## anonymous 5 years ago 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?

1. phi

you want a vector [ x y] pointing in the direction of the tangent let x= 1, and y= 3x^2 [1 3x^2]

2. anonymous

Why is x=1? What is the reason for having x as 1?

3. 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

4. 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.

5. 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.

6. 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?

7. phi

looks good. Any way to check it?

8. 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.

9. 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!

10. 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...

11. phi

Here's something germane. See the very bottom http://tutorial.math.lamar.edu/Classes/CalcIII/GradientVectorTangentPlane.aspx

12. 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?

13. 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?

14. 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

15. anonymous

Based on the check you suggested, <2x,2y,2z> doesn't look perpendicular, does it?

16. 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

17. 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

18. phi

oops on [x y] --> [-y x] (to find tangent from the normal in 2-d)

19. anonymous

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

20. anonymous

thank you so much for explaining. really helped me. i am very slow in math.

21. phi

I'm barely one step ahead of you.. but thanks.