Question

Explain the concept of finding small percentage changes using differentiation with the help of the following problem.
The time period T for a simple pendulum of length l is given by $T=2\pi \sqrt{\dfrac{l}{g}}$ where g is a constant. Find the percentage change in T, when l changes by 6%.

Answer 1

Hint: We must use the definition of differentiation to achieve a formula establishing a relation between the change in dependent variable with the change in independent variable. We must then use this formula after differentiating T, with respect to l, and then dividing the complete equation by T. In this way, we can calculate the percentage change in l.

Complete step-by-step solution:
We know by the definition of differentiation that if y = f(x) is a function of x, then the differentiation of f(x) is defined as
$\dfrac{dy}{dx}=\displaystyle \lim_{\Delta x \to 0}\dfrac{f\left( x+\Delta x \right)-f\left( x \right)}{\Delta x}$
We should note that $f\left( x+\Delta x \right)-f\left( x \right)$ is nothing but change in f(x), i.e., change in y.
Thus, we can write
$\Delta y=f\left( x+\Delta x \right)-f\left( x \right)$
Hence, we have
\[\dfrac{dy}{dx}=\displaystyle \lim_{\Delta x \to 0}\dfrac{\Delta y}{\Delta x}\]
So, let us assume that
$\delta x$ is a small change in the variable x.
And, $\delta y$ is the corresponding change in variable y.
So, now we can write
\[\dfrac{dy}{dx}\approx \dfrac{\delta y}{\delta x}\]
We can arrange the terms to get
\[\delta y\approx \dfrac{dy}{dx}\times \delta x...\left( i \right)\]
This relationship is called the Method of small increments, or the Method of small change.
Using this above formula, we can calculate small percentage change.
In our problem, we are given with the equation
$T=2\pi \sqrt{\dfrac{l}{g}}...\left( ii \right)$
We have the percentage change in l, and we have to find the percentage change in T.
So, let us assume $\delta l$ is a small change in the variable l.
And $\delta T$ is the corresponding change in variable T.
So, by equation (i), we can write
\[\delta T\approx \dfrac{dT}{dl}\times \delta l...\left( iii \right)\]
So, we need to find \[\dfrac{dT}{dl}\] .
Let us differentiate equation (ii) with respect to l,
\[\dfrac{dT}{dl}=\dfrac{2\pi }{\sqrt{g}}\dfrac{d\left( \sqrt{l} \right)}{dl}\]
\[\Rightarrow \dfrac{dT}{dl}=\dfrac{2\pi }{\sqrt{g}}\times \dfrac{1}{2\sqrt{l}}\]
\[\Rightarrow \dfrac{dT}{dl}=\dfrac{\pi }{\sqrt{g}\sqrt{l}}...\left( iv \right)\]
Putting the value from equation (iv) into equation (iii), we get
\[\delta T\approx \dfrac{\pi }{\sqrt{g}\sqrt{l}}\times \delta l\]
We need percentage change, so dividing T on both sides, we get
\[\dfrac{\delta T}{T}\approx \dfrac{\pi }{\sqrt{g}\sqrt{l}}\times \dfrac{\delta l}{T}\]
Putting the value of T from equation (i) on the RHS, we get
\[\dfrac{\delta T}{T}\approx \dfrac{\pi }{\sqrt{g}\sqrt{l}}\times \dfrac{\delta l}{2\pi \dfrac{\sqrt{l}}{\sqrt{g}}}\]
On simplifying the above equation, we get
\[\dfrac{\delta T}{T}\approx \dfrac{1}{2}\dfrac{\delta l}{l}\]
Now, multiplying by 100 on both sides,
\[\dfrac{\delta T}{T}\times 100\approx \dfrac{1}{2}\dfrac{\delta l}{l}\times 100\]
Thus, we get
$\text{Percentage change in }T\approx \dfrac{1}{2}\times \text{Percentage change in }l$
It is given that percentage change in l is 6%.
$\therefore \text{Percentage change in }T\approx \dfrac{1}{2}\times 6\%$
$\Rightarrow \text{Percentage change in }T\approx 3\%$
Hence, the percentage change in Time Period is approximately 3%.

Note: We must keep in mind that this is just an approximate method to find the percentage change in dependent variable, with respect to the change in independent variable. We should also notice here that the variable g is a constant, so we have used derivatives, and not partial derivatives. We must be very clear that \[\text{Percentage change}=\dfrac{\text{Change}}{\text{Original Value}}\times 100\%\] .