91Ó°ÊÓ

Partial derivatives are a fundamental tool in calculus, especially when dealing with functions of multiple variables. They allow us to see how a function changes with respect to one variable while keeping the others constant.
In our problem, we have a function of two variables, \( u(x,t) = \sin(x - kt) \). To solve the wave equation, we need to compute its partial derivatives with respect to \( t \) and \( x \).

• First, we find the first and second derivatives with respect to \( t \):

* For \( \frac{\partial u}{\partial t} = -k \cos(x - kt) \).
* For \( \frac{\partial^2 u}{\partial t^2} = -k^2 \sin(x - kt) \).

• Then, we find the first and second derivatives with respect to \( x \):

* For \( \frac{\partial u}{\partial x} = \cos(x - kt) \).
* For \( \frac{\partial^2 u}{\partial x^2} = -\sin(x - kt) \).
By understanding these derivatives, we can substitute them into our wave equation to find the solution.

Trigonometric functions play a crucial role in this problem, particularly the sine and cosine functions. Both are periodic functions and have derivatives that are also trigonometric functions.
When dealing with expressions like \( \sin(x - kt) \), we often encounter how these functions interact with differentiation.

• The derivative of \( \sin(x - kt) \) is \( \cos(x - kt) \), showing the cyclical nature of these functions.
• Similarly, the derivative of \( \cos(x - kt) \) goes back to \( -\sin(x - kt) \), emphasizing the periodic property.

This cyclical change is vital when solving differential equations like the wave equation. Understanding these derivatives helps us manipulate and substitute into equations more straightforwardly.

The one-dimensional wave equation is a second-order linear partial differential equation commonly expressed as \( \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2} \), where \( c \) is a constant representing wave speed.
In our exercise, this becomes \( \frac{\partial^2 u}{\partial t^2} = 36 \frac{\partial^2 u}{\partial x^2} \). This kind of equation models how waves propagate through a medium in one spatial dimension.
Solving this equation typically involves finding a balance between the time and space derivatives. For \( \sin(x - kt) \) to be a solution, its derivatives must satisfy the equation's relation.
Practically, this means deriving the right \( k \) that aligns the equation correctly, where \( k^2 = 36 \).

Mathematical problem solving is an essential skill that involves logical reasoning and systematic approaches.
In this problem, we follow a step-by-step process:

• Identify the problem: We begin by understanding the components of the wave equation and recognizing \( u(x,t) = \sin(x - kt) \) as our intended solution.
• Compute necessary derivatives: This involves calculating the partial derivatives with respect to both \( t \) and \( x \).
• Substitute and simplify: By placing these derivatives back into the wave equation, we can see if they satisfy the equation.
• Solve for unknowns: Finally, solving \( k^2 = 36 \) gives us potential values for \( k \).

This structured approach makes it easier to tackle complex problems by breaking them into manageable parts.