Approximation of Errors
Consider a rectangle ABCD of length \(l\) and breadth \(b\). Let its breadth is increased and length is decreased to form a new rectangle A'B'C'D'.
Then,
Area of rectangle = \(l\times b \)
Error/change in length \(= \delta l \)
Error/change in breadth \(= \delta b \)
Error/change in Area \(= \delta A \)
\[ \delta A = -\frac{\partial A}{\partial l}\delta l + \frac{\partial A}{\partial b}\delta b \]
Negative sign means decrease in the quantity and Positive sign means increase in the quantity.
So if there is a function \(u\) in two variables \(x\) and \(y\), then change/error in \(u\) is given by:
\[ \boxed{\delta u = \frac{\partial u}{\partial x}\delta x + \frac{\partial u}{\partial y}\delta y} \]
If \(u\) is in three variables \(x,y,z\), then
\[ \boxed{\delta u = \frac{\partial u}{\partial x}\delta x + \frac{\partial u}{\partial y}\delta y + \frac{\partial u}{\partial z}\delta z} \]
For one variable,
\[ \boxed{ \delta y = \frac{dy}{dx}\delta x } \]
Types of Error/Change
1) Absolute Error/ Absolute Change = \(| \delta x| \)
2) Relative Error/ Relative Change = \(\Large \frac{\delta x}{x} \)
3) Percentage Error/ Percentage Change = \(\Large\frac{\delta x}{x} \small \times 100 \)
Here \(x = \) Actual value/ True value
\( \delta x = \) Error/Change in \(x\)
Sample Problem 1
If the radius of base and the height of a cone are measured as 4cm and 8cm with a possible error of 0.04cm and 0.08cm respectively, then calculate the percentage error in calculating the volume of cone.
Solution
Volume of cone = \(V = \large\frac{1}{3}\pi r^2h \)
Given, radius = \( r = 4cm \)
height = \( h = 8cm \)
Error in radius = \(\delta r = 0.04cm \)
Error in height = \(\delta h = 0.08cm \)
Now,
\[ \delta V = \frac{\partial V}{\partial r}\delta r + \frac{\partial V}{\partial h}\delta h \]
\[ = \left(\frac{2}{3}\pi rh \right)(0.04) + \left(\frac{1}{3}\pi r^2h \right)(0.08) \]
\[ \Rightarrow \boxed{\delta V = \frac{384}{300}\pi } \]
Percentage Error in Volume
\[ = \frac{\delta V}{V}\times 100 \]
\[ = \large \frac{\frac{384}{300}\pi}{\frac{1}{3}\pi r^2h} \times 100 \]
\[ = \large \frac{\frac{384}{300}\pi}{\frac{1}{3}\pi (4)^2(8)} \times 100 \]
\[ = \boxed{3\%} \]
SAmple Problem 2
The time \(T\) of a complete oscillation of a simple pendulum of length \(L\) is governed by equation \[ T = 2\pi \left(\frac{L}{g} \right)^{\large\frac{1}{2}} \] Find maximum error in \(T\) dur to possible errors upto 1% in \(L\) and 2% in g.
Solution
\[ T = 2\pi \left(\frac{L}{g} \right)^{\large\frac{1}{2}} \]
Percentage Error in \(L = 1\%\)
\[ \Rightarrow \frac{\delta L}{L}\times 100 = 1 \]
Percentage Error in \(g = 2\% \)
\[ \Rightarrow \frac{\delta g}{g}\times 100 = 2 \]
In such type of equations, it is better to take log on both sides as it will simplify the calculation
\[ \log{T} = \log{2} + \log{\pi} + \frac{1}{2}\log{L} - \frac{1}{2}\log{g} \]
Partially Differentiating,
\[ \frac{1}{T}\delta T = 0 + 0 + \frac{1}{2}\frac{\delta L}{L} - \frac{1}{2}\frac{\delta g}{g} \]
\[ \frac{\delta T}{T} = \frac{1}{2}\frac{\delta L}{L} - \frac{1}{2}\frac{\delta g}{g} \]
\[ \frac{\delta T}{T}\times 100 = \frac{1}{2}\frac{\delta L}{L}\times 100 - \frac{1}{2}\frac{\delta g}{g}\times 100 \]
\[ \frac{\delta T}{T}\times 100 = \frac{1}{2} - \frac{2}{2} \]
\[ \boxed{\frac{\delta T}{T}\times 100 = -0.5\%} \]
