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I Do normal differentiation rules apply to vectors?

  1. Apr 13, 2017 #1
    (I am not very sure if this is a high-school level question or a undergraduate level question. Sorry.)

    Does our normal differentiation rules, like the product rule and quotient rule apply to vectors?

    Say for example, differentiate ##r \times \dot r##

    ##r## is radius vector, ##\dot r## is the time derivative of the radius vector (i.e. velocity vector), and you cross both vector (thus the product is a vector, not a scalar as in dot multiplication.)

    Now in normal case, assume that ##r## and ## \dot r## are both scalar, not vector, we will apply product rule, i.e. differentiate the first variable (##r##), retain the second (##r##), multiply both, plus differentiate the second variable (##r##) and add with the first variable (##r##)

    So, your result is:

    (##r \times \ddot r ##) + (## \dot r \times \dot r##)

    But what about vectors? If both ##r## and ## \dot r## are vectors and you cross multiply or dot multiply them, will you differentiate them the same way like you do as if they were scalars?
     
  2. jcsd
  3. Apr 13, 2017 #2

    jedishrfu

    Staff: Mentor

  4. Apr 13, 2017 #3
    Hi mentor jedishrfu.

    Ok, so this apply to both dot multiplication and cross multiplication of vector?

    And what about if there is an angle between the vectors? How would you differentiate them?

    Thanks for answering.

    Zheng Tien.

    Edit: I have seen your posted link. But I am just a high school graduate, the maths in your posted link are too tough for me.
     
  5. Apr 13, 2017 #4

    jedishrfu

    Staff: Mentor

  6. Apr 13, 2017 #5

    jedishrfu

    Staff: Mentor

    Also if you check out the site mathispower4u.com they have a course on Calculus three and there should be a video on this to help you along.

    Also this wiki article on cross product shows the distributive rule

    https://en.m.wikipedia.org/wiki/Cross_product

    In your case, the r' x r' term is zero since the vectors are parallel leaving the r x r" term
     
    Last edited: Apr 13, 2017
  7. Apr 13, 2017 #6
    $$\boldsymbol a(t+h)=\boldsymbol a(t)+h\dot{\boldsymbol a}(t)+o(h),\quad \boldsymbol b(t+h)=\boldsymbol b(t)+h\dot{\boldsymbol b}(t)+o(h),\quad h\to 0;$$
    $$\boldsymbol a(t+h)\times \boldsymbol b(t+h)=\boldsymbol a(t)\times \boldsymbol b(t)+h\big(\boldsymbol a(t)\times\dot{\boldsymbol b}(t)+\dot{\boldsymbol a}(t)\times\boldsymbol b(t)\big)+o(h).$$
     
  8. Apr 14, 2017 #7
    Thank you, jedishrfu!

    I don't understand what zwierz is writing. Can anyone explain what he is writing to me? Sorry and thank you.
     
  9. Apr 14, 2017 #8

    jedishrfu

    Staff: Mentor

    @zwierz was showing you the limit way of proving the cross product time derivative with h being delta t but he could explain it better

    He used a Taylor series expansion of the a and b vector functions and then applied the cross product them.
     
  10. Apr 14, 2017 #9
    Sorry for my late reply, I had my lunch break just now.

    Ah, sorry, I haven't learnt Taylor series expansion yet, even though I know a little about it. Is it stated in the links of your posts above, mentor jedishrfu? Thanks mentor jedishrfu, I will study that in detail and ask further questions here if required. Thanks you zwierz as well.

    Zheng Tien.
     
  11. Apr 14, 2017 #10
    i would rather refer the definition of derivative than the Taylor expansions
     
  12. Apr 14, 2017 #11
    Hmm. I see. You don't prefer your own method? The definition of derivative is fine for me, basically just differentiation by first principle. I have learned that in high school, and I just have to change the variables into vectors. I have checked product rule for wikipedia, and yes it is stated there. Thank you for all of your kind help. :smile:.

    upload_2017-4-14_20-4-25.png
     
  13. Apr 14, 2017 #12
    If this were so you would understand that there are several equivalent definitions of derivative. I used one of them
     
  14. Apr 14, 2017 #13
    But Dr. zwierz (Oh, I discovered that you hold a PhD, I apologize on behalf of my childish behaviour just now, sorry), if you check the stackexchange link that mentor jedishrfu posted just now, you will discover some funny thing.

