Equation of Tangent Plane with Vector-Valued Function

atm1993
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Homework Statement


Find an equation of the tangent plane to the vector valued function at the origin, (0,0,0).

Homework Equations


The Attempt at a Solution


I don't really know how to start. I've been reading and searching around for quite a bit. I know how to do the problem with a regular function, using the partial derivatives of a function and such, but I'm not really sure where to go with this vector-valued function.
 
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atm1993 said:

Homework Statement


Find an equation of the tangent plane to r(u,v) = uvi + ue^vj + ve^uk at the origin, (0,0,0).



Homework Equations





The Attempt at a Solution


I don't really know how to start. I've been reading and searching around for quite a bit. I know how to do the problem with a regular function, using the partial derivatives of a function and such, but I'm not really sure where to go with this vector-valued function.

The first step is to figure out what u and v give R(u,v) = <0,0,0>.

Then use the fact that Ru X Rv evaluated at that point (u,v) gives a normal vector to the plane. You can use that to write its equation.
 
Alright, so I worked it out and I want to see if I'm on the right track following what you said.

For the function to = <0,0,0>, (u,v) would have to be (0,0). I then found the partials of the function with respect to u and v.


ru(u,v) = vi + e^vj + ve^uk
and
rv(u,v) = ui + ue^vj + e^uk

I then evaluated both of those at (0,0), resulting in

<0, 1, 0> and <0, 0, 1>.

Crossing those resulted in <1, 0, 0>, which then leaves me with an equation of a plane as x=0, using the point (0, 0, 0).

I'm not completely sure if I did that correctly.
 
atm1993 said:
Alright, so I worked it out and I want to see if I'm on the right track following what you said.

For the function to = <0,0,0>, (u,v) would have to be (0,0). I then found the partials of the function with respect to u and v.


ru(u,v) = vi + e^vj + ve^uk
and
rv(u,v) = ui + ue^vj + e^uk

I then evaluated both of those at (0,0), resulting in

<0, 1, 0> and <0, 0, 1>.

Crossing those resulted in <1, 0, 0>, which then leaves me with an equation of a plane as x=0, using the point (0, 0, 0).

I'm not completely sure if I did that correctly.

That looks correct to me. It just says your surface is tangent to the yz (x=0) plane at the origin.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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