Find parametric equations for the tangent line to the curve at t=1

  • Thread starter DeadxBunny
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Original question:

a) Say r'(t) = 3t^2 i - cost j + 2t k, and r(0) = i + k. Find r(t).
b) Find T(t).
c) Find parametric equations for the tangent line to the curve at t=1.

I have done parts a and b and got the following results:

a) r(t) = t^3 + 1 i - sint j + t^2 + 1 k
b)T(t) = r'(t)/|r'(t)| = (3t^2 i - cost j + 2t k)/(sqrt(9t^4 + cos^2(t) + 4t^2))

Are these answers correct so far? I'm unsure about my answer for T(t) because the denominator seems so messy.

Also, and most importantly, how would I do part (c)?

Thanks!!
 

Answers and Replies

  • #2
Galileo
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DeadxBunny said:
Original question:

a) Say r'(t) = 3t^2 i - cost j + 2t k, and r(0) = i + k. Find r(t).
b) Find T(t).
c) Find parametric equations for the tangent line to the curve at t=1.

I have done parts a and b and got the following results:

a) r(t) = t^3 + 1 i - sint j + t^2 + 1 k
b)T(t) = r'(t)/|r'(t)| = (3t^2 i - cost j + 2t k)/(sqrt(9t^4 + cos^2(t) + 4t^2))
The component of the k-vector should be a primitive function of 2t. I`m sure this was a silly mistake, since you got it right for the i-vector.

T(t) is correct. It's just a matter of plugging in your equations.

For c), use T(t) and plug in t=1. That gives you the unit tangent, so you know the direction of the line. Then given that the line passes through the point r(1) you have enough info to determine the line.
 
  • #3
HallsofIvy
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It would be a good idea to put in parentheses:

r(t)= (t3+ 1)i- (sin t)j+ (t2+ 1)k

Assuming that by T(t) you mean the unit tangent vector, then your answer is correct- divide r' by its own length.

(c) is easy: you don't need the unit tangent vector, just "a" tangent vector and r' works nicely:

x= 3t2+ 1
y= -cos t
z= 2t+ 1

The "+1" terms are, of course, from r(0)= i+ k.
 
  • #4
Help

I don't know how to make new threads! Well anyway, could someone help me with my parametric equation?
 
  • #5
HallsofIvy
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jazz_invincible said:
I don't know how to make new threads! Well anyway, could someone help me with my parametric equation?

Go back to the list of threads. At the bottom is a "new thread" button. As far as your "parametric equation" is concerned, we can't help you if you don't post it!
 
  • #6
HallsofIvy
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HallsofIvy said:
It would be a good idea to put in parentheses:

r(t)= (t3+ 1)i- (sin t)j+ (t2+ 1)k

Assuming that by T(t) you mean the unit tangent vector, then your answer is correct- divide r' by its own length.

(c) is easy: you don't need the unit tangent vector, just "a" tangent vector and r' works nicely:

x= 3t2+ 1
y= -cos t
z= 2t+ 1

The "+1" terms are, of course, from r(0)= i+ k.

Of course, those should be

x= 3t+ 1
y= -cos(1)t
z= 2t+ 1
 

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