Using Chain Rule and Gradient to find Derivative

[e^x]

Plugging the derivatives of x and y into the chain rule formula, I got

dw/dt = [w][/x] (-sint) + [w]/[y] (cost).

I'm stuck on what to do next. I'm thinking that since the derivative of y (sint) is x (cost), that would explain why x is the j component of the vector. But why is the i componet y? Isn't the derivative of x (cost) negative y (so - sint)?

 

[blamocur]

I find your notations somewhat misleading, but otherwise it looks like you are going in the right direction. You do realize that [imath]\nabla w[/imath] is the vector with components [imath]\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}[/imath], don't you? (For some reason you write those components as [w][/x] and [w]/[y].)

 

[Dr.Peterson]

View attachment 29747
Plugging the derivatives of x and y into the chain rule formula, I got

dw/dt = [w][/x] (-sint) + [w]/[y] (cost).

I'll assume @blamocur is right and you meant that [math]\frac{dw}{dt} = \frac{\partial w}{\partial x}(-\sin t) + \frac{\partial w}{\partial y}(\cos t)[/math]

I'm stuck on what to do next. I'm thinking that since the derivative of y (sint) is x (cost), that would explain why x is the j component of the vector. But why is the i component y? Isn't the derivative of x (cost) negative y (so - sint)?

Do you realize that f(x,y) is unrelated to x(t) and y(t), so that [imath]\nabla w[/imath] doesn't depend on the latter? There is no reason "why x is the j component of the vector", or "why ... the i component [is] y". They are both arbitrary.

The way w depends on x and y defines some surface; the ways x and y depend on t define some path on that surface. The derivative of x with respect to t is not related to the derivative of w with respect to x.

I can see why you might imagine a connection, since part of it does look related, but that is accidental.

Now, you need to replace x and y with their values in terms of t, in order to find [imath]\frac{dw}{dt}[/imath] as a function of t and answer the question.

 

[e^x]

I'll assume @blamocur is right and you meant that [math]\frac{dw}{dt} = \frac{\partial w}{\partial x}(-\sin t) + \frac{\partial w}{\partial y}(\cos t)[/math]
Do you realize that f(x,y) is unrelated to x(t) and y(t), so that [imath]\nabla w[/imath] doesn't depend on the latter? There is no reason "why x is the j component of the vector", or "why ... the i component [is] y". They are both arbitrary.

The way w depends on x and y defines some surface; the ways x and y depend on t define some path on that surface. The derivative of x with respect to t is not related to the derivative of w with respect to x.

I can see why you might imagine a connection, since part of it does look related, but that is accidental.

Now, you need to replace x and y with their values in terms of t, in order to find [imath]\frac{dw}{dt}[/imath] as a function of t and answer the question.

Yes, that's what I meant. Sorry, I tried to use the formatting tools but clearly they didn't work.

So what I'm stuck on now is what partial x and y for w are. How would i know that if I don't know what w is as a function of x and y?

 

[blamocur]

I am not 100% sure what "yi+xj", but my best guess is [imath]\nabla w = (y,x) = \left(\frac{\partial w}{\partial x},\frac{\partial w}{\partial y}\right)[/imath]. If my guess is right then you don't need to know [imath]w(x,y)[/imath] to compute the partials since they are already given to you explicitly. Does this make sense?