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### Course: AP®︎/College Calculus AB>Unit 4

Lesson 5: Solving related rates problems

# Related rates: shadow

Let's take a look at a problem involving an owl hunting a mouse and the shadow it casts. We can use calculus to determine how fast the shadow moves as the owl dives towards its prey. It's a real-world application of related rates that brings the concept to life! Created by Sal Khan.

## Want to join the conversation?

• As Sal points out near the end of the video, the shadow is moving quite fast compared to the bird. When I first solved a similar problem (diver leaping from 40m platform above water in normal earth gravity with a light source 15m from the diver at same height as diver; how fast is the diver's shadow moving after 1.5 seconds?) I wondered why it was moving so fast compared to the diver. After thinking about it, I realized that the shadow is actually moving at infinite velocity, not some finite velocity coupled to the simple geometry of the light source-diver-diver's shadow system. With the diver on the platform at the same height as the light source, isn't the shadow at infinity? If so, the distance the shadow moves is infinite. If it moved this infinite distance in 1.5 seconds, shouldn't the velocity also be infinite? v = d/t => v=infinity/1.5 => v=infinity!! Or am I thinking about this in the wrong way?
(31 votes)
• With the light source the same height as the diver, the shadow is parallel to the ground and never touches, so the shadow's velocity is infinity, but only in that instant. It's only an instantaneous velocity. That's why the number can be so large at the start. It's velocity slows down, approaching zero, as the falling object moves closer to the ground, also approaching zero. The velocity of the falling owl is constant at 20 ft/sec, but the velocity of 160 ft/sec is not constant. If it were, the 30 feet between the initial shadow and the mouse would be covered in 30/160 = 0.1875 seconds. We know the owl takes 0.75 seconds to fall (15/20 = 0.75), so the shadow can't possibly cover the 30 feet in less time.
(50 votes)
• I tried to solve the same question using my own intuition before Sal began,
I used `f(t)` as the distance between the light post and the shadow, defining it as
`f(t) = 10 + 10h/(20-h)`
`h` is a function of time (`h(t)`), and refers to the distance between the bird and the ground/mouse. As you can see, the function works fine at `h = 15, giving f = 30`. After differentiation, I got `df/dt = 0 + (10 * dh/dt * (20-h) - dh/dt * 10h)/(20-h)^2`
Substituting `dh/dt = 20 ft/s and h = 15 ft`, you get
`df/dt = (10 * 20 * (20-15) - 20 * 10 * 15)/(20-15)^2`
`df/dt = 200(5 - 15)/5^2`
`df/dt = -10 * 200/25 => -10 * 8`
`df/dt = -80`
As you can see, I am off by a factor of two. Can someone please explain where I made a mistake?
Thanks,
Anitej Banerjee
(3 votes)
• You have a wrong sign in your differentiation, it should be `df/dt = 0 + (10 * dh/dt * (20-h) + dh/dt * 10h)/(20-h)^2`. With that you get the same result as in the video.

I would point out that you choose a very complicated form of your equation to work with. You could have defined your formula as `f(t) = 200/(20-h)`, which would have been much easier to differentiate into `df/dt = 200/(20-h)² dh/dt`. Working with simpler formulas also reduced the risk of errors when differentiating.
(5 votes)
• Taking the derivative with respect to time of the equation x^2+y^2=1125, (1125 being obtained by adding the squares of x and y once both their values are known), seems to yield a different result = 10ft/sec. (?) Anyone knows why taking the derivative of the above equation (which seems pretty valid to take...) is not the right way to solve the problem?

