TITLE: Force as a partial derivative
QUESTION [0 upvotes]: I'm trying to calculate the average power for a 1D sine wave (traveling through a rope with tension $T'$)
$$y(x,t)=A \sin(kx+ \omega t)$$
I start by calculating the instant power: I know that $P=Fv$ and by Newton's law $F=ma=m\frac{\partial ^2y}{\partial t ^2}$ and $v=\frac{\partial y}{\partial t}$ ($v$ is the velocity of an element $dx$ of the rope).
Calculating the partial derivatives I end up with:

$$\frac{\partial y}{\partial t}=A \omega \cos(kx+ \omega t)$$
$$\frac{\partial ^2y}{\partial t ^2}=-A \omega ^2 \sin(kx+ \omega t)$$

And I really don't know how to continue, so I looked at the solution and the professor came up with this:
$$P=Fv=\left(T' \frac{\partial y}{\partial x}\right)\left(\frac{\partial y}{\partial t}\right)$$
I don't understand why $F=T' \frac{\partial y}{\partial x}$ ?

REPLY [2 votes]: We obtain this result from trigonometry and small-angle approximations. Each point in the rope is moving only vertically. For small angles, the vertical force $F$, or vertical component of the tension, is the product of the tension $T'$ and the slope (i.e., $F=T'\sin\theta\approx T'\tan\theta=T'\frac{\partial y}{\partial x}$):

Source: Morin, Transverse Waves on a String, p. 20. Obtained from the fifth result of an online search for "power" sinusoid rope "tension" "force".
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