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Mathematics 6 Online
OpenStudy (raffle_snaffle):

Find the unit normal vector to the space curve r(t)= at t=0.

OpenStudy (raffle_snaffle):

this is a helix, right?

OpenStudy (kainui):

Awesome, this is a fun question! So what's your approach? What's your best guess? I'll help steer you in the right direction.

OpenStudy (kainui):

Yes exactly, it definitely is.

OpenStudy (raffle_snaffle):

Hold on here. Let me see if I can figure this out as far as approaching it.

OpenStudy (raffle_snaffle):

so first we take the derivative of the vector function?

OpenStudy (kainui):

Yes, that's right.

OpenStudy (raffle_snaffle):

r'(t)=<cos(t), 1, -sin(t)> =====> Then we ge r'(o)=<1,1,0>

OpenStudy (raffle_snaffle):

So we need an initial point, right?

OpenStudy (raffle_snaffle):

We use |dw:1385706851928:dw|

OpenStudy (kainui):

Nope, you don't need an initial point. The unit normal vector doesn't depend on an initial point, only on wherever you are in the curve, right?

OpenStudy (raffle_snaffle):

okay

OpenStudy (kainui):

First off, we need to get a unit tangent vector, which we will then use to get a unit normal vector.

OpenStudy (raffle_snaffle):

we are not looking for the tangent line to a space curve right?

OpenStudy (raffle_snaffle):

we obtain a unit tangent vector by taking out vector function and dividing it by its own magnitude>

OpenStudy (kainui):

If we can find the unit tangent vector function (this changes depending on where we are in the curve right?) then we can find the change of the unit tangent function, and that will be normal to the curve, and then we divide by the magnitude to make it a unit vector function... I hope that makes sense.

OpenStudy (raffle_snaffle):

Okay so how do we obtain the unit vector function? By dividing it by its own magnitude?

OpenStudy (raffle_snaffle):

|dw:1385707478892:dw|

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