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# Analyzing related rates problems: equations (trig)

AP.CALC:
CHA‑3 (EU)
,
CHA‑3.D (LO)
,
CHA‑3.D.1 (EK)
,
CHA‑3.D.2 (EK)

## Video transcript

a 20 meter ladder is leaning against a wall the distance X of T between the bottom of the ladder and the wall is increasing at a rate of 3 meters per minute at a certain instant T Sub Zero the top of the ladder is a distance Y of T Sub Zero of 15 meters from the ground what is the rate of change of the angle theta of T between the ground and the ladder at that instant so what I'm going to do is draw this out and really the first step is to think about well what equation will be helpful for us to solve this problem and then we might just go ahead and actually solve the problem so a 20 meter ladder is leaning against a wall so let me draw ourselves a wall here that is my wall now let me draw our 20 meter ladder so maybe it looks something like that so that is 20 meters they say the distance X of T between the bottom of the ladder and the wall so it's this distance right over here this is this distance right over here is X of T they say it's increasing at a rate of 3 meters per minute so we know that we could either say X prime of T which is the same thing as DX DT is equal to 3 meters I'll write it out because it's hard if I just said M / M it might not be that clear meters for a minute so they give us that piece of information so the rate of change of X with respect to time they give us that at a certain instant T Sub Zero the top of the ladder is a distance of 15 meters so the top of the ladder so let's make this very clear so this distance right over here is y of T Y of T and they say at time T sub 0 Y of T is 15 meters let me just write it here Y of T sub 0 is equal to 15 meters and so maybe let me write this right over here this is Y of T let's just assume that we're drawing it at that moment T Sub Zero because I think that's going to be important why at T Sub Zero is equal to 15 meters so they want to know what is the rate of change of the angle theta between the ground and the ladder and this is the same theta is also going to change with respect to time there's gonna be a function of time between the ground and the ladder at that instant so theta let me get a new color here theta is this angle right over here this is Theta and it's also going to be a function of time so what we'll always want to do in these related rates problems is we want to set up an equation and really an algebraic equation maybe a little bit of trigonometry involved that relates the things that we care about and then we're likely to have to take the derivative of both sides of that in order to relate the related rates so let's see we want to know we want to know the rate of change of the angle between the ground and the ladder at that instant so what we need to figure out we want to figure out theta Prime at T sub zero this is what we want to figure out now they've given us some interesting things they've given us I guess our rate of change of X with respect to time is constant at 3 meters per minute and we know what Y is at that moment so let's see can we create a relationship because they gave us DX DT it'll actually be more useful to find a relationship between X and theta and then take the derivative of both sides and then then use this information possibly to figure out what the appropriate value of x or theta is at that moment so let's do that so how does X relate to theta well we use a little bit of a trigonometry right over here if you took the hypotenuse times the cosine of theta you would get X so let me write this right over here X of T X of T is equal to the hypotenuse 20 meters that's the length of the ladder times the cosine cosine of theta and I could say the cosine of theta of T just to make it clear that this is a function of time this comes straight out of trigonometry actually our basic trig trigonometric function and definitions now why is this useful why do I think this is useful well let's think about what happens when I take the derivative of both sides using the chain rule on the left hand side I am going to have an X prime of T and then that's going to be equal to what do I end up on the right-hand side well using the chain rule first I'll take the derivative with respect to theta and so that's just going to be negative 20 sine of theta of T and then I need to multiply that times theta prime of T so what I could do is say hey look at T sub 0 I know what X prime of T is I could try to figure out what sine of theta of T is and then I'll just solve for this right over there so let's do that so at T sub 0 so at t is equal to T sub 0 what we're gonna have X prime of T well that's at every time it's 3 meters per minute we'll assume that our rates are in meters per minute and just our values are in meters when we're talking about distance and our angles are in radians so this is going to be equal to 3 is equal to negative 20 times sine of theta of T times the derivative of theta with respect to time so how do we figure out what sine of theta of T is going to be well let's just use that other information they gave us and I'm gonna scroll down a little bit get a little bit more real estate so sine of theta let me write it over here sine of theta at time T sub naught that's what we care about t is equal to T sub naught what's that going to be well sine is opposite over hypotenuse so that's going to be Y at T sub naught over our hypotenuse of 20 meters well that's going to be equal to that's going to be equal to they tell us Y of T sub naught is 15 meters over 20 meters which is the same thing as 3/4 so by this yellow information they actually told us that this right over here is going to be equal to 3/4 so this times 3/4 times the rate of change with of theta with respect to T and so now we just solve for this and we're done so this is going to be what's negative 20 times 3/4 that is negative 15 that is negative 15 if we divide both sides by negative 15 we get theta prime of T is equal to 3 over negative 15 3 over negative 15 which is the same thing as being equal to negative 1/5 and this the units here would be in radians per minute because our rates are all in per minute so if I wanted to I could write it radians per minute ideally I would write it right over here but there you go we were able to figure out it this is an interesting one because they give you that information on Y but really use that information of Y to figure out what sine of theta of T is but the equation you set up involves X and theta