AP®︎/College Calculus BC
- Related rates intro
- Analyzing problems involving related rates
- Analyzing related rates problems: expressions
- Analyzing related rates problems: expressions
- Analyzing related rates problems: equations (Pythagoras)
- Analyzing related rates problems: equations (trig)
- Analyzing related rates problems: equations
- Differentiating related functions intro
- Worked example: Differentiating related functions
- Differentiate related functions
Analyzing related rates problems: equations (trig)
Finding the rate of change of an angle that a falling ladder forms with the ground.
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- At7:30, the results are presented in radians per minute. How do we know that the results are in radians and not some other unit of measure of angles (degrees perhaps)?
Typically it's not an issue because I would either input a figure measured in radians into a formula or tell my calculator to give me results in radians, however in this case all of the inputs are in meters and minutes and the rate of change of the angle just pops out in radians. At what point did we introduce radians, as the unit of measure of the angle, into the equation?(33 votes)
- That's right. When we say the derivative of cos(x) is -sin(x) we are assuming that "x" is in radians. In degrees it would be "(d/dx)cos(x) = -sin(x)(π/180)" because the "x" in degrees increases in a rate 180/π times faster than in radians.(28 votes)
- At4:07, it says "if you take the hypotenuse times the cosine of theta, you would get x." I'm not sure how that was worked out. How does cosine relate to x?(15 votes)
- cos θ(t) = x(t) / 20 because cosθ = adjacent/hypotonuse
in order set that equal to x(t), you multiple both sides by 20, it cancels out on the right, leaving 20 * cos θ(t) = x(t)(18 votes)
- I think I just broke all the rules using arccos to solve this problem. Are steps 1 and 3 I used valid? Is there a more concise way to use arccos that I overlooked? Thanks!
1. Because x’(t) = 3, let x(t) = 3t
2. x(t0)= 5sqrt(7) Pythagorean theorem
3. Combining steps 1 and 2, t0 becomes 5/3*sqrt(7) minutes
4. d/dt arccos(3t/20) = -3/sqrt(400 - 9*t^2)
5. Set t = 5/3 sqrt(7): -3/sqrt(400 - 9*(5/3 sqrt(7))^2) = -1/5(6 votes)
- You lost me at step 1. It appears you took the antiderivative of x'(t) = 3 to get x(t) = 3t. The antiderivative of x'(t) = 3 is actually 3t + C. From there, I understand step 2, but how did you get step 3, specifically? In what way did you combine steps 1 and 2? Step 4 looks like it would take an application of the chain rule, which I don't see evidence of. I didn't even look at step 5.
Can you please review how you got step 3 and 4, explicitly?(5 votes)
- Could this question be answered using tan instead of cos for the initial equation set up?
For instance, instead of x(t) = 20(cos(θ(t))),
could you do x(t) = 15/(tan(θ(t)))?
I was only able to get as far as x'(t) = 15(sec^2(θ(t))) * (θ'(t)) because when I simplified, one of the variables was x(t0) which was not one of the given constants. If someone can give me a step-by-step solution of this if possible that would be great(5 votes)
- You can use tangent but 15 isn't a constant, it is the y-coordinate, which is changing so that should be y(t).
When you solve for θ you'll get
θ = arctan(y(t)/x(t))
then to get θ', you'd use the chain rule, and then the quotient rule.
During the quotient rule you'll get a y'(t), which isn't given, so then you'll have to set up another related rates equation between y and x to get y', and then plug that back in, etc.
It would take a lot lot more work.
You want to minimize the number of variables in the equation, so to avoid using x AND y, Sal used the relationship between variable x and the CONSTANT 20.(1 vote)
- What does sub zero mean?(3 votes)
- sub anything indicates a subscript, a small number on the lower right of a number, an exponent is a superscript, a small number or letter in the upper right.
subscripts are usually used to label things, especially in an order. so if you have multiple variables a lot of times people will label them like x sub 0, x sub 1, x sub 2 and so on. To write it out in typing it's usually done like this, x_0 with an underscore.
