Finding the Direction of P sub 0 in the Direction of A

[bobsmith76]

Homework Statement

find the direction of P sub 0 in the direction of A

see second post for attachment, I forgot to place it on this one.

The Attempt at a Solution

I'm only worried about the part that says g_y(x,y,z) = -3ze^xsin yz. I also don't understand the conversion of g^x and g^z.

1. why does cos change to sin? obviously its the derivative but why isn't the derivative being used in g_x and why is it also used in g_z

2. if they're taking the derivative then it doesn't look like the product rule is being used. why not?

3. I would think they would be using u substitution with the x in e^x but I'm better at using u substitution with integrals than with derivatives. I'm pretty sure this is the explanation for why there is a -3z and a -3y in g_y and g_z. I understand where the negative sign comes from, it's because the derivative of cos in -sin. But I don't understand where the z and y comes from.

 

[bobsmith76]

attachment is here.

 

[SammyS]

attachment is here.

What is the problem that this is the solution for ?

What id the function, g(x,y,z) ?

 

[bobsmith76]

see attachment for question.

the function is 3e^xcos xy

 

[Dick]

see attachment for question.

the function is 3e^xcos xy

It's actually 3*e^x*cos(yz). If you want the partial derivative of that with respect to y then you don't have to use the product rule because e^x is a constant. You do have to use the chain rule on the cos(yz) part.

 

[bobsmith76]

well if I use the chain rule on 3e^xcos yz, then why is the answer for g_x the same as the question?

 

[Dick]

well if I use the chain rule on 3e^xcos yz, then why is the answer for g_x the same as the question?

I really don't understand that. Just use the chain rule and tell me how you can get -3*e^x*z*sin(yz).

 

[bobsmith76]

well if use the chain rule for 3e^xcos yz

I'm pretty sure the derivative of e^x is xe^x but I'm not 100% certain, I'll have to look it up. that would make the derivative with respect to x

-3e^x sin yz

not 3e^xcos yz like the book says.

I'm not sure what to do about the yz part, sometimes it plays a factor in the answer sometimes it does not. Clearly in the derivative with respect to y and z, it does play a role, but I'm not sure what operation to use because when you take the derivative with respect to y, the y disappears and when taking it with respect to z, the z disappears.

 

[Dick]

The derivative of e^x is not x*e^x. The rest of what you are saying makes even less sense.

 

[bobsmith76]

If i use chain rule on

3e^xcos yz

why does cos not turn into -sin?

What is the status of the yz? Why does it not factor into the answer for the derivative of

g_x but does play a role in _y and g_z ?

i've already said all that but i'll say it again.

 

[SammyS]

If i use chain rule on

3e^xcos yz

why does cos not turn into -sin?

What is the status of the yz? Why does it not factor into the answer for the derivative of

g_x but does play a role in _y and g_z ?

I've already said all that but I'll say it again.

When you take

$\displaystyle \frac{\partial}{\partial x}\left(3e^x \cos(yz)\right)$​

you treat y and z as if they're constants, so cos(yz) is treated as a constant times the constant 3 times e^x .

Therefore,

$\displaystyle \frac{\partial}{\partial x}\left(3e^x \cos(yz)\right)=3\cos(yz)\frac{d}{dx}e^x\,.$​

 

[bobsmith76]

ok, seems a bit weird, but that will work as an answer for now. thanks for your help.