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]

- anonymous

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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## More answers

- anonymous

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.

- anonymous

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?

- anonymous

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!

- anonymous

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

- anonymous

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?

- anonymous

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

- anonymous

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)

- anonymous

ohh...wow...thanks, phi! :)

- anonymous

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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