Question about the application of the chain rule

In summary: But in the end, the chain rule is just something that you have to accept based on experience. So, in the original problem, the volume is a function of the radius which is a function of time. Therefore, to find the rate of change of volume with respect to time, we use the chain rule to multiply the rate of change of volume with respect to radius by the rate of change of radius with respect to time. In summary, to determine the rate at which the radius of a spherical balloon is increasing when the diameter is 20 cm, we use the chain rule to find the rate of change of volume with respect to time by multiplying the rate of change of volume with respect to radius by the rate of change of radius with
  • #1
Astrum
269
5
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.
 
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  • #2
##\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
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
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
Vorde said:
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.

I noticed this about the canceling out.

[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
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
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
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 explanation works as long as you don't look too deep and if you do look deep then you can proof it rigorously.
 

1. What is the chain rule and how is it applied in mathematics?

The chain rule is a mathematical rule used to find the derivative of a composite function. In other words, when a function is composed of two or more functions, the chain rule allows us to find the derivative of the overall function by breaking it down into smaller parts and applying the rule.

2. How do you know when to use the chain rule?

The chain rule is used when the function is composed of two or more functions. For example, f(g(x)) or f(h(x)). Additionally, the chain rule is used when finding the derivative of a function that cannot be easily differentiated using other rules, such as the power rule or product rule.

3. Can you provide an example of how the chain rule is applied?

Sure! Let's say we have the function f(x) = (x^2 + 3)^4. We can rewrite this function as f(g(x)) where g(x) = x^2 + 3. To find the derivative of f(x), we first find the derivative of g(x) which is g'(x) = 2x. Then, we substitute g'(x) into the chain rule formula which is f'(x) = g'(x) * (f(g(x)))' = 2x * 4(x^2 + 3)^3 = 8x(x^2 + 3)^3.

4. Is the chain rule only used in calculus or can it be applied in other areas of mathematics?

The chain rule is primarily used in calculus to find derivatives, but it can also be applied in other areas of mathematics, such as physics and engineering. In these fields, the chain rule is used to find rates of change and to solve differential equations.

5. What are some common mistakes when applying the chain rule?

Some common mistakes when applying the chain rule include forgetting to take the derivative of the inner function, not simplifying the expression after applying the rule, and incorrectly applying the power rule or product rule within the chain rule formula. It is important to carefully follow the steps and practice to avoid these mistakes.

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