Implicit Differentiation
UBC Calculus Online Course Notes

Applications of The Chain Rule

There are many examples in science and in day-to-day life in which quantities associated with some process or situation are linked through a relationship of some kind. For example, the volume of a cylinder depends on the radius and the height of the cylinder,


\[ 
V= \pi r^2 h 
\]

Thus, if one or more of the basic quantities changes (for example, the cylinder grows longer, or gets fatter), the others will change as well. The chain rule can be applied to determining how the change in one quantity will lead to changes in the other quantities related to it. (In some books, this topic is treated in a special chapter called "Related Rates", but since it is a simple application of the chain rule, it is hardly deserving of title that sets it apart.)


Example 1: The Expanding Cylinder (alias Party Balloon)
Question:

(a) A party balloon is inflated at a constant rate. If the ratio of the length of the balloon to its radius stays fixed, at what rates do the length and the radius of the balloon change?

Solution: We will make an assumption to get started: we will suppose that the balloon is roughly cylindrical. (This is not exactly true, since balloons have nice round ends, but it will do to give us some handle on the problem.) We will also creatively "interpret" the wording of the problem to mean that the volume of the balloon is being increased at a constant rate. (Even though this is not given explicitely, it is the only sensible interpretation since only three variables, $  V, h, r $ are related, and two of them $ h, r  $ are to be determined.

Now here is what we know:

(1) The relationship of volume to radius and height (or "length") in a cylinder


\[ 
V= \pi r^2 h 
\]

(2) The additional fact that the radius and height of the "cylinder" are proportional. (The height is just the length of the balloon, and this is just another way of saying that their ratio is fixed.) We could use the symbol k for the constant relating length and radius.


\[ 
l=k r 
\]

(3) The fact that the Volume increases at a constant rate


\[ 
\frac{dV(t)}{dt} = F 
\]

where $ F $ is the constant ("flow rate") of air into the balloon.

Putting together what we know, we deduce that


\[ 
V(t)=\pi r^2 l = \pi r^2 ( k r) = \pi k r^3 
\]

Thus


\[ 
\frac{dV(t)}{dt}= \frac{ d}{dt} (\pi k r^3) 
= \pi k \frac{d}{dt}(r^3) 
\]

At this point we must remember to use the chain rule to calculate the derivative in the above expression, since $ r(t) $ is a function of time which is then itself cubed in the expression. Thus,


\[ 
\frac{dV(t)}{dt}= \pi k \frac{d}{dr}(r^3) \frac{dr}{dt} 
= \pi k (3 r^2) \frac{dr}{dt} 
\]

We can now use (3) to answer the question posed, plugging in


\[ 
F = \pi k (3 r^2) \frac{dr}{dt} 
\]

we find that


\[ 
\frac{dr}{dt}= \frac{F}  {3 \pi k r^2} 
\]

Observe that the rate of increase of the radius is not constant ! When the radius is still small (e.g. $ r= 1~cm $ , its rate of increase is relatively large, $  dr/dt = F/(3 k \pi) $ . But when the radius is already quite large, (e.g. $  r= 5~cm $ ) its growth is much smaller, $  dr/dt = F/( 75 k \pi)  $ .


For your consideration:

(a) Explain why the radius seems to be changing more quickly when it is small than when it is larger.

(b) Finish the solution to this problem by determining the rate of change of the length of the balloon. (Hint there is a simple way to do this, using the relationship between length and radius.)



Example 1: The Expanding Cylinder, Continued

Question (b): Suppose that we are not told anything about the relationship of the radius and the length in the cylinder. What can we say about their relative rates of growth at the instant that the radius is 5 cm and the length is 20 cm ?

Solution: In this case, we must assume that the quantities are time dependent, so that


\[ 
V(t)= \pi r^2(t) l(t) 
\]

(Note: $ r^2(t) $ means $ r^2 $ where $ r $ is a function of $ t $ .) Thus, to proceed, we must use the Product Rule of Differentiation,


\[ 
\frac{dV(t)}{dt}= \pi \frac {d}{dt} (r^2(t) l(t)) = 
\pi (r^2(t) \frac{d}{dt} l(t) + l(t) \frac{d}{dt} r(t)^2) 
\]

We must apply the chain rule to the expression involving $  r^2(t) $ , and we get


\[ 
\frac{dV(t)}{dt}= \pi (r^2(t) \frac{d l}{dt} + l(t) \frac{d}{dr} r(t)^2 
\frac{dr}{dt}) 
\]

Simplifying this expression, and using the fact that $  dV/dt = F $ we find that


\[ 
F= \pi (r^2(t) \frac{d l}{dt} +  2 l(t) r(t) \frac{dr}{dt}) 
\]

Thus, at the instant that $  r=5 ~cm, l= 20~cm $ , the relationship between the rates of change of the radius and the length of the balloon are:


\[ 
F= \pi (25 \frac{d l}{dt} +  200 \frac{dr}{dt}) 
\]

(Thus, if we know $ F $ and we know one of the rates in the above expression, we can find the other.)


