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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 4

Lesson 5: Solving related rates problems

# 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.

## Want to join the conversation?

• Can you give us the dimension analysis of how you get the unit of m/min? • 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
• At , you take the derivative of h/500. Aren't you supposed to use the quotient rule for that? • 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? • 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.
• at : Why is this angle changing? • Is it possible to solve this without calculus? • 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 ? • 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? • At Why does h/500 become 1/500 (dh/dt)? Can someone walk me through the steps? • 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).
• 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?  