# Homework Help: Product rule

1. Feb 6, 2012

### danny_manny

1. The problem statement, all variables and given/known data

Look up, figure out, or make an intelligent guess at the product rule for the scalar
product. That is, a rule of the form

d/dt [a(t).b(t)] =?+?

Verify your proposed rule on the functions
a(t) = ti + sin(t)j + e^(t)k and b(t) = cos(t)i - t^(2)j - e^(t)k:

2. Relevant equations

3. The attempt at a solution

I'm thinking that i can differentiate each function separately and then take the scalar product of these two vectors?

which gives me d/dt (a) =i - cos(t)j - e^(t)k
and d/dt (b) = sin(t)j - 2tj - e^(-t)k

and then i would take the scalar product of these two vectors? but how do i verify that this result is correct?

Thanks Danny

Last edited: Feb 6, 2012
2. Feb 6, 2012

### Staff: Mentor

The above is almost right. What's the derivative of et?
Two of the terms above are incorrect.
First off, you need to "look up, figure out, or make an intelligent guess" about what the derivative of the dot product of two vectors is.
Second, you are given two example vector funtions, a(t) and b(t). Take their dot product, and then differentiate that product.

Compare your result with what you found, figured out, guessed in the first part. If they are the same, then good.

3. Feb 6, 2012

### danny_manny

whoops,
a'(t) = i + cost(t)j -e(t)k
b'(t) = -sin(t)i -2tj-e^(-t)k

Last edited: Feb 6, 2012
4. Feb 6, 2012

### Staff: Mentor

Now you're just guessing. You'll never be able to get this problem until you have the basic differentiate rules down pat.

5. Feb 6, 2012

### danny_manny

Can you point out the terms that are incorrect after i differentiated.

6. Feb 6, 2012

### Staff: Mentor

In each of a'(t) and b'(t), the exponential term is wrong. What is d/dt(et)?

7. Feb 6, 2012

e^(t)

8. Feb 6, 2012

### Staff: Mentor

Yes, so why are you showing a'(t) = <omitted terms> -et?
And why are you showing b'(t) = <omitted terms> -e-t?

9. Feb 6, 2012

### danny_manny

Ah no! I typed the question in wrong my apologies.

a(t) = ti+sin(t)j-e^(t)k
b(t) = cos(t)i-t^(2)j+e^(-t)

10. Feb 6, 2012

### danny_manny

So taking the dot product of the two derivatives i get
a'(t) . b'(t) = -sin(t)-2tcos(t)-1

11. Feb 6, 2012

### Staff: Mentor

Well, that makes a difference. You and I were working different problems. With that change, your derivatives look fine.

That's NOT what I said back in post #2. Your formula is for the derivative of the dot product, which may or not be the same as the dot product of the derivatives.

12. Feb 6, 2012

### danny_manny

Ok well doing that I get,

cos(t)-tsin(t)-2tsin(t)-t^(2)cos(t)-t

Which does not match the other way.

13. Feb 6, 2012

### Staff: Mentor

You have a mistake above . What did you get for a(t)$\cdot$b(t)?

14. Feb 6, 2012

### danny_manny

tcos(t)-t^(2)sin(t)-1

15. Feb 6, 2012

### Staff: Mentor

Yes, now what do you get for the derivative of that dot product?

16. Feb 6, 2012

### danny_manny

cos(t)-tsin(t)-2tsin(t)-t^(2)cost(t)-t

17. Feb 6, 2012

### Staff: Mentor

You are getting the relatively complicated stuff right, but are completely falling down on some really simple stuff. What is d/dt(-1)?

18. Feb 6, 2012

### danny_manny

ah mustn't be my day :(

19. Feb 6, 2012

### danny_manny

So the answer is,
cos(t)-tsin(t)-2tsin(t)-t^(2)cos(t)

20. Feb 6, 2012

### danny_manny

nvm I looked it up.