# A formula for power [duplicate]

My question is this, $$P= dW/dt.$$

dW=F.dr

That leads to P= F.v But how?!

The book (Schaum's theoretical mechanics) stated that this formula is true. But I think it is true just in case that F is constant, otherwise we have to apply the rule of the product of derivatives for dot product!

Can any one help me and illustrate the bug picture for me?

• $F$ is assumed constant during each slice of time ${\rm d}t$. – John Alexiou Apr 29 at 0:28
• But can we apply this formul for variable force fields? – Sohaib Ali Alburihy Apr 29 at 0:37
• Sorry I meant formula* not formul – Sohaib Ali Alburihy Apr 29 at 0:50

The work $$W_{\text{by }\vec{\bf F}}$$ by some force $$\vec {\bf F}$$ over a path $$C:\vec {\bf r}(t)$$ is $$W_{\text{by }\vec{\bf F}}=\int_C \vec {\bf F} \cdot \mathrm d\vec {\bf r}.$$

Recall $$\vec {\bf{v}} = \dfrac{\mathrm d \bf\vec {r}}{\mathrm dt}$$, then, for some time interval from $$t_0$$ to $$t$$, the work is

$$W_{\text{by }\vec{\bf F}}=\int_{t_0}^{t} \vec {\bf F} \cdot \vec {\bf v}\ \mathrm dt'$$

where the $$t'$$ is used simply for notation (since we can't have the variable of integration as an integral bound).

We know power is the time derivative of work, therefore,

$$P=\dfrac{\mathrm d}{\mathrm dt}\int_{t_0}^{t} \vec {\bf F} \cdot \vec {\bf v}\ \mathrm dt',$$

and, by the fundamental theorem of calculus, we have that $$P=\vec {\bf F} \cdot \vec {\bf v}.$$

• So I understand that this formula apply even when the force is not constant, am I right?, thank you for your help. – Sohaib Ali Alburihy Apr 29 at 0:33
• @SohaibAliAlburihy yes. we never had to impose any restrictions on the force – user256872 Apr 29 at 1:22