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## Differential Calculus

### Unit 4: Lesson 5

Solving related rates problems- Related rates intro
- Related rates (multiple rates)
- Related rates: Approaching cars
- Related rates: Falling ladder
- Related rates (Pythagorean theorem)
- Related rates: water pouring into a cone
- Related rates (advanced)
- Related rates: shadow
- Related rates: balloon

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# Related rates: shadow

AP.CALC:

CHA‑3 (EU)

, CHA‑3.E (LO)

, CHA‑3.E.1 (EK)

Sal solves a related rates problem about the shadow an owl casts as it's hunting a mouse. 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?(30 votes)
- Calefornia's answer above, was quite nice. I'll just add some intuition backed by the equation we get from this problem.

Let h(t) be the height of the bird at time t and d(t) the distance between the shadow and the (freezed) mouse, also at time t.

If, as Sal did in the video, one uses similar triangles, one gets an equation relating h(t) to d(t). (This is the main objective when working on related rates problems.) After that, by differentiation (maybe implicit as Sal chose to do in this case), one gets the relation involving the rates that matter. Solving for the instantaneous rate of change of d(t), one obtains:

d'(t) = 200*h'(t) / (20 - h(t))^2

where:

d'(t) = the instantaneous rate of change of the distance between the shadow and the mouse at time t.

h'(t) = the instantaneous rate of change of the height of the bird at time t.

If, as Calefornia said, you assume that h'(t) is a constant negative number (that is, the speed at which the bird approaches the mouse doesn't change), you can see that d'(t) increases (becomes less negative) as the height decreases. That means that the speed (absolute value of d'(t)) decreases as the height decreases. (Be careful with the distinction between speed and velocity!)

Given the assumptions, that means, the closer the bird is from the mouse, the slower its shadow approches the prey. You can verify that empirically! (Well, maybe under different conditions and using inanimate objects...)

Technical remark: you can check that positive relationship between the speed of the shadow and the height of the bird (within that range), by differentiating the above equation**with respect to h**:

d/dh[d'(t)] = 400*h'(t) / (20 - h(t))^3

which, assuming h'(t) < 0 and constant, yields

d/dh[d'(t)] < 0, if 0 < h(t) < 20

That means, if h increases, d' decreases (even more negative), or speed increases.(5 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)

- 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?(12 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.(5 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.(7 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)

- At2:45: 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)

- I have done this problem by identifying functions that change in respect to time and linked them together in a way that they are affecting each other.

We have constant of 10 ft from lamp to bird. And we have 20 -15= 5 as the length of opposite side of smallest (upper) right triangle between owl and tip of lamp. That side changes like this 5+20t since it increases as bird dive. So we can calculate length of this side at that time "t". Important thing to see here is that angle of this smallest upper triangle (near bird) is same as angle (near shadow) of the larger triangle shadow-bird-ground. We can calculate**angle**of larger triangle at that time simply by taking`arctan((5+20t)/10)`

, (because it is same as angle of smaller triangle).

Here (5+20t)/10 represent opposite/adjacent side of that smallest triangle. We also have the model of diving bird trajectory, that is`15-20t`

as bird is diving. That model(function) represent opposite side of that larger triangle that shadow is forming with the bird and ground.

Now we have means to calculate (to know) opposite side and angle of that larger triangle. By taking`sine(arctan((5+20t)/10 ))`

we have proportion between hypotenuse and opposite side of larger triangle or**sine**of that angle. With this we can calculate:

`legth of opposite side (larger triangle) / proportion of sine of that triangle = hypotenuse of largest triangle`

.

Adjecent side of interest(shadow approaching side) = sqrt(hypotenuse^2-oppositeSide^2 ), looking like this:`sqrt( ((15-20t)/sin( arctan(5+20t ))^2 - (15-20t)^2 )`

the derivative of this can be long and tedious and final answer is:`640t-480/(4t+1) |4t-3| |4t+1|`

when we substitute t=0 at that moment we have`-480/3 = -160`

Sal's method compered to this is much more elegant and uses implicit differentiation, are there some parts of science or general problem solving that could use this longer method (function composition) of calculation and what would they be?(3 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.(2 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.