first thing.. that is your "C"

Answer 13

I did parameterize, but after getting \[\int\limits_{0}^{6} \sqrt(4 + 4t^2 + t^4)\]
I don't know where the 1/2 comes from.

Answer 14

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

Answer 15

marvelous parametering ;)

Answer 16

all hail distance formula

Answer 17

@amistre64 that was your idea though

Answer 18

Ya, then I plugged into the formula. But it has a 1/2 i believe... the normal distance formula doesn't have a 1/2..right?

Answer 19

\[d(t)=\sqrt{x'^2+y'^2+z'^2}\]

Answer 20

x(t) = t = 6
y(t) = t^2/2 = 18
z(t) = t^3/6 = 36

so from t=0, to t=6

Answer 21

Still don't see where the 1/2 comes from...

Answer 22

z'(t) = t^2/2

Answer 23

x'(t) = 1
y'(t) = t
z'(t) = t^2/2

\[\int_{0}^{6}\sqrt{1+t^2+\frac14t^4}~dt\]

Answer 24

then how do you take that complex integral??

Answer 25

\[\int_{0}^{6}\sqrt{1+t^2+\frac14t^4}~dt\]
\[\int_{0}^{6}\sqrt{\frac14(4+4t^2+t^4)}~dt\]
\[\frac12\int_{0}^{6}\sqrt{4+4t^2+t^4}~dt\]
if need be

Answer 26

yep

Answer 27

(2+t)^2

Answer 28

i mean (2+t^2)^2

Answer 29

ohhh I see where the 1/2 comes from. and then you just factor.

Answer 30

soo\[\frac12\int_{0}^{6}t^2+2~dt\]doesnt seem so complex to me

Answer 31

:)

Answer 32

haha, ya, didn't think to factor. Thanks!