solve-the-problems-in-related-rates-the-shadow-of-a-24-mathrmm-high-building-is-increasing-at-the-ra

Question

Solve the problems in related rates.
The shadow of a $24-\mathrm{m}$ high building is increasing at the rate of $18 \mathrm{cm} / \mathrm{min}$ when the shadow is $18 \mathrm{m}$ long. How fast is the distance from the top of the building to the end of the shadow increasing?

Will Erickson · Accepted Answer

either. So in this problem, we have, uh the problems actually illustrated for us. So let&#x27;s copy that picture down. First of all, we have a five meter lamp. Oh, so this is always five. You have a man whose height is 1.8. So this is always 1.8, and the lamppost is casting a shadow on the ground here as he&#x27;s walking. So as the book, the book is calling this distance X with this red distance here because calling X and then this blue part, which is the shadow, This is the part we care about. We&#x27;re calling that. Why? So we have a right triangle. Um good. And the other thing we know is that he&#x27;s walking at a speed of 1.2 meters per second. Mr, In other words, of this red side here is increasing at 1.2 meters per second, so we can actually translate that into math. If he is walking that quickly, that means his distance away from the lamppost X is changing at that rate. So what they&#x27;ve actually given us is DX DT. And it&#x27;s positive 1.2 because, uh, because X is growing at 1.2 meters per second. So they have told us that. Okay. What we want to get is how fast the shadows growing. We want d y t t the rate of growth of side. Why? Okay, what&#x27;s our equation? This is trickier than a lot of them because it&#x27;s not a Pythagorean theorem. Ah, problem. We don&#x27;t know much about the high pot news at all here. So what&#x27;s the one thing we do know? Well, the one thing we know is that these triangles are similar triangles. The small triangle here a lot lining green into the small triangle and the large triangle, right? They&#x27;re similar triangles. So what do you know about similar triangles? Well, don&#x27;t think what we know about the side lengths, but we know that five is Thio 1.8. As you know, you&#x27;re you have to be a little careful. What is the overall bottom? While the bottom of the big triangle is X plus y? And the bottom of the small triangle is just why? So make sure you can see this. This is just similar Triangles have proportionate sides. That&#x27;s all we used here. Okay, so now at this point. Is there anything that we can do to make the math easier before we take derivatives? Um, I would say so. So let&#x27;s see. Why don&#x27;t we actually multiply both sides by 1.8 and then multiply both sides by why? Or if you prefer to stay cross products So right five times y equals 1.8 x close 1.8. Why, okay. Or, in other words, you can do even better. And if we combine our I like terms five minus 1.8 gives us 3.2. Why equals 1.8 x? Okay, so we have a nice equation relating X and y. Now, when we take the derivative so we&#x27;ll take both sides derivative with respect to t and see what we get. So you have the derivative of 3.2. Why, it&#x27;s just 3.2. Then we need to multiply radio I d. T on the right. Likewise, the derivative of 1.8 x is 1.8, but we need two more fly by DX DT. And since we want to solve for D Y e t, all we have to do is divide by 3.2 and we&#x27;ll get d Y t t by itself. 1.8 over three points. You d x t t Remember, we know that&#x27;s 1.2. So in the end, we have it. We just have to calculate this out. So we need to d&#x27;oh! 1.8 times one point to you and then divide that answer by 3.2, we should get 0.675 Now, let&#x27;s just make sure we have the right units. We&#x27;re measuring distance and this problem distances of meters and a unit of time for our rate in this problem. This seconds. So we get points 675 meters per second. Kind of a tricky problem. Hopefully, this helped

