The study of frost penetration in highway engineering, the temperature T at time t hours and depth x feet can be approximated by, T = T₀e⁻λˣsin(ωt − λx) where T₀, ω, and λ are constants. The period sin(ωt − λx) is 24hours. Show that T satisfies ∂T/∂t = k * ∂²T/∂x² where k is a constant

$$T = T_0 e^{-\lambda x} \sin(\omega t - \lambda x)$$ $$\frac{\partial T}{\partial t} = T_0 e^{-\lambda x} \frac{\partial}{\partial t}(\sin(\omega t - \lambda x))$$ $$\frac{\partial T}{\partial t} = T_0 e^{-\lambda x} (\omega \cos(\omega t - \lambda x))$$ $$\frac{\partial T}{\partial t} = \omega T_0 e^{-\lambda x} \cos(\omega t - \lambda x)$$

Using the product rule:

$$\frac{\partial T}{\partial x} = \frac{\partial}{\partial x}(T_0 e^{-\lambda x}) \sin(\omega t - \lambda x) + T_0 e^{-\lambda x} \frac{\partial}{\partial x}(\sin(\omega t - \lambda x))$$ $$\frac{\partial T}{\partial x} = (-\lambda T_0 e^{-\lambda x}) \sin(\omega t - \lambda x) + T_0 e^{-\lambda x} (-\lambda \cos(\omega t - \lambda x))$$ $$\frac{\partial T}{\partial x} = -\lambda T_0 e^{-\lambda x} (\sin(\omega t - \lambda x) + \cos(\omega t - \lambda x))$$

Again, using the product rule:

$$\frac{\partial^2 T}{\partial x^2} = \frac{\partial}{\partial x}(-\lambda T_0 e^{-\lambda x}) (\sin(\omega t - \lambda x) + \cos(\omega t - \lambda x)) + (-\lambda T_0 e^{-\lambda x}) \frac{\partial}{\partial x}(\sin(\omega t - \lambda x) + \cos(\omega t - \lambda x))$$ $$\frac{\partial^2 T}{\partial x^2} = (-\lambda)(-\lambda T_0 e^{-\lambda x}) (\sin(\omega t - \lambda x) + \cos(\omega t - \lambda x)) + (-\lambda T_0 e^{-\lambda x}) (-\lambda \cos(\omega t - \lambda x) + \lambda \sin(\omega t - \lambda x))$$ $$\frac{\partial^2 T}{\partial x^2} = \lambda^2 T_0 e^{-\lambda x} (\sin(\omega t - \lambda x) + \cos(\omega t - \lambda x)) + \lambda^2 T_0 e^{-\lambda x} (\cos(\omega t - \lambda x) - \sin(\omega t - \lambda x))$$

Factor out $\lambda^2 T_0 e^{-\lambda x}$:

$$\frac{\partial^2 T}{\partial x^2} = \lambda^2 T_0 e^{-\lambda x} [\sin(\omega t - \lambda x) + \cos(\omega t - \lambda x) + \cos(\omega t - \lambda x) - \sin(\omega t - \lambda x)]$$ $$\frac{\partial^2 T}{\partial x^2} = \lambda^2 T_0 e^{-\lambda x} [2\cos(\omega t - \lambda x)]$$ $$\frac{\partial^2 T}{\partial x^2} = 2\lambda^2 T_0 e^{-\lambda x} \cos(\omega t - \lambda x)$$

Substitute the expressions for $\frac{\partial T}{\partial t}$ and $\frac{\partial^2 T}{\partial x^2}$ into the given equation:

$$\omega T_0 e^{-\lambda x} \cos(\omega t - \lambda x) = k \left( 2\lambda^2 T_0 e^{-\lambda x} \cos(\omega t - \lambda x) \right)$$

Assuming $T_0 e^{-\lambda x} \cos(\omega t - \lambda x) \neq 0$, we can divide both sides by this common factor:

$$\omega = k (2\lambda^2)$$

Solve for $k$:

$$k = \frac{\omega}{2\lambda^2}$$

Since $\omega$ and $\lambda$ are constants, $k$ is also a constant.
Therefore, $T = T_0 e^{-\lambda x} \sin(\omega t - \lambda x)$ satisfies the one-dimensional heat equation with $k = \frac{\omega}{2\lambda^2}$.