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## Differential Calculus

### Unit 4: Lesson 5

Solving related rates problems- 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

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# Related rates: balloon

AP.CALC:

CHA‑3 (EU)

, CHA‑3.E (LO)

, CHA‑3.E.1 (EK)

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.

## Want to join the conversation?

- Can you give us the dimension analysis of how you get the unit of m/min?(38 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:

http://mathforum.org/library/drmath/view/64034.html(23 votes)

- 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]

Simplified; 500*0.2*0.2*2sec(theta)=40*sec(theta)

The symbol " * " has been used to represent multiplication throughout this answer.(9 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`hyp`

to match the definition of the function. So I found the`hyp`

by using cos(θ) function;`cos(θ)= adj/hyp`

-->`hyp= adj/cos(θ)`

. Since`θ= π/4`

, the hyp of the angle is`hyp= 500/cos(π/4)`

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).(5 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)] * 500`

will give you the result whose unit is`meters`

, not`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)

## 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 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.