Thanks in advance for any clarifying answer.
(6 votes)
• 1125 is not constant, so you cannot put that in before you derive. You would need a third variable, z, where z is 1125 and you don't know dz/dt. That would take you even further away from where you are trying to get to.
(8 votes)
• The bird is moving downwards at 20 ft/sec from 15 ft above the mouse which means it reaches the mouse in 3/4 of a sec. The 160 ft/sec answer to the rate of movement of the shadow doesn't appear right as in 3/4 sec the shadow would travel a distance of 120 ft! What am I not getting??
(3 votes)
• The velocity of 160 ft/sec is an instantaneous velocity, i.e., it is only 160 ft/sec at that precise moment. As the owl continues its descent, the velocity of the shadow will drop rather quickly until it reaches the ground, at which point the velocity would equal zero. Make sense?
(13 votes)
• hmm, something seems wrong. so in half a second the bird will dive 10ft, but it's shadow will have already moved 80ft to the left (smashing through the light post). Was there a mistake or did I miss something?
(1 vote)
• You missed something. We've determined the instantaneous rate of change in the position of the shadow, which is -160 ft/sec, but that figure changes dramatically as the bird moves closer to the ground (and the mouse). When the height of the bird is 10 ft, for example, the shadow is moving only -40 ft/sec, and at the height of 5 ft the shadow moves less than 20 ft/sec. So the shadow doesn't move 80 ft in the half second it takes for the bird to dive from 15 ft to 5 ft. If my arithmetic is right, it moves only 26-2/3 ft.
(12 votes)
• I very unsuccessfully tried to solve it before Sal. Can someone explain where I went wrong ?

If x=30 and y= 15 then,
x^2 + y^2 = 1125 || d/dt(x^2 + y^2) = d/dt(1125) || d/dt(x^2) + d/dt (y^2) = 0 | |2x•dx/dt + 2y•dy/dt = 0 || dx/dt = (-2y•dy/dt)/2x = 15•-2 •-20/60 = 10 ans.

Please answer it took me a long time to type all this.
(4 votes)
• It's going to take the owl .75 seconds to travel -15 feet to the ground. Since the shadow will reach the same point in the same amount of time, won't the shadow travel -30 feet in .75 seconds, making the rate of its leftward movement just 40 feet per second? If the shadow is traveling at a rate of -160 feet per second as the solution suggests, I think it would travel the -30 feet to the mouse in just .1875 seconds, which is much less than the .75 it takes the owl to reach the same point. I must be missing something obvious.
(2 votes)
• You're assuming that the shadow moves at a constant rate. It doesn't. If the owl rose back up to 20ft at a constant rate, the shadow would shoot off to infinity, which it can't do at a constant rate.

What you're calculating is the average speed of the shadow over the next 0.75 seconds (assuming the owls speed remains constant). We're looking for the speed of the shadow at the moment that the owl is 15ft from the ground and moving at 20 ft/s.
(5 votes)
• At : Why is there shadow on that point Sal describes?
(2 votes)
• At any time, you can draw a straight line from the light and through where the owl is. Wherever that line winds up hitting the ground, that's where the shadow is.
(6 votes)
• I am having a tough time trying to solve this algebraically. I would be much appreciated if anyone could show me step by step how to solve it. Thanks.
-2sin(2x)sin(x) + cos(2x)cos(x) = 0, x = ?
(2 votes)
• There are a number of ways to solve this. The simplest I can come up with is this:
-2sin(2x)sin(x) + cos(2x)cos(x) = 0
2sin(2x)sin(x) = cos(2x)cos(x)
2sin(2x)sin(x) = cos(2x)cos(x)
2 sin(2x)/cos(2x) = cos(x)/sin(x)
2 sin(2x)/cos(2x) = cot(x)
`Identity: cot x = sin(2x)/ [1- cos(2x)]`
2 sin(2x)/cos(2x) = sin(2x)/ [1- cos(2x)]
`sin (2x) cancels provided x ≠ ½πk, where k is any integer`
2 /cos(2x) = 1/ [1- cos(2x)]
`take reciprocal`
½ cos (2x) = 1 - cos(2x)
½ cos (2x) + cos(2x) = 1
³⁄₂ cos (2x) = 1
cos (2x) = ⅔
arccos(cos(2x)) = arccos(⅔)
2x = arccos(⅔)
x = ½ arccos(⅔)
This can also be solved in terms of arctan and arcsin, but the arccos is a little bit simpler.

