# Time derivatives of sin and cos phi

## Homework Statement

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

## 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?

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Homework Helper

## Homework Statement

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

## 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'.