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## Solving related rates problems

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0 energy points

# Related rates: Falling ladder

AP.CALC:

CHA‑3 (EU)

, CHA‑3.E (LO)

, CHA‑3.E.1 (EK)

## Solving related rates problems

## Video transcript

So I've got a 10 foot ladder
that's leaning against a wall. But it's on very slick ground,
and it starts to slide outward. And right when it's-- and
right at the moment that we're looking at this ladder,
the base of the ladder is 8 feet away from the
base of the wall. And it's sliding outward
at 4 feet per second. And we'll assume that the top
of the latter kind of glides along the side of the wall,
it stays kind of in contact with the wall and
moves straight down. And we see right over here, the
arrow is moving straight down. And our question
is, how fast is it moving straight
down at that moment? So let's think about
this a little bit. What do we know and
what do we not know? So if we call the distance
between-- let's call the distance between the
base of the wall and the base the ladder, let's call that x. We know right now x
is equal to 8 feet. We also know the
rate at which x is changing with respect to time. The rate at which x is
changing with respect to time is 4 feet per second. So we could call this dx dt. Now let's call the distance
between the top of the ladder and the base of the latter h. Let's call that h. So what we're really
trying to figure out is what dh dt is,
given that we know all of this other information. So let's see if we can come up
with the relationship between x and h and then take the
derivative with respect to time, maybe using
the chain rule. And see if we can
solve for dh dt knowing all of this information. Well, we know the
relationship between x and h at any time because of
the Pythagorean theorem. We can assume this
is a right angle. So we know that x
squared plus h squared is going to be equal to the
length of the ladder squared, is going to be equal to 100. And what we care
about is the rate at which these things
change with respect to time. So let's take the
derivative with respect to time of both sides of this. We're doing a little bit of
implicit differentiation. So what's the derivative with
respect to time of x squared? Well the derivative of x
squared with respect to x is 2x. And we're going to
have to multiply that times the derivative of
x with respect to t, dx dt. Just to be clear, this
is the chain rule. This is the derivative of x
squared with respect to x, which is 2x, times dx
dt to get the derivative of x squared with
respect to time. Just the chain rule. Now similarly, what's
the derivative of h squared with respect to time? Well that's just going
to be 2h, the derivative of h squared with
respect to h is 2h times the derivative of h
with respect to time. Once again, this right
over here is the derivative of h squared with respect
to h, times the derivative of h with respect to time, which
gives us the derivative of h squared with respect to time. And what do we get on
the right-hand side of our equation? Well the length of our
ladder isn't changing. This 100 isn't going to
change with respect to time. Derivative of a constant
is just equal 0. So now we have
it, a relationship between the rate of change
of h with respect to time. The rate of change of
x with respect to time. And then at a given
point in time, when the length of
x is x and h is h. But do we know what h is
when x is equal to 8 feet? Well, we can figure it out. When x is equal to 8 feet, we
can use the Pythagorean theorem again. We get 8 feet squared,
plus h squared is going to be equal to 100. So 8 squared is 64. Subtract it from both sides, you
get h squared is equal to 36. Take the positive square
root, a negative square root doesn't make sense because
then the ladder would be below the ground, it
would be somehow underground. So we get h is equal to 6. So this is something
that was essentially given by the problem. So now we know. We can look at this original
thing right over here, we know what x is,
that was given. Right now x is 8 feet. We know the rate of change
of x with respect to time. It's 4 feet per second. We know what h is
right now, it is 6. So then we can solve for the
rate of h with respect to time. So let's do that. So we get 2 times 8
feet, times 4 feet per second, so times
4, plus 2h, is going to be plus 2 times, our
height right now is 6, times the rate at which
our height is changing with respect to t is equal to 0. And so we get 2 times
8 times 4 is 64. Plus 12 dh dt is equal to 0. We can subtract 64 from
both sides, we get 12. 12 times the derivative
of h with respect to time is equal to negative 64. And then we just have to
divide both sides by 12. And so now we get a
little bit of a drum roll. The derivative,
the rate of change of h with respect to time is
equal to negative 64 divided by 12. It's equal to
negative 64 over 12, which is the same
thing as negative 16 over 3, yeah that's right. Which is equal to--
let me scroll over to the right a little
bit-- negative 5 and 1/3 feet per second. So we're done. But let's just do
a reality check, does that make sense that we
got a negative value right over here? Well our height is decreasing. So it completely makes sense
that its rate of change is indeed negative. And we're done.

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