Kinetic energy density in a string (derivative)

[boomdoom]

Homework Statement

In our physics course, we were studying one dimensional waves in a string. There, our teacher stated that the kinetic energy in a small piece of a string is $$dK=\frac{1}{2}μdx\frac{\partial y}{\partial t}^2$$ were μ is the linear density of the string, so he claimed that $$\frac{\text{d}K}{\text{d}x}=\frac{1}{2}μ\frac{\partial y}{\partial t}^2$$ which is impossible for me to understand, since K is a function of both x and t.

The Attempt at a Solution

Shouldn't we find the derivative of K=K(x,t) as $$\frac{\partial K}{\partial x}=\frac{\partial }{\partial x}(\frac{1}{2}μx)\frac{\partial y}{\partial t}^2+\frac{\partial }{\partial x}(\frac{\partial y}{\partial t}^2)\frac{1}{2}μx$$
What am i missing?

 

[Dr.D]

I am going to assume that K is your symbol for kinetic energy (T is the more classical choice).

The statement for dK is correct. This is a differential mass multiplied by the square of the velocity and divided by 2.

The kinetic energy (K) is a single time depenndent quantity, not a function of position. To find K(t), you would need to integrate the expression for dK over the length of the string.

Why does he even bother to write dK/dx? Does he really mean a total derivative, or perhaps a partial derivative?

 

[boomdoom]

Thanks for replying,
it makes sense that kinetic energy is only a function of time physically, however by looking the equation above since y=y(x,t), shouldn't ∂y/∂t be also a function of both position and time?

 

[Dr.D]

With y = y(x,t), then partial y wrt t is a function of both x and t, so that much is correct.