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

• Feb 17th 2010, 04:23 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.
• Feb 17th 2010, 04:56 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 $\displaystyle \frac{2\pi}{6}$ into your derivatives

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

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

$\displaystyle z = 1$

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

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

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

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

now we put all this together

$\displaystyle x = \frac{1}{2} -\frac{\sqrt{3}}{2}t, y = \frac{\sqrt{3}}{2} +\frac{1}{2}t, z = \frac{2\pi}{6} +t$
• Feb 17th 2010, 05:19 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.
• Feb 17th 2010, 05:24 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
• Feb 17th 2010, 05:31 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.
• Feb 17th 2010, 05:57 PM
palmaas
Wow, I figured it out. I was making a really stupid mistake. Thanks a lot for your help!
• Dec 9th 2010, 09:52 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 $\displaystyle \frac{2\pi}{6}$ into your derivatives

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

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

$\displaystyle z = 1$

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

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

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

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

now we put all this together

$\displaystyle x = \frac{1}{2} -\frac{\sqrt{3}}{2}t, y = \frac{\sqrt{3}}{2} +\frac{1}{2}t, z = \frac{2\pi}{6} +t$

• Dec 9th 2010, 09:53 AM
ianorourke
Whoops, 2π/6