Would anyone know how to begin a problem where a function z(x,y) is given and x(t),y(t),z(t).

Thanks!

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- Apr 28th 2013, 05:06 PMtopsaanormal line point on surface
Would anyone know how to begin a problem where a function z(x,y) is given and x(t),y(t),z(t).

Thanks! - Apr 28th 2013, 07:30 PMjpritch422Re: normal line point on surface
the normal line to the surface is expressed as:

**N**$\displaystyle = \displaystyle\frac{\partial{z}}{\partial{x}}$$\displaystyle \space$**i**$\displaystyle + \displaystyle\frac{\partial{z}}{\partial{y}}$$\displaystyle \space$**j**$\displaystyle - $$\displaystyle \space$**k**$\displaystyle \displaystyle\frac{\partial{z}}{\partial{x}} = y$; $\displaystyle \space$ $\displaystyle \displaystyle\frac{\partial{z}}{\partial{y}} = x$

**N**$\displaystyle = y$$\displaystyle \space$**i**$\displaystyle + x$$\displaystyle \space$**j**$\displaystyle - $$\displaystyle \space$**k**from the given line we have the symmetric equations:

$\displaystyle \displaystyle\frac{x - 3}{-2} = \displaystyle\frac{y - 4}{5} = \displaystyle\frac{z - 3}{3}$

which gives us:

$\displaystyle \space$ $\displaystyle -2$**i**$\displaystyle \space$$\displaystyle +$$\displaystyle \space$$\displaystyle 5$**j**$\displaystyle \space$$\displaystyle +$$\displaystyle \space$$\displaystyle 3$**k** - Apr 28th 2013, 08:14 PMtopsaaRe: normal line point on surface
so we would solve for where <y,x,-1> is equal to <-2,5,3>, wouldn't that be never because of k, -1 cannot equal 3

- Apr 28th 2013, 08:55 PMjpritch422Re: normal line point on surface
what this is saying is that the given line coincides with vector:

$\displaystyle -2$$\displaystyle \space$**i**$\displaystyle + 5$$\displaystyle \space$**j**$\displaystyle + 3$$\displaystyle \space$**k**

the normal line to the surface coincides with the vector:

$\displaystyle y$$\displaystyle \space$**i**$\displaystyle + x$$\displaystyle \space$**j**$\displaystyle - $**k**

since the two vectors are parallel, that would make the two lines parallel. So then,

$\displaystyle y$$\displaystyle \space$**i**$\displaystyle + x$$\displaystyle \space$**j**$\displaystyle - $**k**$\displaystyle = \displaystyle\frac{2}{3}$$\displaystyle \space$**i**$\displaystyle - \displaystyle\frac{5}{3}$$\displaystyle \space$**j**$\displaystyle - $$\displaystyle \space$**k**