Product Rule for Scalar Product: Verifying the Functions

[danny_manny]

Homework Statement

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:

Homework Equations

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

 

[Mark44]

Homework Statement

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:

Homework Equations

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

The above is almost right. What's the derivative of e^t?

and d/dt (b) = sin(t)j - 2tj - e^(-t)k

Two of the terms above are incorrect.

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

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.

 

[danny_manny]

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

 

[Mark44]

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

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

 

[danny_manny]

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

 

[Mark44]

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

 

[danny_manny]

e^(t)

 

[Mark44]

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

 

[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)

 

[danny_manny]

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

 

[Mark44]

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)

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

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

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.

 

[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.

 

[Mark44]

Ok well doing that I get,

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

Which does not match the other way.

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

 

[danny_manny]

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

 

[Mark44]

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

 

[danny_manny]

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

 

[Mark44]

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

 

[danny_manny]

ah mustn't be my day :(

 

[danny_manny]

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

 

[danny_manny]

nvm I looked it up.