Finding equation of a tangent line using parametric equations in 3D

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February 17th 2010, 05:23 PM
palmaas

Finding equation of a tangent line using parametric equations in 3D

Hello everyone!
I have been having some trouble with this problem and was hoping that you guys could help me out.

Find parametric equations for the tangent line at the point
http://webwork2.asu.edu/webwork2_fil...16c65457d1.png on the curve http://webwork2.asu.edu/webwork2_fil...4400b3a0c1.png

I dont really know where to start, other than finding the derivative, which is x=-sin(t), y=cos(t), and z = 1. Thanks for any help.
February 17th 2010, 05:56 PM
11rdc11

Quote:

Originally Posted by palmaas

Hello everyone!
I have been having some trouble with this problem and was hoping that you guys could help me out.

Find parametric equations for the tangent line at the point
http://webwork2.asu.edu/webwork2_fil...16c65457d1.png on the curve http://webwork2.asu.edu/webwork2_fil...4400b3a0c1.png

I dont really know where to start, other than finding the derivative, which is x=-sin(t), y=cos(t), and z = 1. Thanks for any help.

hope this helps.

You are almost finish

Next step would be to plug in your t value of $\frac{2\pi}{6}$ into your derivatives

$x = -\sin{\frac{2\pi}{6}}= -\frac{\sqrt{3}}{2}$

$y = \cos{\frac{2\pi}{6}} = \frac{1}{2}$

$z = 1$

Now what you also need are your original values for x,y,z

$x = \cos{\frac{2\pi}{6}} = \frac{1}{2}$

$y = \sin{\frac{2\pi}{6}} = \frac{\sqrt{3}}{2}$

$z = \frac{2\pi}{6}$

now we put all this together

$x = \frac{1}{2} -\frac{\sqrt{3}}{2}t, y = \frac{\sqrt{3}}{2} +\frac{1}{2}t, z = \frac{2\pi}{6} +t$
February 17th 2010, 06:19 PM
palmaas

Thanks a lot! I have another one like it that is giving me trouble as well.
Find the parametric equations for the tangent line to the curve
x = t^{2} - 1, y = t^{3} + 1, z = t^{1}
at the point (8, 28, 3). Use the variable t for your parameter.

Any help would be appreciated.
February 17th 2010, 06:24 PM
11rdc11

Quote:

Originally Posted by palmaas

Thanks a lot! I have another one like it that is giving me trouble as well.
Find the parametric equations for the tangent line to the curve
x = t^{2} - 1, y = t^{3} + 1, z = t^{1}
at the point (8, 28, 3). Use the variable t for your parameter.

Any help would be appreciated.

Try the approach I worked above
February 17th 2010, 06:31 PM
palmaas

I tried and it come out incorrect. I end up with x = 16t+63, y = 1587t+12166, and z =t+3. That doesnt seem right. I must be doing something wrong.
February 17th 2010, 06:57 PM
palmaas

Wow, I figured it out. I was making a really stupid mistake. Thanks a lot for your help!
December 9th 2010, 10:52 AM
ianorourke

This is great, but what do you do when 2π/3 is not given?

Quote:

Originally Posted by 11rdc11

hope this helps.

You are almost finish

Next step would be to plug in your t value of $\frac{2\pi}{6}$ into your derivatives

$x = -\sin{\frac{2\pi}{6}}= -\frac{\sqrt{3}}{2}$

$y = \cos{\frac{2\pi}{6}} = \frac{1}{2}$

$z = 1$

Now what you also need are your original values for x,y,z

$x = \cos{\frac{2\pi}{6}} = \frac{1}{2}$

$y = \sin{\frac{2\pi}{6}} = \frac{\sqrt{3}}{2}$

$z = \frac{2\pi}{6}$

now we put all this together

$x = \frac{1}{2} -\frac{\sqrt{3}}{2}t, y = \frac{\sqrt{3}}{2} +\frac{1}{2}t, z = \frac{2\pi}{6} +t$
December 9th 2010, 10:53 AM
ianorourke

Whoops, 2π/6