- #1

bugatti79

- 792

- 1

## 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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter bugatti79
- Start date

- #1

bugatti79

- 792

- 1

[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

- #2

Mark44

Mentor

- 36,342

- 8,297

Definition of the derivative of a vector-valued function?## 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?

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

- #3

bugatti79

- 792

- 1

[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]

- #4

I like Serena

Homework Helper

MHB

- 16,346

- 250

[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]

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.

- #5

bugatti79

- 792

- 1

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.

Yes, I meant the cross product! I didnt know the correct LaTex :-)

Will look at it tomorrow. Thanks!

- #6

bugatti79

- 792

- 1

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]

Let [itex]\vec u =(x(t),x_2(t),x_3(t)...x_n(t))[/itex] and similarly for v hence

[itex]\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))[/itex]

[itex]=x(t) y'(t)+y(t) x'(t), x_2(t)y_2'(t)+y_2(t)x_2'(t).....[/itex]

[itex]=(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)....)[/itex]

[itex]=(x_n(t)y_n'(t))+ (y_n(t)x_n'(t))[/itex]

[itex]\displaystyle =\frac{d}{dt} (x_n(t) \times y_n(t))[/itex]

[itex]\displaystyle =\frac{d}{dt} (\vec u(t) \times \vec v (t))[/itex]..........?

- #7

Mark44

Mentor

- 36,342

- 8,297

The cross product is defined only for vectors in R^{3}.

- #8

I like Serena

Homework Helper

MHB

- 16,346

- 250

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

- #9

bugatti79

- 792

- 1

The cross product is defined only for vectors in R^{3}.

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

- #10

I like Serena

Homework Helper

MHB

- 16,346

- 250

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.

[tex]\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}[/tex]

Did you mean this vector product?

- #11

Mark44

Mentor

- 36,342

- 8,297

- #12

bugatti79

- 792

- 1

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.....

- #13

- 22,178

- 3,305

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.....

What do you mean with [itex]\times[/itex]?? How is it defined??

- #14

I like Serena

Homework Helper

MHB

- 16,346

- 250

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?

- #15

bugatti79

- 792

- 1

What do you mean with [itex]\times[/itex]?? How is it defined??

According to my notes, its the cross product....

- #16

bugatti79

- 792

- 1

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...?

- #17

I like Serena

Homework Helper

MHB

- 16,346

- 250

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:

[itex]\vec a \vec b = a_x b_x \vec i+a_y b_y \vec j+a_z b_z \vec k[/itex]

- #18

bugatti79

- 792

- 1

So is #6 correct?

- #19

I like Serena

Homework Helper

MHB

- 16,346

- 250

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.

- #20

bugatti79

- 792

- 1

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]

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...?

- #21

I like Serena

Homework Helper

MHB

- 16,346

- 250

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:

- #22

bugatti79

- 792

- 1

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!

- #23

I like Serena

Homework Helper

MHB

- 16,346

- 250

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!

- #24

bugatti79

- 792

- 1

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]

[itex]=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}[/itex]

Is this attempt correct so far?

- #25

I like Serena

Homework Helper

MHB

- 16,346

- 250

What you wrote it correct.

- #26

bugatti79

- 792

- 1

What you wrote it correct.

So I just evaulate this, tidy it up and show that it equals the RHS of question?

Cheers

- #27

I like Serena

Homework Helper

MHB

- 16,346

- 250

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).

- #28

bugatti79

- 792

- 1

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).

Hmm...How would you evaluate the LHS? I thought the RHS matrix could be simplified further to arrive at something that resembles the LHS :-)

- #29

Mark44

Mentor

- 36,342

- 8,297

- #30

bugatti79

- 792

- 1

Thank you Mark and I.L.S

Share:

- Last Post

- Replies
- 4

- Views
- 455

- Replies
- 7

- Views
- 519

- Replies
- 9

- Views
- 583

- Replies
- 5

- Views
- 372

- Replies
- 20

- Views
- 433

- Last Post

- Replies
- 1

- Views
- 475

- Replies
- 19

- Views
- 370

- Last Post

- Replies
- 7

- Views
- 900

- Replies
- 90

- Views
- 3K

- Replies
- 30

- Views
- 702