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Finding parametric equations for the tangent line 
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#1
Oct205, 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 = 1t, y = t, z = 1t Thanks. 


#2
Oct205, 04:33 PM

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P: 1,021

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) = (1t,t,1t) 


#3
Oct205, 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?



#4
Oct205, 05:00 PM

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P: 1,021

Finding parametric equations for the tangent line
Because the zcoördinate (e^(t)) can only by 1 for t = 0. You can check for x & y too



#5
Oct205, 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?



#6
Oct205, 05:04 PM

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Correct, but since t = 0 was the only value that was correct for the zvalue, it HAD to be correct for x and y too (if not, the point wouldn't have been on the curve).



#7
Oct205, 05:06 PM

P: 1,629

Awesome, thanks for the explantion! it helped greatly!



#8
Oct205, 05:07 PM

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P: 1,021

No problem



#9
Oct605, 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. 


#10
Oct605, 11:15 AM

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P: 1,021

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] 


#11
Oct605, 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 


#12
Oct605, 12:09 PM

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P: 1,021

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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