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

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In summary, to find the equation of the tangent line at the point P = (2,4,8) on the curve r(t) = (t, t^2, t^3) in R3, you need to first find the direction of the tangent line, represented by the vector 1i + 4j + 12k at t=2. Then, using the point-slope formula, the equation of the tangent line can be written as v = (2,4,8) + k(1i + 4j + 12k), where k is an arbitrary constant.
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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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  • #2
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.
 
  • #3
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'?
 
  • #4
Yes,the point (2,4,8) is on the graph of the original curve,iff t=2...

Daniel.
 
  • #5
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.
 
  • #6
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?
 
  • #7
That is the direction of the tangent when [itex]t=2[/itex], not the equation of the tangent. Read my last post (carefully :smile:).
 

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

1. What is the tangent equation at point P?

The tangent equation at point P is the equation of the line that is tangent to the curve at point P. It represents the slope of the curve at that particular point.

2. How do you find the tangent equation at point P?

To find the tangent equation at point P, we first need to find the derivative of the curve at point P. Then, we substitute the coordinates of point P into the derivative to find the slope of the tangent line. Finally, we use the point-slope form of a line to write the equation of the tangent line.

3. What is the purpose of solving the tangent equation at point P?

Solving the tangent equation at point P allows us to find the slope of the curve at that point, which can be useful in many applications, such as finding the maximum or minimum point of a curve or determining the rate of change of a function.

4. Can the tangent equation at point P have multiple solutions?

No, the tangent equation at point P will only have one solution, as it represents the slope of the curve at that specific point.

5. How is the tangent equation at point P related to the original curve?

The tangent equation at point P is related to the original curve by representing the slope of the curve at that point. It can also be used to approximate the behavior of the curve at nearby points.

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