the-answers-to-most-of-the-following-exercises-are-in-terms-of-logarithms-and-exponentials-a-calcu-5

Question

The answers to most of the following exercises are in terms of logarithms and exponentials. A calculator can be helpful, enabling you to express the answers in decimal form.
The intensity $L(x)$ of light $x$ feet beneath the surface of the ocean satisfies the differential equation
$$\frac{d L}{d x}=-k L$$.
As a diver, you know from experience that diving to $18 \mathrm{ft}$ in the Caribbean Sea cuts the intensity in half. You cannot work without artificial light when the intensity falls below one-tenth of the surface value. About how deep can you expect to work without artificial light?

Bobby Barnes · Accepted Answer

So we have this differential equation that describes how the intensity of light changes as we are ex feet below the surface of the ocean. We&#x27;re told that 18 feet below the surface would give us half of our intensity that we would get from the surface. And we want to determine at what point will must diving down reach 1/10 the value of the surface level. But I have 1/10 of that shack Should be 1/10 l not right. So it doesn&#x27;t matter the l knots you&#x27;ll see as we go through the steps so cancel out, but just kind of writing get out the way they, uh, state and maybe should also write that l of zero is l&#x27;m not. That will be our initial surface. So let&#x27;s go ahead and start solving this differential equations. So first, we&#x27;re gonna divide each side by l. So we&#x27;re gonna get DEA one over L d O is equal to negative K DX. If we were toe, multiply the DX over. Now, Once we have this different this incontestable equation, we could go ahead and integrate each side. The left hand side should integrate to the natural log of the absolute value of, though is equal to then on the right hand side. Negative K X, plus some constancy. No, Since they don&#x27;t want us to solve for the differential equation, all they want us to do is figure out. When does this hold? We need to start plugging everything into here. Don&#x27;t stoop to that. So let&#x27;s use our first condition. Oh, that l of zero is zero to El Ni so we can plug that into give so l of zero Oh, so the absolute value of el not will intensity should be positive. So absolute value. That should just be l&#x27;m not so we don&#x27;t have anything to worry about that if we just dropped the absolute value and then this is going to be equal to negative k time zero, which is gonna be zero plus our constancy. So we just solved to see that C is equal to the natural log of the absolute or the natural log of l not. All right, so that&#x27;s seems pretty good so far. Now that shoes are second condition here, help us all. For what K is so remember. This is saying X is equal to 18 and l is gonna want half l not. So we&#x27;re gonna have a natural log of 1/2. No, not is equal to the negative of K times 18 plus the natural log of L. Not now. We want to get Kay by itself. So we&#x27;re gonna want to subtract hell. Not over. Er the natural log a bell knots. That&#x27;s gonna give us natural log of 1/2 l not minus natural. Log of El Not. And then in doing that, we would just be left with K times. Negative eight teams Such divide this by negative 18 while we&#x27;re at it. And this should give us our value for Kay. Now, notice when we have to natural logs being subtracted like this, we can write it as the division of the insides. So that would actually give us that. Kay is even too negative. Won over 18 of the natural log of 1/2. All right, so we have this snow and let&#x27;s get rid of that negative outfront. Remember, we can do that by pulling it in to be the power. Oh, 1/2 of the natural log by the power rule. So there&#x27;s gonna be 1/18 natural log of two. Hurry, Majola line a little bit better now let&#x27;s just plug everything in and write out our equation. This start. So we have that Ellen of the natural log of L. There&#x27;s going to be equal to negative Ex Natural log up to you over 18 plus the natural log. Oh, l not. So this is the equation that we have. Well, let&#x27;s go ahead and plug in to figure out when this should be good to 1/10 of l not. So we want to solve for X this time. So we&#x27;re gonna have Ellen is equal to 1/10. Well, not is equal to negative X Natural. Log on to over 18 plus natural log of el Not so we&#x27;re going to subtract over. So, just like before, the natural log should combine into one. And the l knots would just cancel out, which would give us Ellen of 1/10. So this is why I was saying at the start, the natural logs are the l knots Don&#x27;t really matter all that much. So we have negative X natural log. 02 all over 18. Now just solving for X we should end up with X is equal to And I&#x27;m gonna rewrite this as negative natural log of 10. So we&#x27;re going to multiply by 18 and then divide by negative natural log to sew these negatives here counts out with each other. And then So this would be our exact answer and feet bullets actually approximate this because I have no idea what that number is even supposed to mean. So natural Aga 10 is about 2.3. Multiply that by 18 get about 41 then divide that by the natural log of two. And this gives about 59 27 9 beat. So around this here is wind. We would need to use some kind of artificial life.

