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## AP®︎/College Calculus AB

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

Lesson 5: Solving related rates problems

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? •  With the light source the same height as the diver, the shadow is parallel to the ground and never touches, so the shadow's velocity is infinity, but only in that instant. It's only an instantaneous velocity. That's why the number can be so large at the start. It's velocity slows down, approaching zero, as the falling object moves closer to the ground, also approaching zero. The velocity of the falling owl is constant at 20 ft/sec, but the velocity of 160 ft/sec is not constant. If it were, the 30 feet between the initial shadow and the mouse would be covered in 30/160 = 0.1875 seconds. We know the owl takes 0.75 seconds to fall (15/20 = 0.75), so the shadow can't possibly cover the 30 feet in less time.
• 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 • 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.
• 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? • 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?? • 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?
• 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.
• 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. • At : Why is there shadow on that point Sal describes? • It's going to take the owl .75 seconds to travel -15 feet to the ground. Since the shadow will reach the same point in the same amount of time, won't the shadow travel -30 feet in .75 seconds, making the rate of its leftward movement just 40 feet per second? If the shadow is traveling at a rate of -160 feet per second as the solution suggests, I think it would travel the -30 feet to the mouse in just .1875 seconds, which is much less than the .75 it takes the owl to reach the same point. I must be missing something obvious. • You're assuming that the shadow moves at a constant rate. It doesn't. If the owl rose back up to 20ft at a constant rate, the shadow would shoot off to infinity, which it can't do at a constant rate.

What you're calculating is the average speed of the shadow over the next 0.75 seconds (assuming the owls speed remains constant). We're looking for the speed of the shadow at the moment that the owl is 15ft from the ground and moving at 20 ft/s.
• 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 = ? • 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. 