Lapse rates in the atmosphere Refer to Example 2. Concurr...

Question

Lapse rates in the atmosphere Refer to Example 2. Concurrent measurements indicate that at an elevation of 6.1 km, the temperature is -10.3° C and at an elevation of 3.2km , the temperature is 8.0°C . Based on the Mean Value Theorem, can you conclude that the lapse rate exceeds the threshold value of 7°C/ km at some intermediate elevation? Explain.

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says consider the function G of X, which is equal to the quantity of X minus 2 in quantity multiplied by the quantity of X + 1 in quantity squared on the closed interval from -1 to 2. Determine if Rohl's theorem applies to this function on the given interval. If it does, find the points guaranteed by Rohl's theorem. We have 4 possible choices as our answers. For choice A, we have X is equal to -1, and X is equal to 1. For choice B, we have X is equal to 1. And for choice D, we have X is equal to -1 and X is equal to 0. And for choice D, we have the Rose theorem does not apply to GFX on the given interval. Now, the first part of this question wants us to determine if R's theorem is applicable to our function on the given interval. So there are 3 criteria for Roll's theorem to be applicable. So recall for the first one that we need our function GFX here in this case to be continuous on the closed interval from -1 to 2. Now GFX here is a polynomial, and polynomials are continuous everywhere on the real line. So GFX is continuous on the closed interval from -1 to 2. The second condition is that our function G of X is differentiable on the open interval from -1 to 2. And here again, since we have a polynomial, it is differentiable everywhere on the real line, so therefore, our function G of X is differentiable on the open interval from -1 to 2. And finally, the last condition is that our function G of X evaluates to the same value at the end points of our interval. So we need to calculate those values. So we need to calculate G minus 1 for the lower bound, as well as G2. So we'll start off with G minus 1. So this is going to be equal to using our function for G of X. This is going to be the quantity of -1, -2. In quantity, multiplied by the quantity of minus 1 plus 1 in quantity squared, which is equal to minus 3 multiplied by 0 squared. Which is equal to 0. So now we need to calculate G of 2. Which is going to be equal to the quantity of 2 minus 2 in quantity, multiplied by the quantity of 2 plus 1 in quantity squared, which is equal to 0 multiplied by 3 squad, which is equal to 0. So, here, both of our um values or G minus 1 and G of 2, evaluate to zero. So, our third condition for Roll's theorem um is um true. So that means that Ruhl's theorem is applicable. So now we need to find the points guaranteed by Ruhl's theorem. And so recall for Rose's theorem that says that there is a point X. Such that G of X. is equal to 0 in this case, on the open interval from -1 to 2. So, first thing we need to do is calculate RG X, and then we'll set equal to 0 and solve for X. So calculating G of X. That is going to be equal to the derivative with respect to X. Of the quantity Of X minus 2 in quantity, multiplied by the quantity of X plus 1 in quantity squared. And so here to take this derivative, we're going to have to apply our product rule. So this is going to be equal to, and here we call your product rule. So we'll need to take for our product rule, the derivative of the first term in our product. So we'll take the derivative with respect to X of the quantity of X minus 2. When I take that derivative, I get 1, and that gets multiplied by the second term in our product, which is going to be the quantity of X + 1 in quantity squared. Then we'll have plus our first term in our product, which is the quantity of X minus 2 in quantity, multiplied by the derivative with respect to X of the second terminal quantity in our product, sorry, um, which is the quantity of X + 1 in quantity squared. So we're going to take a derivative of that. We'll get 2 multiplied by the quantity of X + 1 in quantity multiplied by 1. So here we've applied both our power rule as well as our chain rule. So, we're gonna need to simplify this expression, that this is going to be equal to. For the first term, I'll multiply out our binomial, so we'll get X2 + 2 X plus 1, then we'll have plus. 2, multiplied by the quantity, and here we'll multiply our two terms that have X in them, and then we'll get X2 minus X minus 2 in quantity. So distributing the two, this is going to be equal to. X2 + 2 X + 1 + 2 X2 minus 2X minus 4. So collecting like terms? This is going to be equal to. For X2 terms, we have X2 + 2 X2, which gives me 3 X2. For the linear terms, I'll have 2 X minus 2 X, which gives me zero, and then for our constant terms, we'll have plus 1. -4, which will give me -3 for a constant term. So now we need to set this equal to 0 and solve for X. So to solve this equation for X, we need to add 3 to both sides, so we'll have 3, multiplied by X2 is equal to 3, then we'll divide both sides by 3 and we'll get X2, is equal to 1, and then we'll take the square root of both sides. So that means I will have X equal to Plus and minus 1. So here we have to be careful about which answer we actually choose. To recall that Rose's theorem, the solution needs to be within the open interval of what we are given. So here we're looking for the open interval of -1 to 2. So in this case, -1 is not included in that open interval. So that means that our answer here is just going to be X equal to 1. So we ignore the X equal to -1 result because it falls outside of the open interval from -1 to 2. So this is the answer that we're actually looking for, and when we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice B. So I hope that this explanation has been useful, and I'll see you in the next video.

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