Time derivatives of sin and cos phi

[A2Airwaves]

Homework Statement

By using chain rule of differentiation, show that:
$$ \frac{\mathrm{d} sin\phi }{\mathrm{d} t} = \dot{\phi} cos\phi , \frac{\mathrm{d} cos\phi }{\mathrm{d} t} = -\dot{\phi} sin\phi , $$

Homework Equations

The Attempt at a Solution

I got this right for a homework problem, but I'm still confused about why the ##\dot{\phi}## comes out. Does the ##\phi## come out because we are doing:
$$ \frac{\mathrm{d} sin \phi }{\mathrm{d} \phi} \frac{\mathrm{d} \phi }{\mathrm{d} t} $$

Also, when do you know if you're working with cartesian unit vectors or ##r## and ##\phi## unit vectors..?
They have nothing to do with time derivatives right?

 

[Dick]

Homework Statement

By using chain rule of differentiation, show that:
$$ \frac{\mathrm{d} sin\phi }{\mathrm{d} t} = \dot{\phi} cos\phi , \frac{\mathrm{d} cos\phi }{\mathrm{d} t} = -\dot{\phi} sin\phi , $$

Homework Equations

The Attempt at a Solution

I got this right for a homework problem, but I'm still confused about why the ##\dot{\phi}## comes out. Does the ##\phi## come out because we are doing:
$$ \frac{\mathrm{d} sin \phi }{\mathrm{d} \phi} \frac{\mathrm{d} \phi }{\mathrm{d} t} $$

Also, when do you know if you're working with cartesian unit vectors or ##r## and ##\phi## unit vectors..?
They have nothing to do with time derivatives right?

Well, yes. ##\dot{\phi}## means the same thing as ##\frac{\mathrm{d} \phi }{\mathrm{d} t}##. It doesn't really matter what the symbols mean. 'Dot' just usually means 'time derivative'.