    Here is the maths that was uploaded and do you notice that there is a plus sign at the end of the 1st and 3rd line, and then suddenly an equal sign at the end of the 2nd line. Curious...

    Also, I don't know how to complete the derivative, doctor. The OP says that at the end of his post: now divide by delta t and take the limit as delta t approaches zero.

    I look at his equation and I go through that equation by myself. If I divide the 4th line by delta t, that means I will only divide u(t) by delta t. Then, I still won't get the correct answer.

    (Below is the derivation shown by the OP in stackexchange.)

    \begin{aligned}\vec{u}\left(t+\delta t\right)\times\vec{v}\left(t+\delta t\right)-\vec{u}\left(t\right)\times\vec{v}\left(t\right) & =\vec{u}\left(t+\delta t\right)\times\vec{v}\left(t+\delta t\right)-\vec{u}\left(t\right)\times\vec{v}\left(t+\delta t\right)+\\
    & =\vec{u}\left(t\right)\times\vec{v}\left(t+\delta t\right)-\vec{u}\left(t\right)\times\vec{v}\left(t\right)=\\
    & =\left[\vec{u}\left(t+\delta t\right)-\vec{u}\left(t\right)\right]\times\vec{v}\left(t+\delta t\right)+\\
    & =\vec{u}\left(t\right)\times\left[\vec{v}\left(t+\delta t\right)-\vec{v}\left(t\right)\right]
    \end{aligned}

    So, Doctor, can you help me by showing me the the correct way to derive the product rule for cross product of vector.

    Thank you, Doctor.
     
  15. Apr 14, 2017 #14

    jedishrfu

    Staff: Mentor

    In the future, you can dispense with the titles mentor and Doctor, there is no need to use them here. Just use the @jedishrfu tag and I'll be sure to see an alert to your post when ever I sign on to PF.
     
  16. Apr 14, 2017 #15
    Ok, @jedishrdu (Can you professional experts choose a better name, I have problem remembering "jedishrfu" and "zwierz", just kidding. :smile:) Why not you help me to answer the question and derive the equation for me, @jedishrfu ? (oh, I see you have a master degree as well. Hi Master jedishrfu. )

    You know what, Physics Forum is the most efficient forum I have ever seen. The mentors act swiftly to every problem and they are very kind to teenagers (like me, I am 18 years old. :smile:) I will definitely recommend Physics Forum to others. All of you professional experts have done extremely well.

    Keep it up, guys! (And please don't forget to help me to derive my equation...)
     
  17. Apr 14, 2017 #16

    Mark44

    Staff: Mentor

    It's inadvisable to use ##\times## when you mean ordinary multiplication. In your expression above, the natural tendency would be to assume that you're talking about the cross product, which is defined only for certain kinds of vectors.
     
  18. Apr 14, 2017 #17

    Mark44

    Staff: Mentor

    Regarding explaining it better, I agree.
    Your formulas appear to use the first order terms of a Taylor expansion of a(t + h) and b(t + h). Also, without further explanation, it would be difficult for some to follow what you are doing in light of the question that was asked. Since the OP has not studied Taylor series yet, your explanation was of no help to him. Try to gauge the level of your help with the abilities of the person asking the question.
     
  19. Apr 14, 2017 #18
    Knowledge about Taylor expansion is not needed.
    Definition. Assume that for some fixed point ##x\in(a,b)## a function ##f:(a,b)\to \mathbb{R}## represents as follows
    $$f(x+h)=f(x)+Ah+o(h)$$ as ##h\to 0## and ##A## is a constant. Then the constant ##A## is referred to as a derivative of ##f## at the point ##x##.
    It is the same for the vector-functions just after evident words about ##o(h)## and ##A## is a vector constant.
     
  20. Apr 14, 2017 #19

    Mark44

    Staff: Mentor

    But your definition uses Taylor expansion, truncating it at the first-order term, and adding an error term to compensate.
    And I'm guessing that the OP has no idea about ##o(h)##, either.

    Again, try to gauge the level of the poster when you give an answer. Providing an answer that is way over the head of a poster is not helpful.
     
  21. Apr 14, 2017 #20
    rather the Taylor expansion uses this definition as the base case of induction
    ok
     
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