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How can I find the tangent vector direction to a function \[y=x^3\] I took the derivative of it and then I got this: \[y'=3x^2\] But this is just the equation to the gradient. I am trying to find a vector in terms of x \[\begin{bmatrix} x\\ y \end{bmatrix}\] such that the vector gives the direction of the tangent vector. How do I do so?

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  • phi
you want a vector [ x y] pointing in the direction of the tangent let x= 1, and y= 3x^2 [1 3x^2]
Why is x=1? What is the reason for having x as 1?
  • phi
you want a vector [dx dy], and you know dy/dx = 3x^2 so dy = dx 3x^2 you can pick anything you like for dx (except 0). 1 is nice

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I just find it weird because 3x^2 is just the gradient. And how come by setting x=1, the whole thing would just become the direction vector. If I picked x equals to other number, it wouldn't work.
  • phi
If you use x= 2" [2 6x^2] is the vector at x=1, the tangent has slope 3. [2 6] has a slope of 3.
So say in a 3 dimension one, with a function: \[x^2+y^2+z^2-5=0\] I take the partial derivative with respect to x, y and z: \[\frac{\partial f}{\partial x}=2x\] \[\frac{\partial f}{\partial y}=2y\] \[\frac{\partial f}{\partial z}=2z\] So is the direction tangent vector be:\[\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 2x\\ 2y\\ 2z \end{bmatrix}\] Is this right?
  • phi
looks good. Any way to check it?
I am not sure how to check it because this is 3-dimensional and I can hardly visualise it. But is what I have done right? I have a feeling that I am wrong.
  • phi
You have a sphere with radius sqrt(5) If we were at the top of the sphere, (0,0,sqrt(5)), the tangent vector would be [a b 0] where a, b are arbitrary (points in any direction in the x-y plane) On the other hand, vector [2x 2y 2z] gives [0 0 2sqrt(5)]. So this can't be correct. Time to call in the experts!
I am thinking maybe I should take just partial derivative of x and y? Say....give me a sec. I compute it now and try...
  • phi
Here's something germane. See the very bottom http://tutorial.math.lamar.edu/Classes/CalcIII/GradientVectorTangentPlane.aspx
Weird...on its Example1 on the very bottom, its delta f, which is the directional vector, is \[\end{bmatrix}=\begin{bmatrix} 2x\\ 2y\\ 2z \end{bmatrix}\] I thought this shouldn't be it?
Weird...on its Example1 on the very bottom, its delta f, which is the directional vector, is <2x,2y,2z> I thought this shouldn't be it?
  • phi
He's has this at the top: The gradient vector is orthogonal (or perpendicular) to the level curve f(x,y) at the point (xo,y0). same for 3-d so your <2x,2y,2z> is perpendicular to the surface
Based on the check you suggested, <2x,2y,2z> doesn't look perpendicular, does it?
  • phi
It works for your y= x^3 : f(x,y)= x^3 -y f'x = 3x^2 f'y = -1 perpendicular is [3x^2 -1] at x=0, we get [0 -1], which points straight down. in 2-d, we can get the tangent from the normal: [x y]-> [-x y] so the tangent would be [1 3x^2] in 3-d we need to find the equation of a plane
  • phi
Yes [2x,2y,2z] is perpendicular. At the top of the sphere, we want a vector pointing straight up (or down). we get [0 0 2sqrt(5)] which points straight up
  • phi
oops on [x y] --> [-y x] (to find tangent from the normal in 2-d)
ohh...wow...thanks, phi! :)
thank you so much for explaining. really helped me. i am very slow in math.
  • phi
I'm barely one step ahead of you.. but thanks.

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