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Course: AP®︎/College Calculus AB>Unit 4

Lesson 5: Solving related rates problems

Related rates: balloon

Let's imagine we're at a hot air balloon show, and we're tracking a balloon's ascent. Using our knowledge of angles, distances, and related rates we can apply trigonometry and calculus to find the rate of change in the balloon's height. Created by Sal Khan.

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• 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?
• You don't have a variable in the denominator. That 500 m is a constant. So you would treat it as ( ¹⁄₅₀₀ ) h
• 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?
• the angle is changing because the balloon itself is rising and you want to calculate the rate of change
• Is it possible to solve this without calculus?
• 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.
• 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 ?
• 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.
• 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?
• 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.
• 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).
• 1. at , if the balloon is descending, is the convention Negative of something?

2. at , can you do a dimension analysis on how to get to the meter/minute as the unit? Thanks!
(1 vote)
• 1. Yes. Conventionally. we treat "going down and left" as negative and "going up and right" as positive. You can reverse them if you want to, but you need to stick to whatever convention you're following for the whole problem (i.e. you can't switch signs mid problem)

2. Yep! So, we have dh/dt = sec^2(pi/4) * 0.2 * 500. sec(pi/4) is dimensionless so the dimensions of dh/dt depend on the 0.2 and 500. 0.2 has the dimensions of rad/min and 500 has the dimensions of metres. However, coming to rad/min, radian itself has no dimensions (it's length of arc swept per radius). So, the final dimensions come out to be metres (from 500) per minute (from 0.2)