# Vector help! Curves intersecting with Surfaces!

A curve in space is specied by the one parameter set of vectors x(t). Also given is a
surface in space parameterised by x(u, v):

x(t)= <2+t, -t, 1+3t2>
x(u,v)=(u2 - v + u, u+5, v-2>

A) Show that the curve intersects the surface in exactly two points. Show that
xi = <4 - $$\frac{\sqrt{46}}{2}$$, -2 + $$\frac{\sqrt{46}}{2}$$, $$\frac{95}{2}$$ - 6$$\sqrt{46}$$
is one of them

B) Find the Tangential equation $$\frac{dx}{dt}$$ at xi

C) find the normal vector

n=$$\frac{dx(u,v)}{du}$$ x $$\frac{dx(u,v)}{dv}$$

at xi

D)If $$\phi$$ with 0 $$\leq$$ $$\phi$$ $$\leq$$ 90 denotes the angle between the normal vector and the tangential vector, calculate for the intersection point xi (3 signcant figures).

Ive got an idea what to do with A. Make each part of the matrix eaqual to each other and the solve. But to be honest i dont even know if thats right either.

Thanks for the help in advance.

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That is the correct approach. What difficulty are you having with the rest?

How do you then prove the point xi?
And to be honest i dont really know how to do any of the rest!
And pointers/tips to get me going in the right direction

That point should come from your solution of the system of equations (you have 3 equations in 3 variables; the nonlinearity is the only complication).
The second one is just an ordinary derivative; taking the derivative of the curve will give you a vector tangent to the curve at each value of t.
The second asks you to find the tangent vectors along the u and v directions of the surface (the same way you got the derivative above, except in this case you have a partial derivative. Imagine setting either u or v to be a constant; you would get a curve in the other variable embedded in the surface. this gives you a kind of u-v coordinate system for the surface) and then take the cross product of the result to get a vector orthogonal to the two tangent vectors.
The only calculation required is that of taking the derivative; it is taken on each component separately, that is x = (a, b, c) means dx = (da, db, dc).

HallsofIvy
Homework Helper
A curve in space is specied by the one parameter set of vectors x(t). Also given is a
surface in space parameterised by x(u, v):

x(t)= <2+t, -t, 1+3t2>
x(u,v)=(u2 - v + u, u+5, v-2>

A) Show that the curve intersects the surface in exactly two points. Show that
xi = <4 - $$\frac{\sqrt{46}}{2}$$, -2 + $$\frac{\sqrt{46}}{2}$$, $$\frac{95}{2}$$ - 6$$\sqrt{46}$$
is one of them

B) Find the Tangential equation $$\frac{dx}{dt}$$ at xi

C) find the normal vector

n=$$\frac{dx(u,v)}{du}$$ x $$\frac{dx(u,v)}{dv}$$

at xi

D)If $$\phi$$ with 0 $$\leq$$ $$\phi$$ $$\leq$$ 90 denotes the angle between the normal vector and the tangential vector, calculate for the intersection point xi (3 signcant figures).

Ive got an idea what to do with A. Make each part of the matrix eaqual to each other and the solve. But to be honest i dont even know if thats right either.

Thanks for the help in advance.
Yes, that will give you three equations to solve for t, u, and v (there will be, apparently, two solutions). Use either equation then to determine the x, y, and z components of the points of intersection. One of them should be the value given.

For B, do exacly what it says: find $d\vec{x}(t)/dt$ and evaluate at the t that gives the given point.

For C, again, do what it says: find $\partial\vec{x}(u,v)/\partial u$ and $\partial\vec{x}(u,v)/\partial v$ and take the cross product of those two vectors. Again, evaluate that at the given point.

For D, you can use "$\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)$" where $\theta$ is the angle between the vectors. Do that with the two vectors you found in B and C and solve for $\theta$. (I am assuming you meant "calculate $\phi$ for the intersection point xi".)

(It is mildly annoying that they chose to use the same symbol, x, for the position vectors of both the curve and the surface. Not to mention that one can also use "x" as one of the coordinates. It might have been better if the problem had referred to the curve
$\vec{C}(t)= <2+t, -t, 1+3t^2>$
and the surface
$\vec{S}(u,v)= <u^2- v + u, u+5, v-2>$.)

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