Now we must consider the values we excluded to make the simplification:
For what values of x = ½πk is the equation true that:
-2sin(2x)sin(x) + cos(2x)cos(x) = 0
That would be at values where sin(2x)sin(x) and cos(2x)cos(x) both equal 0, which is at:
x= 2πk ± ½π, where k is any integer
which should be obvious from the periods of cos(x) and sin(2x).
Otherwise you can manually check x= 0, ½π, π and ³⁄₂π to see that only ½π and ³⁄₂π (and their coterminal angles) make the original equation true.
So the final answers are:
x = ½ arccos(⅔)
AND
x= 2πk ± ½π, where k is any integer
(3 votes)
• Can't you just use the slope formula to find the distance x? If the bird is at a height of 15 feet when it is 10 feet over from the lamp, you can continue this over... (20, 10) (30, 5) (40, 0) to find 10 + x = 40 and subsequently x = 30. I'm just wondering if calculus is really necessary for that part. Although I guess you need to eventually derive that formula anyway so it isn't much of a time waster.
(2 votes)
• The slope formula from a linear equation won't work if there is not a straight line. If the bird moves up and down and changes its velocity considerably, you won't be able to use something as simple as y = mx + b.
So, basically, a derivative extends y=mx +b by making m be a function of x instead of a constant. Then it can accommodate curves instead of just a straight line.
(3 votes)