Doruk Isik · Answer

in this problem, working in the D l D X is equal to, uh, thank you, J l were also given that after 18 feet the intensity as one. How offense? Origin. Um value. Since this director is proportional to the, uh, variable itself you don&#x27;t have done, the function will be up to form. I&#x27;m not. This is a difficult mission. Exponential. Okay. To the, uh, ex. Okay, we&#x27;ll use this different condition. Two o&#x27;clock flight. Okay. Celester at El 18 would be people 18. This given a question, we would have eloped. Exponential leg today. T k. And that is ico Ella over to those. No way from this. If you take natural log about science, we see that K is equal to nature. Look to divide by 18 so that our function is unloved. Exponential minus nature. Log off too. S over 18. Okay. No. How much would it take for it to be one time opus? Initial value is what we&#x27;re trying to find so that that is equal to Ella. Exponential of minus or negative. Saliva to x over 18 l a s o l once kids out. So we have one or two and dad is equal. Exponential. Negative. Next look off to x over 18. You&#x27;d be technical of both times. You&#x27;ve got natural off 10 disease of the negative Nixon logo, too. Or 18 eggs, some firmness. We see that X is a book than 18 times Marshall Overton, direct financial, um two. And that that is you conclude about 60 He

Bobby Barnes · Answer

they tell us about how this colony of bacteria can start with one bacterium and doubles every half hour. Ed, it has an exponential growth, or we want to determine by the end of 24 hours How many bacteria will this colony have? So what&#x27;s seat? So since they tell us his exponential growth, we can already know that we&#x27;re going to set this up to be so I&#x27;m using P to be the population of our quality. So there&#x27;s gonna be P A. T is equal to see Time&#x27;s E to the K T. All right, so we have this equation. Well, let&#x27;s go ahead and use our initial condition here to solve for C. So that&#x27;s p of zero is a good one. You got a C E o to the k t eat k zero, which that&#x27;s just gonna b E to zero, which becomes not 01 So then that tells us that see, you should just be warned so we can go ahead. And right now, Petey is just equal to e t k t. Now we need to solve for Kay, and we can use the fact that the population should double every half hour. So if our population started at one that&#x27;s gonna tell us that p of 1/2 should be equal to to get us just double. Forget about two is even too e to the 1/2 K. Now we will take the natural law on each side, and that would cancel out and we&#x27;d have 1/2 K so it multiplied by to get Kay is secretive, two natural to which we can rewrite as the natural log of or using the X golden rule for lager rhythms. All right, so let&#x27;s write out what P a. T is, and then we can just plug in 24. So this is going to be he two. The tee times natural log. This should be e to the natural log. Oh, tee times the natural log for I don&#x27;t know why I was I have a hard time figuring out what You&#x27;re right up there. Now we want to go ahead and plug in 24 but actually first knows that weaken, right? This actually in the following way, we could pool in that tea to get natural aga or to the T and in the E in the natural law caps out with each other and we can write. This is just for to the teeth. So this is gonna be P A t. Now let&#x27;s plug in the fact that we want to check for 24 which would give us or raise to the 24 on Let&#x27;s see what the best numbers or raised to the 24. Also, there&#x27;s a huge number, so it&#x27;s approximately equal to two times two times points or 2.1 eight times 10 to the 14 back till you. So if you want the exact answer, it&#x27;s four to the 24 approximately that 2.81 times 10 to the 14.

