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# Related rates: water pouring into a cone

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

## Video transcript

so we've got a very interesting scenario here I have this conical thimble like cup that is four centimeters high and also the diameter of the top of the cup is also four centimeters and I'm pouring water into this cup right now and I'm pouring the water at a rate of one cubic centimetre one cubic centimeter per second and right at this moment there is a height of two centimeters of water in the cup right now so the height right now from the bottom of the cup to this point right over here is 2 centimeters so my question to you is at what rate we know the rate at which the water is flowing into the cup or give it we're being given a volume per time my question to you is right at this moment right when we are filling our cup at 1 cubic centimeter per second and we have exactly 2 centimeters of water in the cup 2 centum it's 2 centimeters deep of water in the cup what is the rate at which the height of the water is changing what is the rate at which what is the rate at which this height right over here is actually changing we know it's 2 centimeters but how fast is it changing well let's think about this a little bit what are we being given we're given we're being given the rate at which the volume of the water is changing with respect to time so let's write that down we're being given the rate at which the volume of the water is changing with respect to time and where it's being told we were told that this is one cubic centimeter one cubic centimeter per second and what are we trying to figure out what we're trying to figure out how fast the height of the water is changing with respect to time we know that the height right now is 2 centimeters but what we want to figure out is the rate at which the height is changing the height is changing with respect to time if we can figure out this then we have essentially answered the question so one way that we can do this is if we can come up with a relationship between the volume at any moment in time and the height at any moment in time and then maybe take the derivative of that relationship possibly using the chain rule to come up with a relationship between the rate at which the volume is changing and the rate at the height is changing so let's try to do it step by step so first of all can we come up with a relationship between the volume and the height at any given at any given moment well we have also been given the formula for the volume of a cone right over here the volume of a cone is 1/3 times the area of the base of the cone times the height and we won't prove it here although we could prove it later on especially when we start doing solids of revolutions with in in integral calculus but we'll just take it on faith right now but this is how we can figure out the volume of a cone so given this can we figure out volume can we figure out an expression that relates volume to the height of the cone well we could say that volume and I'll do it in this I'll do this blue color the volume of water is what we really care about the volume of water is going to be equal to 1/3 times the area of the surface of the water the area of the surface of the water area of water surface water surface times our height of the water so times H so how can we figure out the area of the water surface preferably in terms of H well we see right over here the diameter across the top of the cone is 4 centimeters and the height of the the whole cup is 4 centimeters and so that ratio is going to be true of any when it will at any depth of water it's always going to have the same ratio between the diameter across the top and the height because these are lines right over here so at any given point the ratio between this and this is going to be the same so at any given point the diameter across the surface of the water if the depth is H the diameter across the surface of the water is also going to be H and so from that we can figure out what it radius is going to be the radius is going to be H over 2 and so the area of the water surface is going to be PI R squared pi times the radius squared H over 2 squared that's the area of the surface of the water and of course we still have the 1/3 out here and we're still and we're still multiplying by this H over here so let me see if I can simplify this so this gives us 1 third times pi h squared PI H squared over four times another H which is equal to we have PI H to the third power H to the third power over 12 so that is our volume now what we want to do is relate the volume how fast the volume is changing with respect to time and how fast the height is changing with respect to time so we care with respect to time let's take care so much about what's happening with respect to time let's take the derivative of both sides of this equation with respect to time to do that and just so I have enough space to do that let me move this over let me move this over to the right a little bit so I just move this over to the right and so now we can take the derivative with respect to time of both sides of this business so the derivative with respect to time of our volume and the derivative with respect to time of this business well the derivative with respect to time of our volume we can just rewrite that as DV DT this thing right over here this is d V DT and this is going to be equal to well we could take the constants out of this this is going to be equal to PI over 12 PI over 12 times the derivative with respect to T times the derivative with respect to T of H of H to the third power of H to the third power and just so that the next few things I do will appear a little bit clearer we're assuming that height is a function of time in fact it's definitely a function of time as time goes on the height will change because we're pouring more and more water here so instead of just writing H to the 3rd power which I could write over here let me write H of T to the third power just to make it clear that this is a function of T H of T to the third power now what is the derivative with respect to T of H of T to the third power now you might be getting a tingling feeling that the chain rule might be applicable here so let's think about the chain rule the chain rule tells us let me rewrite everything else DV with respect to T is going to be equal to is going to be equal to PI over 12 times the derivative of this with respect to T and we want to take the derivative of this with respect to T we have something to the third power so we want to take the derivative of that of something to the third power with respect to something so that's going to be let me write this in a different color maybe an orange so that's going to be three times R something squared three times R something squared times times the derivative of that something with respect to T times D H let me I've already used that pink times times D H DT let's just be very clear this orange term right over here and I'm just using the chain rule this is the derivative of this is the derivative of H of T H of T to the third power with respect to H of T with respect to H of T and then we're going to multiply that times the derivative of H of T the derivative of H of T with respect to T and then that gives us the derivative of this entire thing H of T to the third power with respect to T this will give us the derivative the derivative of H of T to the third power with respect to with respect to D with respect to T which is exactly what we want to do when we apply this operator how fast is this changing how is this changing with respect to time so we can just rewrite this just so it gets a little bit cleaner let me rewrite everything I've done so we got DV the rate at which our volume is changing with respect to time the rate at which our volume is changing with respect to time is equal to PI over 12 is equal to PI over 12 times 3 H of T squared or I could just write that as 3 H squared 3 H squared times the rate at which the height is changing with respect to time times d h dt times d h dt times d h dt and you might be a little confused you might have been tempted to take the derivative the derivative over here with respect to age but remember we're thinking about how things are changing with respect to time so we're assuming we did Express volume as a function of height but we're saying that height itself as a function of time so we're taking the derivative of everything with respect to time so that's why the chain rule came into play when we were taking the derivative of H or the derivative of H of T because we're assuming that H is a function of time now what does this thing right over here get us well we're telling us at the exact moment that we set up this problem we know what DV DT is we know that it's one centimeter cubed per second we know that this right over here is 1 centimeter cubed per second we know what our height is right at this moment we were told it is 2 centimeters our height right over here we know it is 2 centimeters so the only unknown we have over here is the rate at which our height is changing with respect to time which is exactly what we needed to figure out in the first place so we just have to solve for that so we get one cubic centimeter let me make it clear we get one cubic centimeter one cubic centimeter per second I won't write the unit's to save some space is equal to PI over 2 and I'll write this in a neutral color actually let me write it the same color is equal to PI over 2 times 3 times 3 times H squared H is 2 so you're going to get 4 squared centimeters we kept the unit's so 3 times 4 or let me be careful that's that wasn't PI over 2 that was PI over 12 this is a PI over 12 right over your PI over 12 so you get PI over 12 times 3 times 2 squared times dhdt times d h dt all of this is equal to 1 so now I'll switch to a neutral color we get 1 is equal to well 3 times 4 is 12 cancels out with that 12 we get 1 is equal to pi times D h dt to solve for d h dt divide both sides by PI and we get we get our drumroll now the rate at which our height is changing with respect to time as we're putting one cubic centimeter of water per second in and right when our height is 2 centimeters side there rate at which this height is changing with respect to time is one over one over PI and I haven't done the dimensional analysis but this is going to be in centimeters in centimeters per second you could work through the dimensional analysis if you like by putting the dimensions right over here but there you have it that's how fast our height is going to be changing at exactly that moment