Solving Tangent Equation at Point P: r(t) = (t, t^2, t^3)

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Discussion Overview

The discussion revolves around finding the equation of the tangent line at a specific point P = (2, 4, 8) on the curve defined by the parametric equation r(t) = (t, t^2, t^3) in R3. Participants explore different approaches to calculating the tangent line, including the use of the derivative and the concept of direction vectors.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests calculating the unit tangent vector T = r'(t)/||r'(t)||, while another argues that the velocity r'(t) is sufficient for determining the tangent line.
  • There is a question about the choice of t=2, with one participant seeking clarification on how to relate this value to the point (2, 4, 8) in the context of the tangent line.
  • Another participant explains that the point (2, 4, 8) lies on the curve when t=2, implying that this is the correct parameter for the tangent line calculation.
  • A more technical perspective is provided, discussing the need for both a point on the line and a direction vector to define the tangent line, emphasizing the importance of solving r(t) = (2, 4, 8) to find the appropriate t.
  • There is a query about whether it is sufficient to present the velocity at t=2 as the answer for the tangent equation or if it should be expressed in a different form.
  • One participant clarifies that the velocity at t=2 represents the direction of the tangent but does not constitute the full equation of the tangent line.

Areas of Agreement / Disagreement

Participants express differing views on whether the unit tangent vector is necessary and how to properly formulate the tangent line equation. The discussion remains unresolved regarding the best approach to present the tangent line equation.

Contextual Notes

Participants have not reached consensus on the necessity of the unit vector versus the velocity vector, and there are unresolved questions about the proper formulation of the tangent line equation.

Beretta
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Let C be the curve: r(t) = (t, t^2, t^3) in R3.

Find the equation of the tangent line at P = (2,4,8).

In this case should I calculate the unite tanget T= r'(t)/||r'(t)||?
Or just the velocity r'(t) = 1i + 2tj + (3t^2)k? How to calculate the tangent equation at the point P?
 
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U don't need the versor (unit vector) for that drection.Just the parametrical equation for the tangent,which u already found...

Then all u have to do is

[tex]\vec{T}_{(2,4,8)}=\left[1\vec{i}+2t\vec{j}+\left(3t^{2}\right)\vec{k}\right]_{t=2}[/tex]

Daniel.
 
Why did you chose t=2 please? I don't know how to put the point (2.4.8) in the equation. Is there any formula to plug the points into r'?
 
Yes,the point (2,4,8) is on the graph of the original curve,iff t=2...

Daniel.
 
A line is a 1-dimensional linear manifold: in order to specify it, you need two vectors: the first one, call it [itex]p[/itex], can be any point on the line. The second, call it [itex]q[/itex], represents the direction of the line (in technical jargon, it forms a basis for the directing space of the manifold). In this case, the line is the set

[tex]\{ p + kq \ | \ k \in \mathbb{R} \}[/tex]

(assuming your vector space is over [itex]\mathbb{R}[/itex]). Alternatively we just say that the line is specified by the equation

[tex]v = p + kq[/tex]

where [itex]k[/itex] is an arbitrary constant (in the appropriate field, here [itex]\mathbb{R}[/itex]).

in this case, you are given that [itex]p = (2, \ 4, \ 8)[/itex] is on the tangent line. You now need to find the direction. This is what your [itex]1i + 2tj + (3t^2)k[/itex] gives you (it does not matter if the vector is normalized or not, since to make our line we scale take every scalar multiple anyway!). First you need to find the appropriate value for [itex]t[/itex]. You do this by solving [itex]r(t) = (2, \ 4, \ 8)[/itex], since you want the direction to be appropriate to that particular point. As dexter said, you find [itex]t=2[/itex], so the direction is dexter's [itex]\vec{T}_{(2, \ 4, \ 8)}=q[/itex].

Now just put this together with what I said at the beginning to get the equation of the line.
 
Is it enough as an answer to the tangant equation to put the velocity in t=2?
ie: r'(t) = [1i + 2tj + (3t^2)k] at t=2. T = 1i + 4j + 12k or should I leave it in the form of dexter's?
 
That is the direction of the tangent when [itex]t=2[/itex], not the equation of the tangent. Read my last post (carefully :smile:).
 

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