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.Quote:
Originally Posted by palmaas
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
$\displaystyle x = \frac{1}{2} -\frac{\sqrt{3}}{2}t, y = \frac{\sqrt{3}}{2} +\frac{1}{2}t, z = \frac{2\pi}{6} +t$
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 aboveQuote:
Originally Posted by 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.
Wow, I figured it out. I was making a really stupid mistake. Thanks a lot for your help!
This is great, but what do you do when 2π/3 is not given?
Quote:
Originally Posted by 11rdc11
Whoops, 2π/6
