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Finding parametric equations for the tangent line

by mr_coffee
Tags: equations, line, parametric, tangent
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mr_coffee
#1
Oct2-05, 02:58 PM
P: 1,629
Hello everyone, i'm having troubles seeing how this works. The directions are:
Find parametric equations for the tagent line to the curve with the given parametric equations at the specified point.
Here is my work and problem:
http://show.imagehosting.us/show/750..._-1_750696.jpg
The answer they have in the back is:
x = 1-t, y = t, z = 1-t
Thanks.
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TD
#2
Oct2-05, 04:33 PM
HW Helper
P: 1,024
The point (1,0,1) corresponds with the parameter t = 0.
Now, in your derivative, let t = 0 to get (-1,1,-1), this gives the direction.

Combining it with the point where it has to go through will give the line:
(1,0,1) + t (-1,1,-1) = (1-t,t,1-t)
mr_coffee
#3
Oct2-05, 04:56 PM
P: 1,629
Ohhh!! thanks again TD! But i'm alittle confused, from the point they gave you: (1,0,1). How did you know that corresponds to t = 0?

TD
#4
Oct2-05, 05:00 PM
HW Helper
P: 1,024
Finding parametric equations for the tangent line

Because the z-co÷rdinate (e^(-t)) can only by 1 for t = 0. You can check for x & y too
mr_coffee
#5
Oct2-05, 05:02 PM
P: 1,629
ohh i c now, so really u just got guess a t, that corresponds to the point they say? right? so if they had like (0,0,0) u would have to find a t that satisfied all them right?
TD
#6
Oct2-05, 05:04 PM
HW Helper
P: 1,024
Correct, but since t = 0 was the only value that was correct for the z-value, it HAD to be correct for x and y too (if not, the point wouldn't have been on the curve).
mr_coffee
#7
Oct2-05, 05:06 PM
P: 1,629
Awesome, thanks for the explantion! it helped greatly!
TD
#8
Oct2-05, 05:07 PM
HW Helper
P: 1,024
No problem
lost_confused
#9
Oct6-05, 11:10 AM
P: 2
im a bit lost now. i get the general idea but what about this one:

x = cos t, y = 3e^(2t), z = 3e^(-2t) and the point is (1, 3, 3)

in this case t is supposed to be 0, but z does not equal 1 when t is 0

so how do they come up with that?? And also x would be 1, y would be 3.
TD
#10
Oct6-05, 11:15 AM
HW Helper
P: 1,024
But z wouldn't have to be 1, it has to be 3 as well, no?

[tex]\left( {\cos t,3e^{2t} ,3e^{ - 2t} } \right)\mathop \to \limits^{t = 0} \left( {\cos 0,3e^0 ,3e^0 } \right) = \left( {1,3,3} \right)[/tex]
lost_confused
#11
Oct6-05, 11:45 AM
P: 2
so what we really have to do is pick a t that gives us the point given? so in this case the point ( 1, 3, 3) is given or found by plugging in t = 0 into the parametric equations?

i think that makes since now

thanks very much
TD
#12
Oct6-05, 12:09 PM
HW Helper
P: 1,024
Well yes, and if it would happen that you cannot find a t so that the parametric equation gives you a certain point, then that point just isn't part of the curve.


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