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## Solving related rates problems

Current time:0:00Total duration:10:39

# Related rates: shadow

AP.CALC:

CHA‑3 (EU)

, CHA‑3.E (LO)

, CHA‑3.E.1 (EK)

## 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.

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