# Homework Help: Temperature at depth

1. Mar 5, 2015

### bobred

1. The problem statement, all variables and given/known data
Determine at what depth below the surface the temperature never falls
below zero.
2. Relevant equations
$$\theta(x)=\bar{\theta}+\theta_0\cos\left(\omega t-\sqrt{\dfrac{\omega}{2D}}x\right)\exp\left(-\sqrt{\dfrac{\omega}{2D}}x\right)$$
where the average is $\bar{\theta}=2.5$ and amplitude $\theta_0=7.5$
3. The attempt at a solution
At a depth x the temperature will be reduced by a factor
$$\exp\left(-\sqrt{\dfrac{\omega}{2D}}x\right)$$
so for some ratio R
$$R=\exp\left(-\sqrt{\dfrac{\omega}{2D}}x_{min}\right)$$
I'm having trouble figuring out what R is so that I can find $x_{min}$.
Any suggestions?

2. Mar 5, 2015

### bigplanet401

Is theta the temperature? If so, you have something that looks like

$$\text{Temperature} = \text{const} + A \cos (\text{something}) \times e^\text{something else}$$

You never want the temperature to fall below zero. The lowest value of the cosine function is -1. So ask when

$$0 \leq \text{const} - A \times e^\text{something else}$$

3. Mar 5, 2015

### bobred

Hi
Theta is temperature. The temperature ranges from +10 to -5.

4. Mar 5, 2015

### bigplanet401

Thanks for clarifying. At the surface (x=0), the temperature is in that range. The exponential factor in the second term lowers the "swing" in temperature. Now you need to solve for the depth at which this swing is low enough so that the temperature is never negative. Have a look at the last equation in my earlier post.

5. Mar 6, 2015

### bobred

Confusing myself, here is an example I have been looking at

6. Mar 6, 2015

### haruspex

First, the equation in the OP isn't quite right. It should be $\theta(x, t) = ...$, as shown in the later post. Pick some depth x. What is the minimum temperature that it can be at that depth?

7. Mar 9, 2015

### bobred

Oops just noticed, the cos should include $+\phi$