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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: Falling ladder

AP.CALC:

CHA‑3 (EU)

, CHA‑3.E (LO)

, CHA‑3.E.1 (EK)

You're on a ladder. The bottom of the ladder starts slipping away from the wall. Amidst your fright, you realize this would make a great related rates problem... Created by Sal Khan.

## Want to join the conversation?

- Did Sal use implicit differentiation in this example because there is a relationship between
`x`

and`h`

(`x² + h² = 100`

)? If yes, can we change it to`h = sqrt(100 - x²)`

and calculate`d/dt (sqrt(100 - x²))`

instead?(10 votes)- Yes you can use that instead, if we calculate
`d/dt[h] = d/dt[sqrt(100 - x^2)]`

:`dh/dt = (1 /(2 * sqrt(100 - x^2))) * -2xdx/dt`

dh/dt = (-xdx/dt) / (sqrt(100 - x^2))

If we substitute the known values,`dh/dt = -(8)(4) / sqrt(100 - 64)`

dh/dt = -32/6 = -5 1/3

So, we arrived at the same answer as Sal did in this video.(22 votes)

- Can we find the time at which the ladder touches the ground?(7 votes)
- For that we would require to express height
as a function of time**h**. If we did this, then we just plug**t**into the formula and solve for**h=0**. However, we lack information to produce the formula needed.**t**

We have figured out**dh/dt**, which is an*approximation*of our formula. Using this approximation would assume that the rate of change of the height (at THAT MOMENT) stays the same or, in other words, there is*no acceleration*.

The ladder has forces (gravitation and friction with the wall and the floor most importantly) acting on it, hence it should have*acceleration*(if those forces don't balance out). We can't figure the acceleration, because there is no information about the mass of the ladder and the materials that make up the ladder and the wall.

If just for fun, you could make some reasonable assumptions about the mass and materials in question; say that it started to slide with no initial velocity and try to express. But that would definitely require modifying the problem (adding data), not just adding another question.**h(t**)(13 votes)

- I can't seem to find an example of the rate of change of an angle of a falling ladder with respect to time?(7 votes)
- Here is how I managed to solve it in different way. You see that as time changes you cosine proportion (8/10 right at this moment) changes, it is getting bigger. At the same time your angle is decreasing so is your sine, that is the thing you want find rate of in respect to time. We have cosine and hypotenuse data. We see that whatever change occures hypotenuse is constant. That is a nice thing so we can use arccos() function to find
**angle**at that moment so we could plug it in sine function(angle decreases over time, sine decreases). So far so good. So, arccos(8+4t/10), here 8+4t represents change of adjacent side(at that moment the value is**8**) over time and 10 represents constant lenght of hypotenuse. So`sine(arccos(8+4t/10))`

. Also we see that our hypotenuse is 10 times bigger that radius of unit circle. So we multiply whole expression by 10 because argument of arrcos() function is making things in unit circle proportion and we have triangle that ladder is forming 10 times bigger.

Looks like this:`10*sine(arccos(8+4t/10)) = Model of our event`

d/dt [ 10*sine(arccos(8+4t/10) ) ] = 10*d/dt [ sine(arccos(8+4t/10) )] =

10* cos( arccos(8+4t/10) ) * d/dt [ arccos(8+4t/10) ] =

10*cos( arccos(8+4t/10) ) * -1/sqrt( 1-(8+4t/10)^2 ) * d/dt [ 8+4t/10 ] =

finally derivative is:

10*cos( arccos(8+4t/10 ) ) * -1/sqrt( 1-(8+4t/10)^2 ) * 4/10 .

Exatly at that moment we know that adjacent side is 8 and velocity is 4ft* time. That moment means that instant, what is time at an instant? Well we think it's infinitesimally close to zero, so we substitute in derivative t=0:

10*cos( arccos(8/10) ) * -1/sqrt( 1-(8/10)^2 ) *4/10 = 8 * -4/6 =*-16/3*

I think key thing to understand here is that adjacent side changes over time, that is making angle do change(decrease in our case) over time. And decreasing angle means decreasing sine of that angle over time. Also, time at that moment is like frozen, but don't want to get philosophical. PLEASE comment if you see error in logic or elsewhere.(1 vote)

- how can we find the time at which the ladder loses contact with the wall?(3 votes)
- The ladder never loses contact with the wall in this example. Try to keep in mind that in these problems, the physics is simplified. We don't worry about the mass of the ladder, it never bounces, and so forth.

For the purposes of this problem, the height begins at h=6 and ends at h=0, and x is never greater than the length of the ladder, so x begins at x=8 and ends at x=10. Were the ladder to lose contact with the wall, there would be no h, and the question would become meaningless.(10 votes)

- My intuition is the speed that the ladder moves down is also 4 ft/s (same as slide outward), because the length never changes, someone explains me why it's wrong?(3 votes)
- This intuition would be correct if the length were given by x + h (because x + h = constant would imply that dx/dt = -dh/dt).

