Mass of two bars Two bars of length L have densities ρ₁(x) = 4...

Question

Mass of two bars Two bars of length L have densities ρ₁(x) = 4e^−x and ρ₂(x) = 6e^−2x, for 0≤x≤L.

a. For what values of L is bar 1 heavier than bar 2?

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Let two beams of length M have linear densities in units of mass per unit length of lambda subscript 1 of X is equal to 7 multiplied by the power of negative X, and lambda subscript 2 of X is equal to 10 multiplied by the power of -2 X. Or 0 is less than or equal to X and X is less than or equal to M. for which values of capital M is beam 1 heavier than beam 2. Awesome. So it appears for this particular problem, we're asked to take all the information that is provided to us, and we're asked to determine which values of capital M is beam 1 heavier than beam 2. And with that in mind, let's read off our multiple choice answers to see what our final answer might be, noting that all of our multiple choice answers state that M is greater than some value. So A is the natural log of 2, B is the natural log of 4 divided by 3. C is the natural log of 5 divided by 2, and D is the natural log of 3 divided by 2. Awesome. So our first step that we need to take in order to solve this problem is we need to determine the mass of beam 1. So as we should recall, we can write the let's denote the mass of this beam one to be lowercase m1. So this will be the mass of beam one and to determine the mass of beam one, we will find that it's equal to the integral. From 0 to M of 7 multiplied by E to the power of negative 7 DX. Our second step now is we need to integrate our expression, so we need to take M1, and we we will find when we perform our integration, we need to integrate that M1 will be equal to the integral from 0 to capital M of 7. Multiplied by E to the power of negative XDX, which is equal to square brackets of -7 multiplied by the power of negative X, evaluated from 0 to M. Our third step now is we need to evaluate, so M1 will be equal to square brackets of -7 multiplied by the power of negative X evaluated from 0 to M, which will be equal to -7 multiplied by the power of negative M plus 7. Our fourth step now is we need to determine the mass of beam 2, which we'll call lowercase m2, which will be equal to the integral evaluated from 0 to M of 10 multiplied by E to the power of -2 XDX. Our 5th step is we need to integrate, so we will find that M2 is equal to the integral from 0 to M. So, From 0 to M of 10 multiplied by E to the power of -2XDX, which is equal to square brackets of negative 10 divided by 2 multiplied by E to the power of -2 X evaluated from 0 to M. Which is equal to square brackets of -5 multiplied by E to the power of -2 X evaluated from 0 to M. Our sixth step is we need to evaluate, and when we evaluate, we will find that M2 is equal to square brackets of -5, E to the power of -2 X evaluated from 0 to M. Which will be equal to when we perform the evaluation, -5 multiplied by E to the power of -2 M + 5. Fantastic. So, moving right along here. Our 7th step now is we need to set up the inequality, which we need to note that our inequality that we are setting up will be that M1 is greater than M2, specifically lower case m1 is greater than lower m2. So with that in mind, we can safely write that 7 minus 7 multiplied by the power of negative M is greater than 5 minus 5 multiplied by the power of -2 M. Fantastic. So now our step number 8 is we need to rearrange, and when we rearrange, we will find that 2 is greater than 7 multiplied by E to the power of negative M minus 5 multiplied by E to the power of -2 M. Fantastic. Our ninth step is we need to note that 7 multiplied by E to the power of negative M minus 5 multiplied by E to the power of -2 M is less than 2. So our 10th step is we need to let Y equal E to the power of negative M M to be precise. And we need to note that E to the power of -2 M is equal to Y2. And when we let Y be defined as these values, we can go ahead and write that 7 Y minus 5Y to the power 2 is less than 2. Fantastic. So our step number 11 is we need to note that 5 to 2 minus 7Y + 2 is greater than 0. So that will mean for our 12th step, we need to factor, so parentheses 5 Y minus 2 multiplied by parentheses minus 1 is greater than 0. Fantastic. So, our step 13, we need to note that either parentheses 5 Y minus 2 is greater than 0, and parentheses Y minus 1 is greater than 0. Or, or we can say that parentheses 5 Y minus 2 is less than 0 and parentheses Y minus 1, so Y minus 1 is less than 0. But there's a big but, but we need to note that Y is equal to E to the power of negative M. So 0 is less than Y and Y is less than or equal to 1. So, we also need to note that or so. So, 0 is less than Y and Y is less than or equal to 1, or parentheses Y minus 1 is less than or equal to 0. So as a result, only the later case is valid, or here's another or, or 5 minus 2 is less than 0, and parentheses Y minus 1 is less than 0 are valid. So, with that in mind, our step 14 states that, so, Y is less than 1, and Y is less than 2 divided by 5 are valid. So with that in mind, we need to note that for step 15, so we're on to step 15 now, and our 15th step will state that, as we should recall that Y is less than 1. And E to the power of negative M is less than 1. And the natural log of E to the power of negative M. is less than the natural log of 1. So we could say that capital M is greater than 0, and we can state that Y is less than 2 divided by 5, and we could state that E to the power of negative, M is less than 2 divided by 5. So we could state that M is greater than negative natural log of 2 divided by 5, which is equal to the natural log of 5 divided by 2. So we could safely state that capital M, so once again, M is greater than the natural log of 5 divided by 2. So our 16th step is we need to note that M is greater than 0 and M is greater than the natural log of 5 divided by 2, which yields, so we need to note that this yields M is greater than the natural log of 5 divided by 2. So, for our final step, step 17, we can note that or M is greater than the natural log of 5 halves. We can state that beam one, which I'll call B1, that B1, this B1 value, so B1 is heavier than beam 2. So it's Heavier than B2, which I'll call B2. So B1 is heavier than 2, and we did it. We've solved for this problem officially. So now we could say, without a doubt, looking at our multiple choice answers, so going back to the top to look at our multiple choice answers. Our final answer without a doubt will have to be multiple choice answer letter C, which once again is M is greater than the natural log of 5 divided by 2. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Mass of two bars Two bars of length L have densities ρ₁(x) = 4e^−x and ρ₂(x) = 6e^−2x, for 0≤x≤L.
a. For what values of L is bar 1 heavier than bar 2?

Key Concepts

Density Function and Mass Calculation

Definite Integrals of Exponential Functions

Inequalities Involving Functions of L

Watch next