## Video transcript

It's late at night, and some type of nocturnal predatory bird, maybe this is an owl, is diving for its dinner. So this right over here is a mouse. And it's diving straight down near a street light. And let's get some information about what's going on. So the street light right over here is 20 feet high. So this is a 20 foot high street lamp. And right at this moment, the-- and I haven't drawn it completely to scale-- the owl is 15 feet above the mouse. So this distance right over here is 15 feet. And the mouse itself is 10 feet from the base of the lamp. Let me draw that. So the mouse is 10 feet from the base of the lamp. And we also know-- we have our little radar gun out-- we know that this owl is diving straight down. And right now, it is going 20 feet per second. So right now, this is going down at 20 feet per second. Now what we're curious about is we have the light over here. Light is coming from the street lamp in every direction. And it creates a shadow of the owl. So right now the shadow is out here. And as the owl goes further and further down, the shadow's going to move to the left like that. And so, given everything that we've set up right over here, the question is, at what rate is the shadow moving? So let's think about what we know and what we don't know. And to do that, let's set up some variables. So let me draw the same thing a little bit more geometrically. So let's say that this right over here is the street light. That is 20 feet tall. And then this right over here is the height of the owl right at this moment, so this is 15 feet. The distance between the base of the lamp and where the owl is going, where that mouse is right now. This is 10 feet. And if I were to think about where the shadow is, well, the light's from right over here. And so the owl blocks the light right over there, so the shadow is going to be right over there. So if you just draw a straight line from the source of light through the owl and you just keep going and you hit the ground, you are going to figure out where the shadow is. So the shadow is going to be right over here. It's going to be right over there. And we need to figure out how quickly is that moving. And it's going to be moving in the leftward direction. So let's set up some variables over here. So let's say-- so what's changing? Well, we know that the height of the owl is changing. So let's call that y. Right at this moment, it's equal to 15, but it is actually changing. And let's call the distance between the shadow and the mouse x. Now given this set up, can we come up with a relationship between x and y? And then using that relationship, what we're really trying to come up with is, what is the rate at which x is changing with respect to time? We know what y is right at this moment. We know what dy/dt is right at this moment. Can we come up the relationship between x and y and maybe take the derivative with respect to t so we can figure out what dx/dt is at a given moment in time? Well, both of these triangles-- and when I say both of these triangles, let me be clear what I'm talking about. This triangle right over here, the smaller triangle in green, is a similar triangle to the larger triangle. This is a similar triangle to this larger triangle that I am tracing in blue. It's similar to this larger one. How do I know that? Well, they both have a right angle, right over here. They both share this angle. So all three-- they have two angles in common-- then all three angles must be in common. So they are similar triangles, which means the ratio between corresponding sides must be the same. So we know that the ratio of x to y must be the ratio of this entire base, which is x plus 10, to the height of the larger triangle, to 20. And right there, we have a relationship between x and y. And if we take the derivative of both sides with respect to t, we're probably doing pretty well. Now before taking the derivative with respect to t-- I could do it right over here-- just to simplify things a little bit, let me just cross multiply. So let me multiply both sides of this equation by 20 and y, just so that I don't have as many things in the denominator. So on the left hand side it simplifies to 20x. I don't want to write over it, well, I'll just write-- 20x. And on the left hand side is 20 x. And then on the right hand side-- let's see this cancels with that-- we have xy plus 10y. And now let me take the derivative of both sides with respect to time. So the derivative of 20 times something with respect to time is going to be the derivative of 20 times something with respect to the something, which is just 20. That's the derivative of 20x with respect to x, times dx or the derivative of x with respect to t, is equal to. Now over here we're going have to break out a little bit of the product rule. So first we want to figure out the derivative of x with respect to time. So the derivative of the first thing times the second thing, times y. Plus just the first thing times the derivative of the second thing. So derivative of y with respect to t is just dy/dt. And then finally, right over here. The derivative of 10y with respect to t is the derivative of 10y with respect to y, which is just 10, times the derivative of y with respect to t, which is dy/dt. And there you have it. You have your relationship between dx/dt, dy/dt, and x and y. So let's just make sure we have everything. This is what we're trying to solve for, dx/dt. And let's see here, we have another dx/dt here. We're going to try to solve for that. We know what y is. y is equal to 15. We know what dy/dt is, dy/dt. If we make the convention since y is decreasing, we can say it's -20. So we know what this is. And so if we just know what, so we know what this is. So if we just know what x is we can solve for dx/dt. So what is x right at this moment? Well, we can use this first equation-- we could actually use this one up here, but this one is simplified a little bit-- to actually solve for x. So let's do that, and then we'll substitute back into this thing where we've taken the derivative. So we get 20 times x is equal to x times y. y is 15. And just remember, I could have used this equation, but this is just one step further. We've already crossed multiplied. So it's x times y. y is 15. So it's x times 15, plus 10 times y. Plus 10 times 15. Did I do that right? 20x is equal to x times 15 plus 10 times 15. So let's see if you subtract. So just this is 20x is equal to 15x, plus 150. Subtract 15x from both sides, you get 5x is equal to 30-- sorry, 5x is equal to 150, my brain is getting ahead. 5x is equal to 150. Divide both sides by 5. You get x is equal to 30 feet. x is equal to 30 feet right at this moment. So this distance, just going back to our original diagram. This distance right over here is 30 feet. So let's substitute all the values we know back into this equation to actually solve for dx/dt. So we have-- so let me do it right over here-- we have 20 times dx/dt. I'll do that in orange, we'll solve for that. Actually no, I already used orange. So let's say dx/dt-- I'll use this pink-- 20 times dx/dt is equal to dx/dt times y. y right now is 15 feet. So times 15, times-- I didn't want to do that color. Times 15 plus x, we already know that x is 30. Plus 30, times dy/dt. What is dy/dt? dy/dt, we could say is -20 feet per second. y is decreasing, the bird is the bird is diving down to get its dinner. So times negative-- well, times 20 feet per second. So that's that right over there. Plus 10 times dy/dt. So plus 10 times -20 feet per second. And now we just solve for dx/dt. So let's see, what do we have? We have 20 times-- let's see, let me subtract 15 dx/dt from both sides of this equation-- and we get 5dx/dt's. I just subtracted this from both sides of the equation. This is 15dx/dt this is 20. So that we have 5dx/dt's, is equal to this is, this part right over here is -600. And this part right over here is -200. So it's equal to -800 feet per second-- or -800, and actually this will actually be in feet per second. And so dx/dt is equal to, dividing both sides by 5, 5 times 16 is 80. So this is -160 feet per second. And we're done. And we see the shadow is moving very, very, very, very fast to the left. x is decreasing, and we see that. That's why we have this negative sign here. The value of x is decreasing. It is moving to the left at quite a nice speed here.