However, the length is not given by x + h according to the Pythagorean Theorem, but rather sqrt(x^2 + h^2). So constant length means that x^2 + h^2 = constant^2, which implies that 2x dx/dt = -2h dh/dt.

So the speed at which the ladder moves down is equal to the speed at which the ladder slides outward only when x = h. The ladder moves down faster than it slides outward when x > h, and the opposite occurs when x < h.

Have a blessed, wonderful day!(3 votes)

- Everything checks out mathematically but practically I have a problem. The second after t0, the bottom would have travelled 4 ft and would be at a distance 8+4=12 ft from the wall which means the entire length of the ladder is covered (plus more) and so the ladder has touched the ground.

But according to the solution, the top was at 6 ft from the ground and was falling at the rate of ~5 (less than 6) ft/sec. It doesn't look like the top would touch the ground in the next second.(3 votes)- The velocities of the bottom and top of the ladder cannot both be constant. The instantaneous velocity of the top of the ladder is -5.3
*at the instant where*x=8 and dx/dt=4. Once those values change, the velocity of the top of the ladder changes as well.

Because the velocity of the top of the ladder is not constant, we cannot find the displacement simply by multiplying velocity by time.(3 votes)

- Can we use the tangent function as a relation between both x and h?(2 votes)
- We know nothing about the angles, so tangent wouldn't do us any good.(4 votes)

- Is there a set of rules or a procedure that one should keep in mind when solving a related rates problem (since they're all different), or do you just have to solve each problem intuitively?(2 votes)
- Let's say you have a rectangle and you know that side "x" is 3 feet long and side "y" is 2 feet long. Also, you know that side "x" is growing 1 foot every minute.

x = 3 ft

y = 2 ft

dx/dt = 1 ft/min

Now, how fast is side "y" growing?

The answer is that there is simply not enough information--"y" may be growing or shrinking or even remaining the same length.**However**, if I tell you that the area of the rectangle is constant then there*is*enough information for you to determine the rate of change of side "y" with respect to time. In other words, the constant area of the rectangle acts as a**constraint**because:

1.) We know something about the Area (namely, that it remains constant)

2.) Both x & y are related to area via the formula Area = x*y

Now, to solve, take the derivative with respect to time of both sides, giving you:

0 = y*(dx/dt) + x*(dy/dt)

And then doing the regular arithmetic operations you get:

dy/dt = (-y/x)*dx/dt

Lastly, you just plug in the values for x, y, & dx/dt giving you:

dy/dt = -2/3 ft/min

which is how fast y is growing at the point in time when x=3 and y=2.

Does this help?(2 votes)

- How do I know which one I need to relate with??

Like how do you form the formula to set the derivative to?

I understand getting the y x and all that, but how do I get the equation to set it to d/dt?

I don't understand the relating part. How do I know which ones to relate to which?]

Thank you!(2 votes)- When you're asked to find the rate of change of something
**at this very moment**, it's the same as being asked to find f'(0). Since you don't care about what happens a time later/before with the height, you need to find value of the derivative at t = 0 with the given data.

Maybe it can be better understood how they relate to each other by using the explicit function notation. Let me use`b(t)`

for the function of the base of the triangle respect to time:`b(t) = 8 + 4t`

`b'(t) = 4`

When you are asked to find the rate of change at this moment, t is zero, and the functions become:`b(0) = 8`

`b'(0) = 4`

A function of the height h with respect to time can be written as:`h(t) = √[10² - (b(t))²]`

← Pythagoras (:`h'(t) = 1/[2*√(10² - (b(t))²]*[-2*b(t)]*[b'(t)]`

← Apply chain rule twice.`h'(t) = 1/[2*√(10² - 8²] * [-2*8]*[4]`

← The value of`b(t)`

and`b'(t)`

at this moment are 8 and 4, respectively. We just need to replace them.`h'(t) = [1/2*6] * [-2*8]*[4]`

← Simplify the square root in the denominator.`h'(t) = [1/6] * [-8]*[4] = [-8*4/6]`

← The 2 cancel.`h'(t) = -16/3`

(2 votes)

- If x is increasing at 4ft/sec after one second x would be 12ft? As the ladder is only 10ft this should have seen the ladder flat on the floor.

However h is decreasing at 16/3ft per second so from the starting point in the question where h=6, after one second it would h would be 2/3rds of a foot? How can this be if x=12, the ladder would have to have stretched to over 12ft long? I guess missing an obvious point here.(2 votes)

## 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.