## Astrophysics one year ago @Empty like my heart and @ganeshie8 of course...hey can you help me figure this "trick" out.

1. Astrophysics

Here is the link: http://www.physics.arizona.edu/~leone/ur/phy141/eqns_motion_derivation.pdf page 3, |dw:1435816385662:dw|

2. Astrophysics

Step 3 here

3. misssunshinexxoxo

Looks tough; University of Arizona is fierce. This is incredible

4. Astrophysics

Haha, it's actually very beautiful, if you do calculus you will realize it!

5. misssunshinexxoxo

Are you majoring for physics to engineer? Astronomy? Aerodynamics? Taking calculus in August

6. Astrophysics

Not decided yet, but I do enjoy learning all of the above :)

7. misssunshinexxoxo

Commutative property of multiplication is what the instructor synthesizing, utilize her empowerment. Why do you feel she is expressing that it works so well?

8. Astrophysics

$\huge \vec a (x) = \frac{ d \left[ \frac{ (\vec v(x))^2 }{ 2 } \right] }{ dx }$

9. Empty

I remember the first time I saw a trick like this I was surprised, but it's essentially just taking the derivative of the integral, which are inverses of each other. Work it backwards, it's basically the reverse of the chain rule:$\large \frac{ d \left[ \frac{ (\vec v(x))^2 }{ 2 } \right] }{ dx }=\vec v(x)\frac{ \vec v(x) }{ dx }$ To make it more clear, suppose we have: $\frac{d}{dx} \left( \frac{1}{2} ( f(x))^2 \right) = \frac{1}{2} 2 (f(x))^{2-1} * f'(x)=f(x)*\frac{d f (x)}{dx}$

10. Astrophysics

Wow, I don't know how I missed that, that's pretty clever.

11. Astrophysics

Thanks a lot! Now I can sleep in peace haha.

12. Astrophysics

Hey, if you want you can derive the kinetic energy here, I know we can do it using the work definition, I remember deriving it a while back, all we need is F = ma and $W = \int\limits \vec F \cdot d \vec r$

13. Astrophysics

I should specify by saying work energy theorem

14. Astrophysics

Actually instead of typing it out, here is a nice one just from google images |dw:1435824066157:dw|