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## Differential Calculus

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? • Calefornia's answer above, was quite nice. I'll just add some intuition backed by the equation we get from this problem.

Let h(t) be the height of the bird at time t and d(t) the distance between the shadow and the (freezed) mouse, also at time t.

If, as Sal did in the video, one uses similar triangles, one gets an equation relating h(t) to d(t). (This is the main objective when working on related rates problems.) After that, by differentiation (maybe implicit as Sal chose to do in this case), one gets the relation involving the rates that matter. Solving for the instantaneous rate of change of d(t), one obtains:

d'(t) = 200*h'(t) / (20 - h(t))^2

where:
d'(t) = the instantaneous rate of change of the distance between the shadow and the mouse at time t.
h'(t) = the instantaneous rate of change of the height of the bird at time t.

If, as Calefornia said, you assume that h'(t) is a constant negative number (that is, the speed at which the bird approaches the mouse doesn't change), you can see that d'(t) increases (becomes less negative) as the height decreases. That means that the speed (absolute value of d'(t)) decreases as the height decreases. (Be careful with the distinction between speed and velocity!)
Given the assumptions, that means, the closer the bird is from the mouse, the slower its shadow approches the prey. You can verify that empirically! (Well, maybe under different conditions and using inanimate objects...)

Technical remark: you can check that positive relationship between the speed of the shadow and the height of the bird (within that range), by differentiating the above equation with respect to h:
d/dh[d'(t)] = 400*h'(t) / (20 - h(t))^3
which, assuming h'(t) < 0 and constant, yields
d/dh[d'(t)] < 0, if 0 < h(t) < 20
That means, if h increases, d' decreases (even more negative), or speed increases.
• 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.
• 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?
• 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? • 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.
• At : Why is there shadow on that point Sal describes? • 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.
x = ½ arccos(⅔)
AND
x= 2πk ± ½π, where k is any integer
• Can't you just use the slope formula to find the distance x? If the bird is at a height of 15 feet when it is 10 feet over from the lamp, you can continue this over... (20, 10) (30, 5) (40, 0) to find 10 + x = 40 and subsequently x = 30. I'm just wondering if calculus is really necessary for that part. Although I guess you need to eventually derive that formula anyway so it isn't much of a time waster. • I have done this problem by identifying functions that change in respect to time and linked them together in a way that they are affecting each other.
We have constant of 10 ft from lamp to bird. And we have 20 -15= 5 as the length of opposite side of smallest (upper) right triangle between owl and tip of lamp. That side changes like this 5+20t since it increases as bird dive. So we can calculate length of this side at that time "t". Important thing to see here is that angle of this smallest upper triangle (near bird) is same as angle (near shadow) of the larger triangle shadow-bird-ground. We can calculate angle of larger triangle at that time simply by taking `arctan((5+20t)/10)`, (because it is same as angle of smaller triangle).
Here (5+20t)/10 represent opposite/adjacent side of that smallest triangle. We also have the model of diving bird trajectory, that is `15-20t` as bird is diving. That model(function) represent opposite side of that larger triangle that shadow is forming with the bird and ground.
Now we have means to calculate (to know) opposite side and angle of that larger triangle. By taking `sine(arctan((5+20t)/10 ))` we have proportion between hypotenuse and opposite side of larger triangle or sine of that angle. With this we can calculate:
`legth of opposite side (larger triangle) / proportion of sine of that triangle = hypotenuse of largest triangle`.

Adjecent side of interest(shadow approaching side) = sqrt(hypotenuse^2-oppositeSide^2 ), looking like this:
` sqrt( ((15-20t)/sin( arctan(5+20t ))^2 - (15-20t)^2 ) `
the derivative of this can be long and tedious and final answer is:
`640t-480/(4t+1) |4t-3| |4t+1| `
when we substitute t=0 at that moment we have` -480/3 = -160`

Sal's method compered to this is much more elegant and uses implicit differentiation, are there some parts of science or general problem solving that could use this longer method (function composition) of calculation and what would they be?  