- #1
bugatti79
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Homework Statement
[itex]\displaystyle \frac{d}{dt} (\vec u (t) \vec v (t))= \vec u (t)' \vec v(t)+\vec u(t) \vec v(t)'[/itex]
I know this is a product rule on the RHS but how does one prove it?
Thanks
Definition of the derivative of a vector-valued function?bugatti79 said:Homework Statement
[itex]\displaystyle \frac{d}{dt} (\vec u (t) \vec v (t))= \vec u (t)' \vec v(t)+\vec u(t) \vec v(t)'[/itex]
I know this is a product rule on the RHS but how does one prove it?
bugatti79 said:OK, I guess I shouldn't have left out the x ie
[itex]\displaystyle \frac{d}{dt} (\vec u (t) x \vec v (t))= \vec u (t)' x\vec v(t)+\vec u(t) x\vec v(t)'[/itex]
I like Serena said:Hmm, what do you mean by [itex]x[/itex]?
Did you mean the cross product (also called outer product):
[itex]\displaystyle \frac{d}{dt} (\vec u (t) \times \vec v (t))= \vec u (t)' \times \vec v(t)+\vec u(t) \times \vec v(t)'[/itex]
Or did you mean the dot product (also called inner product):
[itex]\displaystyle \frac{d}{dt} (\vec u (t) \cdot \vec v (t))= \vec u (t)' \cdot \vec v(t)+\vec u(t) \cdot \vec v(t)'[/itex]
(Or did you mean yet another unusual product? )
Either way, I think the problem asks you to write out the products in the components of the vectors and apply the regular product rule.
I like Serena said:Did you mean the cross product (also called outer product):
[itex]\displaystyle \frac{d}{dt} (\vec u (t) \times \vec v (t))= \vec u (t)' \times \vec v(t)+\vec u(t) \times \vec v(t)'[/itex]
Mark44 said:The cross product is defined only for vectors in R3.
bugatti79 said:Not sure I understand what you are saying I.L.S. Is it not correct now that we know u and v are in R^3
Mark44 said:We are now 10 posts into this problem about vector multiplication (11 with this post), but still don't know what kind of multiplication is intended. Please give us the complete problem statement, which should include the space the vectors are in, and the kind of multiplication that is meant.
bugatti79 said:Suppose that vector u(t) and v(t) take values in R^3. Prove the following identity
[itex]\frac{d}{dt}(\vec u(t) \times \vec v (t))= \vec u'(t) \times \vec v(t)+ \vec u(t) \times \vec v'(t)[/itex]
I think I see my mstake. I should not have x_n and y_n because the vectors are defined in 3-D space...
micromass said:What do you mean with [itex]\times[/itex]?? How is it defined??
I like Serena said:I see. So the real problem seems to be that you're not yet aware of the various vector products.
Can you tell us what the dot product is?
And can you perhaps find the definition (that I already gave) of the cross product in your notes?
bugatti79 said:The dot product is
A= a_x i + a_y j +a_ k, B=b_x i+b_y j+b_z k
A dot B = a_xb_x+ay_by+a_zb_z
The cross product is as your post 10. It looks like I did the dot product...?
bugatti79 said:Are you saying my answer in #6 is using the component wise product...but this is what is in my notes for a similar porblem. I don't know any other easier way.
So is #6 correct?
I like Serena said:Good, you have the dot product down (that also works in n dimensions)!
And no, you did the component-wise product, which isn't usually used:
[itex]\vec a \vec b = a_x b_x \vec i+a_y b_y \vec j+a_z b_z \vec k[/itex]
bugatti79 said:Ok, will try that. but what is the difference between your above and my dot product in #16. They look the same to me except with the addition of the basis i,j and k...?
bugatti79 said:A dot B including the basis is the 'component wise product' method and gives a vector. (new to me!)
I like Serena said:Hmm, what do you mean by [itex]x[/itex]?
Did you mean the cross product (also called outer product):
[itex]\displaystyle \frac{d}{dt} (\vec u (t) \times \vec v (t))= \vec u (t)' \times \vec v(t)+\vec u(t) \times \vec v(t)'[/itex]
I like Serena said:What you wrote it correct.
I like Serena said:Uhh... what you have is the RHS of the question.
What you need to do is take the LHS of the question, evaluate it, and show that it is equal to what you just wrote (aka the RHS of the question).
Mark44 said:The left side is the derivative of the cross product u(t) X v(t). Calculate the cross product, and then take the derivative. If that turns out to be equal to what you have in post #24, you have shown that d/dt(u(t) X v(t)) = u'(t) X v(t) + u(t) X v'(t).
The purpose of proving vector identities is to establish the validity of mathematical equations involving vectors. These identities are essential in solving various problems in physics, engineering, and other scientific fields.
Some common vector identities that are frequently proved include the distributive property, the commutative property, and the associative property. Other important identities involve the cross product, dot product, and vector differentiation.
To prove a vector identity, you must start with one side of the equation and manipulate it using known vector operations and properties until it is equivalent to the other side of the equation. This process involves careful algebraic manipulation and may also require the use of geometric reasoning.
Proving vector identities is important because it ensures the accuracy and consistency of mathematical equations involving vectors. It also allows for a deeper understanding of vector operations and their properties, which is essential in solving complex problems in science and engineering.
One helpful tip for proving vector identities is to break down the equation into smaller, simpler steps. This can make the process more manageable and help identify which vector operations and properties to use at each step. Additionally, it is important to carefully track and simplify each step to avoid errors.