Question

Find the vector equation for the tangent line to the curve of intersection of x^2+y^2=25 and y^2+z^2=20 at (3,4,2)

Answer 1

I would take the gradient of each and go from there, but I would have to take the cross product with multiple variables which could take awhile.

Any other way to do this?

Answer 2

I think the intersections are just hyperbolas in the xz plane

Answer 3

y^2 = 25-x^2
(25 -x^2) +z^2 = 20
z^2 -x^2 =-5

Answer 4

That would be the intersection?

Answer 5

sec...

Answer 6

I started reading about this to help, and the way it appears to be done is exactly what you mentioned earlier -- taking the cross product of the gradients.

Answer 7

I got <5+32t, -24t, sqrt(20)+48t> as the intersection, but I'm not sure where to go from there.

Answer 8

those points... how'd you get them?

Answer 9

I think you should be using (3,4,2)

Answer 10

Instead of doing the cross product in variables I plugged in (3,4,2)...
Found the cross product: <32,-24,48> and found the position vector: <5,0,sqrt(20)>
Then just used the r0+t*r1 equation

Answer 11

your slope is right, but not sure why you're not using the point of tangency for a point on your line...

Answer 12

Ok I'm lost

Answer 13

@Algebraic! the tangent vector passes through the point.

Answer 14

Can you walk me through it? I know how to find the tangent vector once I have the equation for the intersection, but how do I get that?

Answer 15

Here's what I have so far...\[
\ \ \ U(x,y,z) = x^2+y^2=25\\
\ \ \ V(x,y,z) = y^2+z^2=20\\
\text{Determine }\nabla U, \nabla V.\\
\ \ \ \nabla U(x,y,z)=2x\hat i+2y\hat j\\
\ \ \ \nabla V(x,y,z)=2y\hat j+2z\hat k\\
\text{Compute the gradients at }(3,4,2).\\
\ \ \ \nabla U(3,4,2)=6\hat i+8\hat j\\
\ \ \ \nabla V(3,4,2)=8\hat j+4\hat k\\
\text{The tangent }W\text{ is orthogonal to both }\nabla U, \nabla V\text{ at said point.}\\
\ \ \ W=\nabla U\times\nabla V =32\hat i-24\hat j+48\hat k
\]

Answer 16

That's what I got too! But, I'm not sure how to get the tangent vector equation from a parametric equation

Answer 17

Then to determine the actual tangent vector equation you do...\[
\ \ \ T(t)=r_0+tr_1\\
\text{where }r_0=3\hat i+4\hat j+2\hat k\text{ and }r_1=W=32\hat i-24\hat j+48\hat k\\
\ \ \ T(t)=3\hat i+4\hat j+2\hat k+32t\hat i-24t\hat j+48t\hat k\\
\ \ \ \ \ \ \ \ \ =[3+32]\hat i+[4-24t]\hat j+[2+48t]\hat k\\
\text{... I think.}
\]

Answer 18

Thank you!

Answer 19

3+32t rather

Answer 20

Apologies again if its wrong and I know for a fact some of my terminology is off... I have relatively little experience with stuff past real analysis of a single variable and I'm just now in a Calculus I course.

Answer 21

No worries!