- #1
mrguru34
- 11
- 0
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 - [tex]\frac{\sqrt{46}}{2}[/tex], -2 + [tex]\frac{\sqrt{46}}{2}[/tex], [tex]\frac{95}{2}[/tex] - 6[tex]\sqrt{46}[/tex]
is one of them
B) Find the Tangential equation [tex]\frac{dx}{dt}[/tex] at xi
C) find the normal vector
n=[tex]\frac{dx(u,v)}{du}[/tex] x [tex]\frac{dx(u,v)}{dv}[/tex]
at xi
D)If [tex]\phi[/tex] with 0 [tex]\leq[/tex] [tex]\phi[/tex] [tex]\leq[/tex] 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 don't even know if that's right either.
Thanks for the help in advance.
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 - [tex]\frac{\sqrt{46}}{2}[/tex], -2 + [tex]\frac{\sqrt{46}}{2}[/tex], [tex]\frac{95}{2}[/tex] - 6[tex]\sqrt{46}[/tex]
is one of them
B) Find the Tangential equation [tex]\frac{dx}{dt}[/tex] at xi
C) find the normal vector
n=[tex]\frac{dx(u,v)}{du}[/tex] x [tex]\frac{dx(u,v)}{dv}[/tex]
at xi
D)If [tex]\phi[/tex] with 0 [tex]\leq[/tex] [tex]\phi[/tex] [tex]\leq[/tex] 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 don't even know if that's right either.
Thanks for the help in advance.