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# Related rates: balloon

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

## Video transcript

you're watching some type of hot-air balloon show and you're curious about how quickly one hot-air balloon in particular is rising and you have some information at your disposal you know the spot on the ground that is directly below the hot air balloon let's state took off from that point it's just been going straight up ever since and you know you've measured it out that you're five hundred meters away from there so you know that you are five hundred meters away from that and you're also able to measure the angle between the horizontal and the hot air balloon you could do that with I know I'm not exactly a surveyor but I guess a viewfinder or something like that so you're able to and I'm not sure if that's the right tool but there are tools that you can measure the angles between the horizontal and and something that's not on the horizontal so you know that this angle right over here is PI over 4 radians or 45 degrees we're going to keep it PI over 4 because when you take derivatives of trig functions you assume that you're dealing with radians so right over here this is PI over 4 radians and you also are able to measure at which the rate at which this angle is changing so this is changing changing changing at 0.2 radians radians per minute now my question to you or the question that you're trying to figure out as you watch this hot air balloon is how fast is it rising right now how fast is it rising just as the angle between the horizontal and kind of the the line between you and the hot air balloon is PI over 4 radians and the that angle is changing at 0.2 radians per minute so let's think about what we know and what we're trying to figure out so we know a couple of things we know that theta is equal to PI over 4 if we call theta the angle the angle right over here so this is Theta we also know the rate at which data is changing we know D theta let me do this in yellow we know D theta DT is equal to 0.2 radians per minute now what are we trying to figure out well we're trying to figure out the rate at which the height of the balloon is changing so if you call this distance right over here this distance right over here H what we want to figure out is D h d h dt that's what we don't know so what we'd want to come up with is a relationship between d h dt d theta/dt and maybe theta if we need it or another way to think about it if we can come up with a relationship between H and theta that we could take the derivative with respect to T and we'll probably get a relationship between all of this stuff so let's the relationship between theta and H well it's a little bit of trigonometry we know we're trying to figure out H we already know what this length is right over here we know opposite over adjacent that's the definition of tangent so let's write that down so we know that the tangent the tangent of theta tangent of theta is equal to the opposite side the opposite side is equal to H is equal to H over the adjacent side which we know is going to be a fixed 500 is going to be a fixed 500 so there you have a relationship between theta and H and then to figure out to figure out a relationship between theta D theta and D HD or D theta DT the rate at which theta changes with respect to T and the rate at which H changes changes with respect to T we just have to take the derivative of this of both sides of this with respect to T implicitly so let's do that and actually let me move over this H over 500 a little bit so let me move it over a little bit so I have space to show the derivative operator so let's write it like that and now let's take the derivative with respect to T so D DT we take the derivative respect to T on the left we're going to take the derivative with respect to T on the right so what's the derivative with respect to T of tangent of theta well we're just going to apply the chain rule here it's going to be first the derivative of the tangent of theta with respect to theta which is just secant squared of theta times the derivative of theta with respect to T times D theta DT once again this is just D the derivative of the C sorry the derivative of the tangent the tangent of something with respect to that something times the derivative of the something with respect to T derivative of tangent theta with respect to theta times the derivative of theta with respect to T gives us the derivative of tangent of theta with respect to T which is what we'd want when we take the dirt when we use this type of a derivative operator taking the derivative with respect to T not just the derivative with respect to theta fair enough so this is the left-hand side and then the right-hand side becomes well it's just going to be 1 over 500 dhdt 1 over 500 D H DT we're literally saying it's just 1 over 500 times the derivative of H with respect to T but now we have our relationship we have the relationship that we actually care about we have a relationship between the rate at which the height is changing with respect to time and the rate at which the angle is changing with respect to time and our angle at any moment so we can just take these values up here throw it in here and then solve for the unknown so let's do that let's do that right over here so we get secant squared of theta so we get secant squared right now our theta is PI over 4 secant squared of PI over 4 let me write those colors and to show you that I'm putting these values in secant squared of PI over 4 so you can times D theta DT well that is just 0.2 so times 0.2 and then this is going to be equal to this is going to be equal to 1 over 500 and we want to make sure since this is in radians per minute we're going to get we're going to get meters per and this is this is meters right over here we're going to get meters per minute right over here we just want make sure all of our we know what our units are doing I haven't written the unit's here for to save some space but we get 1 over 500 times D H DT so if we want to solve for d HD T you can multiply both sides by 500 and you get the rate at which our height is changing is equal to 500 times let's see secant squared of PI over 4 that is 1 over cosine squared of PI over 4 cosine of PI over 4 let me write this over here cosine of PI over 4 is square root of 2 over 2 cosine squared of PI over 4 is going to be equal to 2 over 2 over 4 which is equal to 1/2 and so secant squared of PI over 4 is just 1 over that is equal to 2 so this is going to be equal to let me rewrite this instead of so the secant squared of PI over 4 let me erase this right over here secant squared of PI over 4 all of this business right over here simplifies to 2 so times 2 times 0.2 times 0.2 so what is this going to be this is going to be 500 times 0.4 so this is equal to 500 times let me try to dot instead times 0.4 which is equal to and make sure I get this right this would be with two zeros and we won behind the decimal yep there you go it would be 200 so that rate at which our height is changing with respect to time right at that moment is 200 meters meters per 200 meters per minute dhdt is equal to 200 meters per minute