1. Find the tangent vector

Dear MHF members,

I have the following problem.

Problem. Let the curve (ellipse) $\alpha$ be the intersection of the surfaces $x^{2}+2y^{2}+4z^{2}=10$ and $x-2y=0$. Find the tangent vector of $\alpha$ at the point $(2,1,-1)$.

I was just trying to find the direction of the tangent vector. I tried to find it by using the cross product of the gradients of the surfaces at the given point (as this vector should be in the same direction with the tangent vector), and get the result $(-16,-8,-12)$. But when I try to solve it by using the derivative of the parametric form of the curve at the given point, I get a different result, which is $(2\sqrt{2/3},\sqrt{2/3},\sqrt{3/2})$. Up to me, both of these results should be the same direction. Please help me.

Thanks a lot.
bkarpuz

2. Re: Find the tangent vector

Originally Posted by bkarpuz
Dear MHF members,

I have the following problem.

Problem. Let the curve (ellipse) $\alpha$ be the intersection of the surfaces $x^{2}+2y^{2}+4z^{2}=10$ and $x-2y=0$. Find the tangent vector of $\alpha$ at the point $(2,1,-1)$.

I was just trying to find the direction of the tangent vector. I tried to find it by using the cross product of the gradients of the surfaces at the given point (as this vector should be in the same direction with the tangent vector), and get the result $(-16,-8,-12)$. But when I try to solve it by using the derivative of the parametric form of the curve at the given point, I get a different result, which is $(2\sqrt{2/3},\sqrt{2/3},\sqrt{3/2})$. Up to me, both of these results should be the same. Please help me.

Thanks a lot.
bkarpuz
1. Both vectors are collinear:

Since $\frac23 \cdot \frac94 = \frac32$ you can re-write $\sqrt{\frac32}=\sqrt{\frac23 \cdot \frac94} = \frac32 \cdot \sqrt{\frac23}$

2. Extract the factor (-8) from the 1st vector. You'll get: $\langle 2,1,\frac32 \rangle$

Extract the factor $\sqrt{\frac23}$ from the 2nd vector and you'll get: $\langle 2,1,\frac32 \rangle$

Thus both vectors have the same direction.

3. Re: Find the tangent vector

Originally Posted by earboth
1. Both vectors are collinear:

Since $\frac23 \cdot \frac94 = \frac32$ you can re-write $\sqrt{\frac32}=\sqrt{\frac23 \cdot \frac94} = \frac32 \cdot \sqrt{\frac23}$

2. Extract the factor (-8) from the 1st vector. You'll get: $\langle 2,1,\frac32 \rangle$

Extract the factor $\sqrt{\frac23}$ from the 2nd vector and you'll get: $\langle 2,1,\frac32 \rangle$

Thus both vectors have the same direction.
Thank you earboth, so my result is right. The book says $(2,3,3)$ for the answer. But it is obviously wrong.