For your consideration:

(a) If $ F= 1 litre/min,~~~dl/dt=10 cm/min $ , determine $ dr/dt $ at the instant when the radius is 2 cm and the length is 10 cm. (Note: $ 1~ liter = 1000 ~ cm^3 $ ).

(b) If the balloon was blown up inside a cylindrical sleeve, so that its radius was kept fixed at $  r=5 cm $ , determine the rate at which its length would increase.

(c) If the balloon is sealed, so that its volume is constant, and the "ends" are pushed in at a constant rate, $  dl/dt = -1 cm/sec $ , at what rate would the radius of the balloon change when $  r=4 cm, 
l = 20 cm $ .



Example 2: Chemistry and The Ideal Gas Law
The properties of gases have been studied for centuries, and it has been found that many gases satisfy an approximate relationship called the Ideal Gas Law which states that


\[ 
PV = n R T 
\]
where

 
\begin{eqnarray*} 
 n &=& {\rm number~ of~ moles~ of~ gas~ in~ the~ sample} \\ 
 V &=& {\rm volume~ of~ the~ sample~ in~ liters} \\ 
 T &=& {\rm temperature~ in~ degrees~ Kelvin} \\ 
 P &=& {\rm pressure~ in~ atmospheres}\\ 
 R &=& {\rm gas~ constant} = 0.08205~ {\rm liter~atm~deg}^{-1}~{\rm mole}^{-1}\\ 
\end{eqnarray*}

Question:

(a) Suppose that a 1 litre size gas cylinder containing 100 moles of oxygen gas has been carelessly placed near a radiator, so that its temperature rises at the rate of 2 degrees per minute. At what rate will the pressure build up in the cylinder?

Solution: From the description of the situation, we note that the amount of oxygen in the cylinder, and the volume of the cylinder (as well as the value of the gas constant, $ R $ ) are all kept fixed. Thus, rearranging the expression above,


\[ 
P(t) = \frac{ n R}{V} T(t) 
\] 
 (Notice that we have indicated which of the variables depend on time). Thus,


\[ 
\frac{dP(t)}{dt} = \frac{ n R}{V} \frac{dT(t)}{dt} = 
\frac{100 \times 0.08205}{1}\times 2 =16.4~ {\rm atmospheres~ per~ minute} 
\]


For your consideration:

You should check the units to make sure that they are consistent !


Example 2: Continued
Question (b):A sample of gas in a chamber is compressed by a piston so that its volume changes at a constant rate. Assuming that the temperature is held fixed, at what rate does the pressure change?

We now have a different situation since the volume is time dependent, but the temperature is not. Our relationship might indicate this as follows:


\[ 
P(t) = \frac{ n R T}{V(t)} 
\]

Thus


\[ 
\frac{dP(t)}{dt} =  n R T \frac{ d}{dt} \left( \frac{1}{V(t)}\right) 
\]

Applying the chain rule to calculate the derivative shown in this expression, we have


\[ 
\frac{dP}{dt} =  n R T \frac{ d}{dV} \left( \frac{1}{V(t)}\right)\frac{dV}{dt} 
= n R T (- \frac{1}{V^2}) \frac{dV}{dt} 
\]

We can simplify this further using additional information given. Since we also know that the volume is changing at a constant rate, we can say more about the pressure. Since we are told that the gas is being compressed by the piston, its volume must be decreasing, so that


\[ 
\frac{dV}{dt} = - k 
\]

(where $ k $ is some positive constant measuring the rate of compression). If the volume was initially $ V_0 $ we can even "guess" what the volume should be at some later time, namely


\[ 
V(t)= V_0 - k t 
\]

(Can you see why? This is the simplest dependence on time that has a constant rate of decrease, k, and satisfies the initial value $ V(0)=V_0 $ ). We can use the expressions for volume and for its rate of change in our result to express pressure as a function of time alone, simply by "plugging in" for $  V(t) $ and its derivative:


\[ 
\frac{dP}{dt} = = - n R T k \left( \frac{1}{(V_0-k t)^2}\right) 
\]

Thus, we have a description of exactly how the pressure should change with time.


For your consideration:

(a) In mathematics, as in any other human endeavour, it is wise to retain a strong sense of skepticism, and question every "result" to determine whether it makes sense. For example, do you believe that the above expression makes sense for all values of the time t? Can you see what problem might be encountered when the volume shrinks all the way to zero ? For what values of t is this result valid?


(b) A related application of the chain rule to a chemical situation concerns the relationship between a measure of the average speed of molecules in a gas and the temperature. The kinetic theory of gases develops the theory that links a quantity called the "root mean square speed" - we will just call it $ v $ - to the temperature of the gas $ T $ . The relationship is:


\[ 
v = \sqrt{\frac{3 R T}{M}} 
\]

where $ M $ is the molecular weight of the gas, $ R= 
8.315 \times 10^7 ~{\rm ergs~ deg~}^{-1}~{\rm mole}^{-1} $ is the gas constant written in the units appropriate to this situation. (Note: $  1~ erg = 1~ gm~cm^2 sec^{-2} $ .) Using this relationship, determine the rate of change of the average speed, $ dv/dt $ that occurs if the temperature increases at a constant rate.