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Question about the application of the chain rule 
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#1
Dec2412, 04:16 AM

P: 274

Air is being pumped into a spherical balloon at a rate of 5 cm3/min. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.
So, to solve, I know HOW to do it, I just don't know WHY it's right. [itex]\frac{dv}{dr}=4pi r^{2}[/itex] [itex]\frac{dv}{dt}=5cm^{3}/s[/itex] = [itex]\frac{dv}{dt}=4pir^{2}\frac{dr}{dt}[/itex] Solve for dr/dt Where does dr/dt come from? I can't understand how it makes sense. It seems as if we're just pulling it out of the air. 


#2
Dec2412, 04:51 AM

Sci Advisor
P: 839

##\frac{dv}{dt} = \frac{dv}{dr}\frac{dr}{dt} = 4 \pi r^2 \, \frac{dr}{dt}##
Specifically what part are you having trouble with? 


#3
Dec2412, 03:29 PM

P: 274

I don't understand why we throw [itex]\frac{dr}{dt}[/itex] in there. It confuses me. the radius is a function of time, but why is it that we can do  [itex]\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}[/itex]?
It just seems like "Well, we need to find [itex]\frac{dr}{dt}[/itex], so let's just throw it in there and it'll all work out. I guess I don't understand how all these derivatives relate to each other. 


#4
Dec2412, 04:04 PM

P: 784

Question about the application of the chain rule
Well, you can look at the derivatives like fractions. Look at: ##\frac{dv}{dr} \cdot \frac{dr}{dt}##. Notice that this quantity has a ##dr## in both the numerator and the denominator so they cancel out, meaning ##\frac{dv}{dr} \cdot \frac{dr}{dt} = \frac{dv}{1} \cdot \frac{1}{dt} = \frac{dv}{dt}##
It is important to note that this isn't actually legitimate. Even though the Leibniz notation for derivatives we are using makes them look like fractions, they aren't actually fractions and can't be treated as such. Often for explaining simple ideas you can act like they are fractions to get the point across, but it's important to realize they actually aren't. 


#5
Dec2412, 04:38 PM

P: 274

[itex]\frac{dv}{dt}[/itex], here, the volume is a function of time. Change in volume with respect to time. [itex]\frac{dv}{dr}[/itex], here volume is a function of the radius. Change in volume with respect to radius. [itex]\frac{dr}{dt}[/itex], here radius is a function of time. Change in radius with respect to time? I really don't understand why [itex]\frac{dv}{dr}\frac{dr}{dr}=\frac{dv}{dt}[/itex] If it's not because they're fractions, what's the reason? Take the derivative of volume with respect to radius, multiply by derivative of radius with respect to time, which gives you change in volume with respect to time. 


#6
Dec2412, 07:08 PM

Sci Advisor
P: 839

I'm tempted to say "because that's exactly what the chain rule says, expressed in that notation", but I don't think that's actually useful. So I'll ask you directly, what do you know about the chain rule? (Write it down as if you were teaching someone.)



#7
Dec2412, 07:34 PM

P: 274

To be honest, if I was trying to show someone the chain rule, I'd do an example.
[itex]\frac{dx}{du}\frac{du}{dx}=\frac{dy}{dx}[/itex] [itex]f(x)=sin(x+1)[/itex] [itex]u=x+1[/itex] [itex]\frac{du}{dx}=1[/itex] = [itex]f(u)=sin(u); \frac{dy}{du}=cos(u)[/itex] [itex]\frac{dy}{dx}=cos(x+1)[/itex] So, in words. If y is a function of u, and u is a function of x, [itex]\frac{dx}{du}\frac{du}{dx}=\frac{dy}{dx}[/itex] So, in the original problem, [itex]\frac{dv}{dt}[/itex] is a function of [itex]\frac{dv}{dr}[/itex] which is a function of [itex]\frac{dr}{dt}[/itex]? 


#8
Dec2412, 09:47 PM

P: 784

The idea of the chain rule is that if you have a function f(p) where p is a function of x: p(x), and you want to find the rate at which f changes with respect to a change in x, you take the rate at which f changes with respect to p and multiply that by the rate at which p changes with regards to x.
There is a proof of this but at the high school level you don't encounter it. As I said, the fraction explaination works as long as you don't look too deep and if you do look deep then you can proof it rigorously. 


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