Stephanie Martinez · Answer

the best way to solve related rates. Problem is by first crying a picture of all the information that is given. So we know that the building is 80 feet tall and the shadow that I cast is 60 feet. It&#x27;s all, and there is an angle between the grounds of the sun, which is directly over the building, and that angle is called data and data is increasing at a rate of 0.27 degrees per minute. On the problem says that we need to use inches permanent as a unit for final Answer radiance, and we need to round to the nearest 10th. So let&#x27;s first start off with some conversions has come for a 0.27 degrees to radiance, and we do that through dimensional analysis. So we have 180 degrees at the bottom and pie radiance at the top degrees will cancel out, and it will leave us with an answer in radiance. So take out your calculator, Andi, multiply 0.27 times pi over 1 80 You should get 0.47 radiance permanent, and this is a rate. This is a rate because It&#x27;s a unit of measure over a unit of time, so the way that we represent a rate is de variable over DT. So in this case, it&#x27;s deep data over DT cause it&#x27;s with respect to time and that equals 0.47 Radiant is permanent. Okay, so let&#x27;s go over our given information. So this is D data over DT. And over here we can consider the 60 feet to be X and the 80 feet to be Why So now that we did all of our conversions and everything will convert feet two inches at the end, let&#x27;s figure out, what are we trying to solve? Well, the problem says to find the rate at which the shadows length decreases. So that&#x27;s this. And I have said before the rate is just D whatever variable over DT. So that means that we&#x27;re trying to find d X over d T. And the way that we do that is by first finding an equation which incorporates all of the given information and the best equation to do that is tangent of Fada equals Y over X. And before we start working with this, let&#x27;s substitute any numbers that are constant. Well, this rate right here isn&#x27;t constant because it&#x27;s a rate. It&#x27;s changing. This 60 feet right here is in constant because it&#x27;s changing. It&#x27;s decreasing this 80 feet, though that&#x27;s the height of the building, and the height of the building doesn&#x27;t change. It&#x27;s going to remain constant, so we can substitute 80 for why? So now we have this. And in a related rates problem, the way that we solve it is by finding the equation and then doing implicit differentiation. So we&#x27;re going to take the derivative of both sides, and every time that we take the derivative off a variable, we need to multiply it. By the derivative of that variable, I&#x27;ll show you what I mean. Enough&#x27;s in a moment. So the derivative of Tangent is Seacon square of data, and we just took a derivative of tangent. So that means that we need two more supply by D data over D. C. Let&#x27;s move. Let&#x27;s, uh, differentiate the right side of the equation. What? We&#x27;re gonna have to use a little bit of quotient rule. Remember the quotient rule is you prime. The minus you really prime over the square. So the derivative of you, which is the numerator, is zero V is X minus 80 and we took the derivative of the denominator, which is eggs. So which is the derivative is one. And since we took the derivative of X, we need to multiply by D X over duty and that&#x27;s all over B squared, which is X squared. So now let&#x27;s write this here. It&#x27;s gonna be negative. 80 d x over d teen all over x squared. So let&#x27;s start substituting in all of the information that we know. So we know that we know that this debate over DT is 0.47 radiance per minute. And this right here this DX over DT is but we&#x27;re solving for that&#x27;s the variable which we&#x27;re solving for an ex axes right here. Excess 60 feet. So the only piece of information that we&#x27;re missing is this angle. What is this angle? Well, how can we solve it while we have X and why? So let&#x27;s do an inverse tangent. Inverse tangent of y over X, which is 80/60. The injuries tangent of fatum equals 80/16 and when you When you put this into the calculator, you get that data equals 53.13 degrees. So that means that we&#x27;re going to take this Seacon Square a 53.13 So the way that you plug this into your calculator right here, just gonna plug in one over, you&#x27;re gonna type one over co sign of 53.13 square just like that, because number that seeking is one over the co sign. But most calculators don&#x27;t have seeking as a function, so you have to type in one over co sign. So let&#x27;s start crunching the numbers in the calculator. So this right here will give you two point seven are 78 times 0.47 equals negative 80 d x over d t over 3600. And these two numbers right here, gold supply to be 0.13 And now you have to multiply 3600 by 0.13 which will give you 46.99 as a result, the equals negative 80 t x over d t. And now you&#x27;re gonna divide by negative 80 and you will get that D X over D T equals negative 0.5 eight or nine. I was saying line so 0.59 feet per minute, and it makes sense that it&#x27;s negative because the shadow link is decreasing. The only thing is is that we need one more step, and that&#x27;s to converted into inches, because that is what the problem is asking for. So it&#x27;s just simple dimensional analysis fee will supply by one foot equals 12 inches. The feet will cancel, so you&#x27;re going to get negative 7.0 five inches permanent, and you have to multiply that round. I&#x27;m sorry with around to the 10th place, so that&#x27;s around negative 7.1 inches permanent, and that is your final answer.

Matt Just · Answer

Okay, so we have a building with the sun shining on the building. Here&#x27;s our son and thank you. Shadow the length X. Let&#x27;s say I mean, the angle of the shot is making the sun. You know, the height of the building is eighty faint good. And so what do we have? Well, we see that. Okay, so we know that detail. Aditi is twenty seven degrees permanent when we want to convert that to radiance. So that&#x27;s what What is that? Twenty seven degrees is pi over one eighty times point two seven. So to serve in a pie over one eighty, this will be radiance per minute. Okay. And C is that simplify any, uh, twenty seven over eighteen. Something like three over too. Yeah, Ass of three or two. So slicks like three pie over two thousand radiance per minute. No, that&#x27;s not so important. So we want to find t x t t when X equals sixty. Cool. So it looks like we&#x27;re going to use the tension. Excuse O tant data is going to be eighty over X, so that will give us a relationship between data and X. So the secret script data you say that? T it is eighty times minus one over x squared d x d t. And now we needed no seeking anything. Well, when X is sixty, the high partners here will be one hundred sixty. Script was eighty squirts one hundred and so a second data is going to be adjacent of ripeness. That&#x27;s five over three. Yeah, and so see, it&#x27;s great. It&#x27;s gonna be twenty five over nine teeth. Entity Tio, he said, was three pi over thirty thousand time over two thousand him and then this is equal Teo eighty negative, eighty over sixty squared. That&#x27;s thirty, six hundred and the DX DT. So we&#x27;re looking for And so we work all this out. We get the X t t is negative. Eight no negative three high over sixteen, which is about to your point five eight nine Great. Her minutes a little over six inches permanent