Bobby Barnes · Answer

So they give us this differential equation model that is supposed to tell us about the change in pressure with respect to our altitude. And they first wants to solve this initial value problem that they give us. So let&#x27;s go ahead and do that. And then they tell us to Saul or everything in terms of these two actual values that they give us. Let&#x27;s first figure out what we have that we comply everything in from there. All right, so this is what we&#x27;re going to solve. Well, that equation there is a slipper be differential equation. So let&#x27;s go ahead and first divide over by P and multiply by D h. So we&#x27;re gonna have one overpay dp is equal to, and so are constant k Th Now we can integrate each side and we&#x27;re going to take the natural log. Are integrating the left hand side is gonna be the natural log of the absolute value of Pete is equal to and then integrating a constant here. Well, that should just be K h plus from constant C. All right, an actual hold a C one for right now. Now we want to go ahead and exponentially ate. Each side is it won&#x27;t matter about the absolute values, coz in this sense it doesn&#x27;t really make sense to talk about negative pressures. So we&#x27;ll never have to plug that in or ah, pressure of zero doesn&#x27;t really make much sense. Either of those council and there were still have p is equal to e to the K h must see one. And now we could rewrite this cause. Remember, when we have things being added as exponents, it would be e to the C one times e to the K h, but then eat. The C one is also just a constant so we can rewrite. This is just see two times e to the cage. All right, so this is what our solution is going to be. But we&#x27;re also told this initial condition right here, let&#x27;s go ahead and plug that in. So we know when p or when h is equal to zero Misha equal to Pena. So this is gonna be able to see to e to the k time zero so that she&#x27;s gonna be eating zero, which ends up just being one Landau with C two is peanuts. So our equation that we&#x27;re going to have is p of H is equal to it. Looks like p not times e to the K h. So this is just our general solution to it. So right here, general solution. But now they want us to use this information here. We&#x27;ll see Level is when h is equal to zero. So this right here tells us what p not is. So we could just go ahead and plug that in. So we&#x27;re gonna have 1013 that we have e to the K H. Now we want to solve for Kay, and then we can use this one here, so this is going to be a TSH. And this is going to be P S O p. Of h. Now plug and all that, and we should get 90 is equal to 1 13 mmm to the K times 20. And now we just want to first divide and then take the natural log on each side. So that&#x27;s going to be 90/1 13. Naturally, zoom out a little bit, so 90 over 1000 and then we&#x27;re gonna want to take the natural log. Like I was saying, and that should give us K times 20. Then we would divide each side by 20 to get K is equal to 20. Natural log. That should have been 10 13. So natural log of 90 over 1 30 and sold out. Lastly, we can plug everything in to this equation. Here, take it. That p of h is going to be equal to. And actually, let&#x27;s just, uh, approximate this number actual. Plug it indirectly. So 10 13 e raised to the 20 natural log of 90/10 13 each. So this is what it is. Exactly. And if you wanted, you could just approximate our value for Kay here and let&#x27;s see what that is. So natural log 0 90 divided by 10 13 times 20. So this is around negative, wordy. Thanks point. Or so you could just replace that with what we have in the longer if you don&#x27;t have to write all that up. But I would say that is our solution for everything that they give us. Now we want to determine what the pressure is at a height of 50 kilometers, So we&#x27;re gonna plug in 50. So that&#x27;s gonna be P of 50 is equal to. So I&#x27;ll say approximately equal to 10. 13 e to the A, And I&#x27;m gonna use that estimation this time. And then this is times 50. So we&#x27;re gonna multiply that number by 50 and then exponentially ate it and actually give the super small value so it&#x27;s actually rate that out. So this is going to be approximately in 13 times Hey, to the negative to or 30 And even when I plug this into my calculator, this just says it&#x27;s approximately zero. So this is extremely small number. All right, so this may be so far out that the pressure is negligible ended. Lastly, we want to figure out when it would be 50 millibars and let&#x27;s see. So that means we&#x27;re going to set our function equal to 900. And in salt for h that would be 10 13 or actual approximately because I&#x27;m going to use the estimation for it again. So negative 48.4 times teeth. So first up is gonna be the same. We&#x27;re going to go ahead and divided side by 10 13. Take the natural log that I should give us natural log of 900 over 10 13 is approximately equal to negative. 48 went or teeth. Then we would divide each side by negative 48.4. And that&#x27;s gonna give us natural log of 900 over 10. 13 over negative, 48.4 at the yourself approximately here. And then let&#x27;s see what we get for this number here. So natural log of 900 divided by 10. 13 and then divide that by negative 48.4. So this is about 0.0 two. Well, Mr Ned Beatty, that should be, uh, h let me replace on those cause this is supposed to be a height, so this would be in kilometers so you could multiply that by 1000 which would be about 2.4 meters. So this is about where that&#x27;s pressure will occur.