Since a lot of the time the ordering starts with 0, a variable sub zero is often used to say "starting point" so here t_0 is starting time.(4 votes)
- I just don't understand how at4:53, the derivative turns into -20 sin theta. As far as I can tell, there was no -1 exponent to multiply onto the coeffecient of 20 and make it negative. Also how did the sin theta come in?(2 votes)
- Can you give me an equation? Also, the minus might come from cosine.(4 votes)
- At4:17- Sal says that x(t) = 20 * cos(theta),
What videos can I watch to study where he got that? My trigonometry is weak :((2 votes)
- I'd have to look to find the videos, but maybe it will help if I just bring up the three trig relations, SOH CAH TOA. In this case x(t) is the adjacent side to the angle theta(t), and 20 is the hypotenuse. So cos(theta(t)) = adjacent/hypotenuse = x(t)/20 So then just solve for x(t). Does that help?(4 votes)
- Why radians?? when did they even come in(2 votes)
- Radians were made by God (naturally occurring), degrees were made arbitrarily by man. Imagine the line that an angle traces along the unit circle's circumference.
What's the most natural way to describe the size of that angle? Well, it's the proportion of the ENTIRE circumference that the line drawn by the angle takes up.
God made the circumference of all the circles in the universe 2*pi*r right? So the unit circle has a 2*pi*(1) or just 2*pi circumference.
So the "angle" that draws a line that goes around half of that circumference is 1/2 of 2*pi or just pi; a quarter of the whole circumference is 1/4 of 2*pi or pi/2.(3 votes)
- what if we just use
d/dt(θ(t)) = d/dt(arc cos(x(t)/20))
figuring out d/dx(arc cos(x(t)/20)) = -(3/20)(1/sqrt(1-(x(t)/20)^2) and x(t) = sqrt 175
same thing, innit?(2 votes)
- How would you solve for the change in height, given the derivative of the hypotenuse and the height of the triangle, but where the angle is constant?(1 vote)
- so x and y are both growing?
y = hypotenuse / sin(theta)
dy/dt = d (hypotenuse)/dt / sin(theta)(1 vote)
- [Instructor] 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 three 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 want 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, 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 three 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 three meters, I'll write it out because it's hard if I said M per M it might not be that clear. Meters per minute, so they give us that piece of information. So the rate of change of X, with respect to time, they gave us that. At a certain sub 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 zero Y of T is 15 meters. So let me just write it in here, Y of T sub zero is equal to 15 meters. And so, maybe let me write this right over here. This is Y of T sub zero. Let's just assume that we're drawing it at that moment T sub zero, because I think that's going to be important. Y of 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 saying theta is also going to change with respect to time. There's going to be a function of time between the ground and the ladder at that instant. So theta, 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 perspective time is constant at three 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 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 trigonometry right over here. If you took the hypotenuse times the cosine of theta you would get X. So let me write this right up here, X of T X of T is equal to the hypotenuse 20 meters, because that's the length of the ladder, times the cosine cosine of theta. And I can 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 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 with 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 I need to multiple that times theta prime of T. So what I can do is, say hey look, at T sub zero, 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 zero so at T is equal to T sub zero. What we're gonna have X prime of T. Well that, at every time, it's three 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 three is equal to negative 20 times sine of theta of T times the derivative of data 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 to get a little bit more real estate. So sine of theta, let me write it over here. Sine of theta at time T sub not, that's what we care about T is equal to T sub not. What's that gonna be? Well sine is opposite over hypotenuse. So that's gonna be Y at T sub not. 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 not is 15 meters over 20 meters. Which is the same thing as three fourths. So, by this yellow information, they actually told us that this right over here is going to be equal to three fourths. So this times three fourths times the rate of change of theta with respect to T. And so now we just have to solve for this and we're done. So, this is going to be what's negative 20 times three fourths? 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 three over negative 15. Three over negative 15. Which is the same thing as being equal to negative one fifth. 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 radians per minute. Ideally I would write it right over here. But here you go. We were able to figure out and 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.