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

Lesson 5: Solving related rates problems

# Related rates: Falling ladder

Let's explore a thrilling real-world scenario in this video: a ladder slipping away from a wall! We'll use related rates to calculate how fast the top of the ladder falls. It's a fun and practical application of calculus that'll keep us on our toes. 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?
• 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/dtdh/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.
• Can we find the time at which the ladder touches the ground?
• For that we would require to express height h as a function of time t. If we did this, then we just plug h=0 into the formula and solve for t. However, we lack information to produce the formula needed.

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 h(t). But that would definitely require modifying the problem (adding data), not just adding another question.
• Damn i didn't know drawing skills were covered in pre-calc
• I can't seem to find an example of the rate of change of an angle of a falling ladder with respect to time?
• 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 eventd/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.
• how can we find the time at which the ladder loses contact with the wall?
• 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.
• 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?
• 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!
• 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.
• 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.
• Honestly, with all these math problems, why do people want to know these answers in the first place? What's so important about how fast a ladder is falling against the wall? xD
• They want to know what the answer is because they are just curious. God created us with curious minds, and this is how we are using it.
• Can we use the tangent function as a relation between both x and h?
• We know nothing about the angles, so tangent wouldn't do us any good.
• 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!
• 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`