Joshua Eastwood · Answer

the question gives us a building and then the sun is up here. So here&#x27;s my son, and it&#x27;s shining down in the building is here and then we have a shadow from the building. So the question says that the building is 80 feet tall and it gives us an angle fatal here. And it says that decided by D. T is 0.27 degrees her minute and then the questions What rate is a shadow decreasing? So we&#x27;re gonna call this X, and I&#x27;m looking for a DX by d. T. So what I&#x27;m gonna do is I&#x27;m gonna say that TANF data is equal to 80 over X, and it tells being that at this particular time X is 60 feet. Okay, so I&#x27;m gonna take the derivative, and I&#x27;m gonna get Seacon Square data. The fate of I d. T. Is he with a minus 80 over X squared times. The exp i t t because the dirt of one over X is one over X squared minus Rex were Okay, So I get this Yes, by the tea. Sorry, not TX by the X. So the question First of all, he is What&#x27;s data. We need to know what they do is or we need least you know what secret date is. We know what X is. We know what the expert ET is in terms of. That&#x27;s what we&#x27;re looking for. And then we know what defeated by E. T. Is. But we have to be careful because they visit degrees and we need to change its radiance. Okay, we&#x27;re gonna do some of that. So first of all, she can square data. I&#x27;m gonna deal with that. So if I know I go back here that this is 16 that&#x27;s 80. That I know this is 100. Hey, and I know that it&#x27;s 100 because it&#x27;s It&#x27;s a 345 triangle. It&#x27;s just times 20. Okay, So seek in squared is high pot news over adjacent, so it&#x27;s gonna be 100 over 60. So I&#x27;m gonna say sequence. Great. Data is 100 over 60 are seeking. Fate is sorry. Should get rid of this square for a second. Sea containers 100 over 60 which is going to reduce first of all, the 10/6 and then to 5/3. Okay, so that means that seek in squared data is gonna then become, ah 5/3 squares, which is 25/9. So that&#x27;s the first step that first you need to do. And then if I look back here, I&#x27;ve got that. Now I now the to convert that point to 70 res per minute into radiance permitted 700.7 degrees. I&#x27;m gonna multiply by pi and divide by 100 me. Okay, 0.27 divided by 180. Um, you could divide them all. Both about 2.7 by over 1 81st And then you can Redoute by 9.27 Divided by nine is 0.3 pie and then it&#x27;s divided by nine b 20. Okay? And then if you divide that, you didn&#x27;t make a 0.0 15 Why a permanent? So now I can fill in my answers. I&#x27;m gonna jump back here. Seacon Square data we said was 25/9. Are you the, uh, is changing by 0.15 Then it&#x27;s minus 80/60. Spared the ex clears and then the act by any. Okay, so d exp i t t is going to equal 25 overnight, multiplied by 0.15 multiplied by sick be spared over minus 80. All right, And I&#x27;m just gonna do that. 25 divided by nine times 0.15 and then divided by 80. And then time 60 squares. Yeah, To 87 95 teams. Way too big. So just give me one second. We checked my calculations again. And while I&#x27;m doing that, um, you can check them as well and see what you get. Make sure that you&#x27;re getting the same answer. I am. Um, I&#x27;m 60 squared, then divided by Yeah, I&#x27;m getting the wrong number here. So now I should get the right number. That makes a lot more sense. Scratch that, Um and this is what I&#x27;m getting. What? I didn&#x27;t know my calculator a second time. Negative 0.1875 Which makes a lot more sense given the context of the problem. And it&#x27;s in feet her minute. Okay. And that&#x27;s my aunt

Abigail Martyr · Answer

were given a problem where a six with person is walking away. Herman eaten. Put Lamport. The person is currently 12 feet away from the lamppost and moving away out of rid of two feet per second. We were asked to find a redoubt which the length of the shadow is changing. We were given a hint to help solve the problem. Andi, that Hank is basically saying The proportions of the triangle formed by the lamp post has equal the proportion of the smaller triangle formed by the individual under shadow. So since I&#x27;m interested in finding go the rate at which the shadow is changing, I will first simplify this equation so that it is sold for us. So we can cross, multiply and are that will give Times X Plus s pastor equal to 18 s solving for s. We find that uh 12 s is equal to it&#x27;s. And if we take the derivative of this, then what we end up with is that 12 s prying is equal to six. I&#x27;m specs bright. Oh, that s prime is equal to 1/2 expert. So the redoubt was just Shadow is changing. Can be flown by ticking How, um, to think per second, which is one or the length of the shadow is changing at a rate of one put for a second.

Problem

Solve the problems in related rates. The accelera…

Problem 29 Easy Difficulty

Solve the problems in related rates.
The shadow of a $24-\mathrm{m}$ high building is increasing at the rate of $18 \mathrm{cm} / \mathrm{min}$ when the shadow is $18 \mathrm{m}$ long. How fast is the distance from the top of the building to the end of the shadow increasing?

Answer

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Catherine R.

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus