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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

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- Thread starter bugatti79
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[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

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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

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[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

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[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

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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

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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

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

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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

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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

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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

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- #12

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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

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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??

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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?

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

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

- #16

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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...?

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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]

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So is #6 correct?

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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.

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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...?

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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

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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!

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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

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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?

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What you wrote it correct.

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