# Prove the Following Vector Identities Part 1

#### bugatti79

1. The problem statement, all variables and given/known data
$\displaystyle \frac{d}{dt} (\vec u (t) \vec v (t))= \vec u (t)' \vec v(t)+\vec u(t) \vec v(t)'$

I know this is a product rule on the RHS but how does one prove it?

Thanks

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#### Mark44

Mentor
1. The problem statement, all variables and given/known data
$\displaystyle \frac{d}{dt} (\vec u (t) \vec v (t))= \vec u (t)' \vec v(t)+\vec u(t) \vec v(t)'$

I know this is a product rule on the RHS but how does one prove it?
Definition of the derivative of a vector-valued function?

Also, for the sake of clarity, what kind of multiplication is implied here?

#### bugatti79

OK, I guess I shouldnt have left out the x ie

$\displaystyle \frac{d}{dt} (\vec u (t) x \vec v (t))= \vec u (t)' x\vec v(t)+\vec u(t) x\vec v(t)'$

#### I like Serena

Homework Helper
OK, I guess I shouldnt have left out the x ie

$\displaystyle \frac{d}{dt} (\vec u (t) x \vec v (t))= \vec u (t)' x\vec v(t)+\vec u(t) x\vec v(t)'$
Hmm, what do you mean by $x$?
Did you mean the cross product (also called outer product):
$\displaystyle \frac{d}{dt} (\vec u (t) \times \vec v (t))= \vec u (t)' \times \vec v(t)+\vec u(t) \times \vec v(t)'$

Or did you mean the dot product (also called inner product):
$\displaystyle \frac{d}{dt} (\vec u (t) \cdot \vec v (t))= \vec u (t)' \cdot \vec v(t)+\vec u(t) \cdot \vec v(t)'$

(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.

#### bugatti79

Hmm, what do you mean by $x$?
Did you mean the cross product (also called outer product):
$\displaystyle \frac{d}{dt} (\vec u (t) \times \vec v (t))= \vec u (t)' \times \vec v(t)+\vec u(t) \times \vec v(t)'$

Or did you mean the dot product (also called inner product):
$\displaystyle \frac{d}{dt} (\vec u (t) \cdot \vec v (t))= \vec u (t)' \cdot \vec v(t)+\vec u(t) \cdot \vec v(t)'$

(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.
Yes, I meant the cross product! I didnt know the correct LaTex :-)
Will look at it tomorrow. Thanks!

#### bugatti79

Did you mean the cross product (also called outer product):
$\displaystyle \frac{d}{dt} (\vec u (t) \times \vec v (t))= \vec u (t)' \times \vec v(t)+\vec u(t) \times \vec v(t)'$
Let $\vec u =(x(t),x_2(t),x_3(t)...x_n(t))$ and similarly for v hence

$\displaystyle \frac{d}{dt} \vec u (t) \times \vec v(t)=\frac{d}{dt} (x(t) \times y(t)), \frac{d}{dt}(x_2(t) \times y_2(t))......\frac{d}{dt} (x_n(t) \times y_n (t))$

$=x(t) y'(t)+y(t) x'(t), x_2(t)y_2'(t)+y_2(t)x_2'(t).....$

$=(x(t)y'(t), x_2(t)y_2'(t),x_n(t)y_n'(t))+(y(t)x'(t), y_2(t)x_2'(t)....)$

$=(x_n(t)y_n'(t))+ (y_n(t)x_n'(t))$

$\displaystyle =\frac{d}{dt} (x_n(t) \times y_n(t))$

$\displaystyle =\frac{d}{dt} (\vec u(t) \times \vec v (t))$..........?

#### Mark44

Mentor
The cross product is defined only for vectors in R3.

#### I like Serena

Homework Helper
What you did would be for yet another (unusual) vector product defined by component-wise multiplication.
As such, your proof would be correct (although your notation could be better).

But are you sure we're talking about a component-wise product?
(It would fit better with leaving out the multiplication sign completely.)

#### bugatti79

The cross product is defined only for vectors in R3.
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

#### I like Serena

Homework Helper
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
Sorry, but no.

Let me refresh the definition of the cross product of u and v.

$$\vec u \times \vec v = \begin{pmatrix} u_y v_z - u_z v_y\\ u_z v_x - u_x v_z \\ u_x v_y - u_y v_x\end{pmatrix}$$

Did you mean this vector product?

#### Mark44

Mentor
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

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.
Suppose that vector u(t) and v(t) take values in R^3. Prove the following identity

$\frac{d}{dt}(\vec u(t) \times \vec v (t))= \vec u'(t) \times \vec v(t)+ \vec u(t) \times \vec v'(t)$

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

Suppose that vector u(t) and v(t) take values in R^3. Prove the following identity

$\frac{d}{dt}(\vec u(t) \times \vec v (t))= \vec u'(t) \times \vec v(t)+ \vec u(t) \times \vec v'(t)$

I think I see my mstake. I should not have x_n and y_n because the vectors are defined in 3-D space.....
What do you mean with $\times$?? How is it defined??

#### I like Serena

Homework Helper
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

What do you mean with $\times$?? How is it defined??
According to my notes, its the cross product....

#### bugatti79

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?
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...?

#### I like Serena

Homework Helper
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...?
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:

$\vec a \vec b = a_x b_x \vec i+a_y b_y \vec j+a_z b_z \vec k$

#### bugatti79

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 dont know any other easier way.
So is #6 correct?

#### I like Serena

Homework Helper
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 dont know any other easier way.
So is #6 correct?
No... if you're supposed to use the cross product, you have to substitute the definition of the cross product, apply the regular product rule, and write it back into vector form.

#### bugatti79

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:

$\vec a \vec b = a_x b_x \vec i+a_y b_y \vec j+a_z b_z \vec k$
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...?

#### I like Serena

Homework Helper
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...?
Yes... those i, j, and k.

The difference is that the one is a scalar (just a number) and the other is a vector, which are pretty much very different things!

As it happens, the product rule for derivatives holds for all three types of vector products. :tongue2:

#### bugatti79

To summarise

A dot B gives a number.

A dot B including the basis is the 'component wise product' method and gives a vector. (new to me!)

A Cross B gives a vector

I will be back with the right answer ;-) hopefully!

#### I like Serena

Homework Helper
A dot B including the basis is the 'component wise product' method and gives a vector. (new to me!)
Let's not call this "A dot B"... it will confuse everyone (including me)!
As I said, it is an unusual vector product, but there are applications for it.
If you have to, simply call it "A B" and always explain which product you mean!

#### bugatti79

Hmm, what do you mean by $x$?
Did you mean the cross product (also called outer product):
$\displaystyle \frac{d}{dt} (\vec u (t) \times \vec v (t))= \vec u (t)' \times \vec v(t)+\vec u(t) \times \vec v(t)'$
$=u' \times v+u \times v'=\begin{pmatrix} i & j & k\\ u'_1 &u'_2 &u'_3 \\ v_1&v_2 &v_3 \end{pmatrix}+\begin{pmatrix} i & j & k\\ u_1 &u_2 &u_3 \\ v'_1&v'_2 &v'_3 \end{pmatrix}$

Is this attempt correct so far?

#### I like Serena

Homework Helper
What you wrote it correct.

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