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# Related rates: shadow

AP.CALC:
CHA‑3 (EU)
,
CHA‑3.E (LO)
,
CHA‑3.E.1 (EK)

## Video transcript

it's late at night in some type of nocturnal predatory bird maybe this is an owl is diving for its dinner so this right over here is this right over here is a mouse and it's diving straight down near a streetlight now let's get some information about what's going on so the streetlight right over here is 20 is 20 feet high so this is 20 a 20 foot high street lamp and right at this moment the and I haven't drawn it completely to scale the owl is 15 feet above the mouse so this distance right over here is 15 feet and the mouse itself is 10 feet from the base of the lamp let me draw that so the mouse is 10 feet 10 feet from the base of the lamp and we also know we have our little radar gone out we know that this owl is diving straight down is driving straight down and right now it is going 20 feet per second so right now this is going down down at 20 feet per second now what we're curious about is we have the light over here light is coming from the street lamp in every direction and it creates a shadow of the owl so right now the shadow is out here and as the owl goes further and further down the shadow is going to move the shadow is going to move to the left like that and so given everything that we've set up right over here the question is at what rate is the shadow moving so let's think about what we know and what we don't know and to do that let's set up some variables so let me draw the same thing a little bit more geometrically so let's say that this is this right over here is the street light that is 20 feet tall and then this right over here is the height of the owl right at this moment so this is 15 feet the distance between the base of the lamp and where the owl is going where that is right now this is ten feet and if I were to think about where the shadow is well the lights emitting from right over here the lights emitting right over here and so the owl blocks the light right over there so the shadow is going to be right over there so if you just draw a straight line from the source of light through the owl and you just keep going and you hit the ground you're going to figure out where the shadow is so the shadow the shadow is going to be the shadow is going to be right over here it's going to be right over there and we need to figure out how quickly is that is that moving and it's going to be moving in the leftward in the leftward direction so let's set up some variables over here so let's say so what's changing well we know that the height of the owl is changing so let's call that Y right at this moment it's equal to 15 but it is actually changing and let's call the distance let's call the distance between the between the the shadow and the mouse let's call that distance let's call that distance X now given this set up can we come up with a relationship between x and y and then using that relationship what we're really trying to come up with is what is the rate at which X is changing with respect to time we know what Y is right at this moment we know what dy DT is right at this moment can we come up with a relationship between x and y and maybe take the derivative with respect to t so we can figure out what DX DT is at a given moment in time well both of these triangles and when I say both of these triangles let me clear let me be clear what I'm talking about this triangle right over here this smaller triangle in green is a similar triangle to the larger triangle is a similar triangle to this larger triangle that I am tracing in blue it's similar to this larger one how do I know that well they both have a right angle right over here they both share this angle so all three if they have two angles in common than all three angles all three angles all three angles must be in common so they are similar triangles which means the ratio between corresponding sides must be the aim so we know that the ratio of X to Y the ratio of X to y must be the ratio of this entire base which is X plus 10 to the height of the larger triangle 220 and right there we have a relationship between x and y and if we take the derivative of both sides with respect to T we're probably doing pretty well now before taking the derivative with respect to T I could do it right over here just to simplify things a little bit let me just cross multiply so let me multiply both sides of this equation by 20 and Y just so that I don't have as many things in the denominator so on the left hand side it simplifies to 20 X I don't want to write over it well I'll just write 20 X and on the left hand side or sorry on the left hand side it is 20 X and on the right hand side let's see this cancels with that we have XY XY plus plus 10 X y plus 10 Y and now let me take the derivative of both sides with respect to time so the derivative of 20 times something with respect to time is going to be the derivative of 20 times something with respect to something which is just 20 that's the derivative of 20 X with respect to x times DX the derivative of X with respect to t is equal to now over here we're going to have to break out a little bit of the product rule so first we want to figure out the derivative of X with respect to time so the derivative of X with respect to time so the derivative of the first thing times the second thing times y plus just the first thing times the derivative of the second thing so the derivative of Y with respect to T is just dy DT dy DT and then finally right over here the derivative of ten Y with respect to T it's the derivative of ten Y with respect to Y which is just ten times the derivative of Y with respect to T which is dy dy DT and there you have it you have your relationship between between DX DT dy DT and X&Y so let's just make sure we we have everything this is what we're trying to solve for DX DT and let's see we have another DX DT here we're going to try to solve for that we know what Y is is equal to 15 we know what dy DT is dy DT if we make the convention since Y is decreasing we can say it's negative 20 so we know what we know what this is and so if we just know what so we know what this is so if we just know what X is we can solve for DX DT so what is X right at this moment well we can use this first equation we can actually use this one up here but this one is simplified a little bit to actually solve for X so let's do that and then we'll substitute back into this thing where we've taken the derivative so we get 20 times X is equal to x times y x times y y is 15 and just remember I could have used this equation but this is just one step further we've already crossed multiplied so it's x times y y is 15 so it's x times 15 plus 10 times y plus 10 times 15 plus 10 times 15 did I do that at 20 X is equal to x times 15 plus 10 times 15 10 times 15 so let's see if you subtract so just this is 20 X is equal to 15 X plus 150 subtract 15 X from both sides you get 5 X is equal to 30 divided by sorry 5 X is equal to 150 my brain is getting a head 5 X is equal to 150 divide both sides by 5 you get X is equal to 30 feet X is equal to 30 feet right at this moment so this distance just going back to our original diagram this distance right over here is 30 feet so let's substitute all the values we know back into this equation to actually solve for DX DT so we have so let me do it right over here we have 20 20 times DX DT I'll do that in orange we'll solve for that actually I already used orange so let's say DX DT I'll use this pink 20 times DX DT is equal to DX DT times y Y right now is 15 feet so times 15 times now what I didn't want to do that color x times 15 plus ay so we already know that X is 30 plus 30 times dy DT what is dy DT dy DT we could say is negative 20 feet per second Y is decreasing the bird is the bird is diving down to get its dinner so times negative and that's just the well times 20 feet per second so that's that right over there Plus 10 times dy DT so plus 10 times negative 20 feet per second and then we just solve for DX DT so let's see what do we have we have 20 times let's see let me subtract 15 DX DT from both sides of this equation and we get 5 DX DT s5 DX DT so I just subtracted this from both sides of the equation this is 15 DX DT this is 20 so we have 5 DX DT s is equal to this is this part right over here is negative 600 and this part right over here is negative 200 so it's equal to negative 800 feet per sec or negative 800 and actually this will actually be in feet per second and so dxdt is equal to dividing both sides by 5 5 times 16 is 80 so this is negative 160 feet per second and we're done and we see the shadow is moving very very very very fast to the left X is decreasing because and we see that that's why we have this negative sign here the value of X is decreasing it is moving to the left quite quite a quite a nice speed here