AP®︎/College Calculus AB
- Related rates intro
- Related rates (multiple rates)
- Related rates: Approaching cars
- Related rates: Falling ladder
- Related rates (Pythagorean theorem)
- Related rates: water pouring into a cone
- Related rates (advanced)
- Related rates: shadow
- Related rates: balloon
Related rates: balloon
In this example, you are analyzing the rate of change of a balloon's altitude based on the angle you have to crane your neck to look at it. Created by Sal Khan.
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- Can you give us the dimension analysis of how you get the unit of m/min?(39 votes)
- I just looked this up because I was curious as well, and Panathakorn is correct. Radians have no dimension. There are a lot of really convoluted explanations on the web (obviously made by the same math teachers that drove us all to the wonderful clarity of the Khan Academy :-), but I did find this link. The student is asking about degrees versus radians, but the answer goes into what we are discussing. I think it is a pretty clear explanation of what is going on:
- At5:20, you take the derivative of h/500. Aren't you supposed to use the quotient rule for that?(9 votes)
- You don't have a variable in the denominator. That 500 m is a constant. So you would treat it as ( ¹⁄₅₀₀ ) h(43 votes)
- Wait... If the rate of change is measured in radians, and it is a constant rate, wouldn't that mean the balloon was accelerating as it was rising? Wouldn't the balloon have to be going infinitely fast after 7 or 8 minutes?(15 votes)
- Yes. This is a valuable observation you have made.
Because the equation for the derivative of H includes a theta term in addition to the constant rate of change in theta, H does have a rate that varies based on time, and thus has an acceleration function.
Regarding the second piece of you question, yes, it would get going quite fast. This is because the limit as x->pi/2 of sec(x) is infinity. These relates are not intended for use beyond several minutes. Your balloon would rise unreasonably fast neat 3.926 minutes, but then would begin falling afterwards.
At "7 or 9 minutes" the balloon would be in the middle of its fluctuations down towards the earth.
The second derivative (acceleration) of H is 40 sec^2(theta).
500*.2*sec^2(theta)->500*0.2*2sec(theta)[the power rule]*dtheta/dt[chain rule]
The symbol " * " has been used to represent multiplication throughout this answer.(8 votes)
- at1:13: Why is this angle changing?(2 votes)
- the angle is changing because the balloon itself is rising and you want to calculate the rate of change(28 votes)
- Is it possible to solve this without calculus?(4 votes)
- Well in Calculus "Rate of Change" means Derivative. Since it's a Related RATE problem, Calculus must be incorporated into the problem in order to solve it.(20 votes)
- At 6.57 Sal says and writes; "cos (pi/4) = sqrt2/2"
Please correct me if I'm wrong or misunderstanding this somehow;
isn't cos (pi/4) = 1/sqrt2 ?(2 votes)
- 1/sqrt(2) is the same thing as sqrt(2)/2. Multiply both sides of the first fraction by sqrt(2) and you'll see :)
People often like to make the denominator rational, so they do these types of tricks.(14 votes)
- When I solved the question in a different way, I got a different answer since I believe my method is correct mathematically. Sal, solve it by using
tan(θ)function. To use
sin(θ), I need
hypto match the definition of the function. So I found the
hypby using cos(θ) function;
hyp= adj/cos(θ). Since
θ= π/4, the hyp of the angle is
hyp=707.106. By using
sin(θ)= h/707.106, the derivative is
dh/dt = 707.1 * cos(θ) * dθ/dt. Substituting the variable.
dh/dt= 707.11 * cos(π/4) * 0.2
dh/dt= 99.99!why is it different?(4 votes)
- Although you're evaluating θ at ¼π, θ and the hyp are not constants. So, you need to plug in those after differentiating. It would harder to do that way.(10 votes)
- At5:09Why does h/500 become 1/500 (dh/dt)? Can someone walk me through the steps?(4 votes)
- Sal is taking the derivative of the entire equation with respect to t, and on the right that means finding the derivative of h/500 with respect to t. His handwriting is hard to read here, but that's d/dt in white next to the expression [h/500].
The expression h/500 is the same as (1/500)*h. In this expression, 1/500 is a constant, so the derivative of (1/500)*h with respect to t is the same as (1/500) times the derivative of h with respect to t, and that's how we get (1/500)*(dh/dt).(6 votes)
- I know that
tan(theta) = opp/adj, so, if I want to calculate the opposite side (the length traveled by the globe) at the given time, and at the given time+1 min, shouldn't I be able to:
tan( pi/4+.2 ) * 500 - tan( pi/4 ) * 500 = 254.25 meters/min
That all divided by 1 minute of course. However, that clearly doesn't give the expected result, which is
200 meters/min... where am I wrong?(2 votes)
- To calculate
[tan(π/4 + 0.2) - tan(π/4)] * 500will give you the result whose unit is
meters/min. We need to find the rate of change.(4 votes)
- At what rate is the distance between the observer and the balloon changing when the balloon is Xft high. The rate is given but no angles are given.(3 votes)
- Sounds like you would set up the rates using the pythagorean theorem if I'm picturing that right. Could you write what the information is that is given?(1 vote)
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 say it 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 500 meters away from there. So you know that you are 500 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 don't 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 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 the rate at which this angle is changing. So this is changing at 0.2 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 line between you and the hot air balloon is pi over 4 radians, and 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 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 h, what we want to figure out is dh dt. That's what we don't know. So what we'd want to come up with is a relationship between dh dt, d theta dt, and maybe theta, if we need it. Or another way to think about it, if we can come up with the relationship between h and theta, then we could take the derivative with respect to t, and we'll probably get a relationship between all of this stuff. So what'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 of theta is equal to the opposite side-- the opposite side is equal to h-- over the adjacent side, which we know is going to be a fixed 500. So there you have a relationship between theta and h. And then to figure out a relationship between d theta dt, the rate at which theta changes with respect to t, and the rate at which h changes with respect to t, we just have to take the derivative 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. I'm going to take the derivative with 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 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 theta with respect to t gives us the derivative of tangent of theta with respect to t, which is what we want when we use this type of a derivative operator. We're 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 dh dt. So 1 over 500 dh 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 in to show you that I'm putting these values in. Secant squared of pi over 4. Secant times d theta dt. Well, that is just 0.2. So times 0.2. And then 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 meters per. And this is meters right over here. We're going to get meters per minute right over here. We just want to make sure we know what our units are doing. I haven't written the units here to save some space. But we get 1 over 500 times dh dt. So if we want to solve for dh dt, we 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. 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 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 just write a dot instead-- times 0.4, which is equal to-- let me make sure I get this right. This would be with two 0's and one behind the decimal. Yep, there you go. It would be 200. So the rate at which our height is changing with respect to time right at that moment is 200 meters per minute. dh dt is equal to 200 meters per minute.