Bobby Barnes · Answer

So we&#x27;re told that this frozen human mummy was estimated to have died about 5000 years ago. And we want to determine how much of the original carbon 14 remained at the time of his discovery. Well, in example five, they tell us that for C 14 carbon dating, we will have our kay is equal to natural log of two over 5730. And the mount left will follow just the typical exponential decay equation. So for 5000 years ago, that means we&#x27;d want to plug 5000 in to that, so we&#x27;d have why 5000 is equal to. So why not times E to the negative K. So I was going to be natural log of two over 5000 730 and then we multiply this by 5000. So let&#x27;s see what that some natural law goes to times 5000 is about 3465. Then divide by negative 5000 730. So this is going to be why not times e to the negative zero 0.6 about and then exponentially eating. That should give about zero point by for six times. Why not? And so that we multiply it by 100 should be our percent left, which would give us about 54.6 per cent of those C 14 that would be left.

Bobby Barnes · Answer

So we&#x27;re told that something is considered in danger if after 20 years, the population is expected to be extinct or within 20 years. So we&#x27;re told that some wildlife has 1147 members at the current time, and population is following a steady decline exponentially, and it&#x27;s been averaging a rate of 39% over the past seven years. We want to determine if this species is going to be endangered and we&#x27;re gonna explain your answer. So here I write endangered if p of 20 is less than one. So if we have some population function pft Um, since they tell us it&#x27;s exponential decline, we know our equation should look something like this to start. So P of t is equal to p, not times e to the negative Katie, and this is because it&#x27;s declined. We have that negative there. And so we say, strictly less than one, as opposed to equal to zero, because exponential can ever equal to zero, at least with this model here. So that&#x27;s why we&#x27;re looking for less than one now. Let&#x27;s go ahead and figure out what to plug it. Soapy. Not is supposed to be our initial population. And we&#x27;re told initially it is 1147. So this is gonna be peanuts. So it&#x27;s got unplugged batted, and this is just from the equations that they give us in the chapter. So we don&#x27;t need to work harder than we need to, and we just plug everything in that they tell us we can use now. This next part, it says every year for the past seven years has been declining around 39%. So that means, after one year, soapy of one should be equal to zero point 61 of 11 47 which is p not so. We&#x27;ll just go ahead and set this equal to our function over here, and we&#x27;ll have e to the negative K, and we need to plug it one for teeth. Now. The reason why I wrote it out as six point one 4.61 as opposed to multiply and those together is notice that when we would divide by our initial population, those just can&#x27;t stop at this point, we would want to take the natural log on each side and then multiply by negative one and that&#x27;s going to give us. Kay is even too negative. Natural log of 0.61 So we have Kay. Now let&#x27;s go ahead and plug that into our original equation. So we&#x27;re gonna get that population after tea years is going to be a 1147 e to the So the negatives will counts out. So it should just be natural of 0.61 times teeth. Now we want to check to see if P of 20 is less than one. So let&#x27;s plug in 20. And actually, I&#x27;ll be a little bit lazy and we&#x27;ll just replace thes tease right here with 20. And so let&#x27;s see what we get after we plug. Everything is so this is going to be approximately 1107 and in the numerator 20 times. The natural log a point 61 is he to the negative 9.88 about and now exponentially ating. That gives us something around five times 10 to the negative fifth and then multiply that by 1147. So this is going to give us the population of about 0.5 So we did. We can&#x27;t have a fraction of an organism and it&#x27;s considered to be alive. So this is less than one which meets our criteria of being endangered. So we would just say something like, yes, since P of 20 is equal to 0.5 which is less than one.

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The answers to most of the following exercises ar…

Problem 27 Medium Difficulty

Answer

$$59.8 \mathrm{ft}$$

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Joel Hass, Christopher Heil, Przemyslaw Bogacki

University Calculus: Early Transcendentals

Catherine R.

Heather Z.

Kayleah T.

Kristen K.

Integration Review - Intro

Definite vs Indefinite

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$$59.8 \mathrm{ft}$$

University Calculus: Early Transcendentals

Catherine R.

Heather Z.

Kayleah T.

Kristen K.

Integration Review - Intro

Definite vs Indefinite

Joel Hass, Christopher Heil, Przemyslaw BogackiUniversity Calculus: Early Transcendentals

Catherine R.

Heather Z.

Kayleah T.

Kristen K.

Joel Hass, Christopher Heil, Przemyslaw Bogacki

University Calculus: Early Transcendentals