Derivative of vector functions

  • #1
Okay, so I am having trouble with this concept (i actually didnt think i was until i realized i was getting all the wrong answers)...
As an example, if you want to take the derivative of:

d/dt [ r(t) X r'(t)]

I just used the general rule of d/dt [r1(t) X r2(t)] = r1(t)Xr2'(t) + r1'(t)Xr2(t)

I ***thought**** I was doing the problem correctly but apparently not.

I just did the following by using the generic equation above:
d/dt [ r(t) X r'(t)] = r(t)Xr''(t) + r'(t)Xr'(t)

What is wrong with what I did? At the back of my book it says the answer is r(t) Xr''(t) which is the first part of my answer (but doesnt include the second portion). Can someone explain where i'm wrong? I'd really appreciate it.
 

Answers and Replies

  • #2
D H
Staff Emeritus
Science Advisor
Insights Author
15,415
687
For any vector [itex]\vec a[/itex], what is [itex] \vec a \times \vec a[/itex]?
 
  • #3
the cross product?? the determinant of the components?

Sorry i'm really shaky on vector calculus...
 
  • #4
D H
Staff Emeritus
Science Advisor
Insights Author
15,415
687
There is no vector calculus here. I asked what is [itex] \vec a \times \vec a[/itex]. Hint: The answer is the same for all vectors.

Step back: What is the magnitude of the cross product of two vectors [itex] \vec a[/itex] and [itex] \vec b [/itex]? What does that become when you set [itex] \vec b = \vec a[/itex]?
 
  • #5
a^2 ? so how does that help solve the problem?
 
  • #6
okay um nevermind i mean 0!
 
  • #7
D H
Staff Emeritus
Science Advisor
Insights Author
15,415
687
Excellent. So now, what is

[tex]\vec r(t) \times \vec r\;^{\prime\prime}(t) + \vec r\;^\prime(t) \times \vec r\;^\prime(t)[/tex] ?
 

Related Threads on Derivative of vector functions

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
10
Views
5K
Replies
13
Views
755
Replies
13
Views
6K
Replies
2
Views
3K
Replies
3
Views
3K
Replies
6
Views
4K
  • Last Post
Replies
1
Views
696
  • Last Post
Replies
16
Views
4K
Top