Approximation Of Errors Engineering Maths, Btech first year

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:

$$\boxed{{\displaystyle \delta u=\frac{\partial u}{\partial x}\delta x+\frac{\partial u}{\partial y}\delta y}}$$

If $u$ is in three variables $x,y,z$, then

$$\boxed{{\displaystyle \delta u=\frac{\partial u}{\partial x}\delta x+\frac{\partial u}{\partial y}\delta y+\frac{\partial u}{\partial z}\delta z}}$$

For one variable,

$$\boxed{{\displaystyle \delta y=\frac{dy}{dx}\delta x}}$$

Types of Error/Change

1) Absolute Error/ Absolute Change = $|\delta x|$
2) Relative Error/ Relative Change = $\frac{\delta x}{x}$
3) Percentage Error/ Percentage Change = $\frac{\delta x}{x}\times 100$
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 = $V=\frac{1}{3}\pi r^{2}h$
Given, radius = $r=4cm$
height = $h=8cm$
Error in radius = $\delta r=0.04cm$
Error in height = $\delta h=0.08cm$
Now, $$\delta V=\frac{\partial V}{\partial r}\delta r+\frac{\partial V}{\partial h}\delta h$$ $$=\left(\frac{2}{3}\pi rh\right)(0.04)+\left(\frac{1}{3}\pi r^{2}h\right)(0.08)$$ $$\Rightarrow \boxed{{\displaystyle \delta V=\frac{384}{300}\pi }}$$ Percentage Error in Volume $$=\frac{\delta V}{V}\times 100$$ $$=\frac{\frac{384}{300}\pi }{\frac{1}{3}\pi r^{2}h}\times 100$$ $$=\frac{\frac{384}{300}\pi }{\frac{1}{3}\pi (4{)}^2(8)}\times 100$$ $$=\boxed{{\displaystyle 3\mathrm{\%}}}$$

SAmple Problem 2

The time $T$ of a complete oscillation of a simple pendulum of length $L$ is governed by equation $$T=2\pi {\left(\frac{L}{g}\right)}^{\frac{1}{2}}$$ Find maximum error in $T$ dur to possible errors upto 1% in $L$ and 2% in g.

Solution

$$T=2\pi {\left(\frac{L}{g}\right)}^{\frac{1}{2}}$$ Percentage Error in $L=1\mathrm{\%}$
$$\Rightarrow \frac{\delta L}{L}\times 100=1$$ Percentage Error in $g=2\mathrm{\%}$
$$\Rightarrow \frac{\delta g}{g}\times 100=2$$ 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, $$\frac{1}{T}\delta T=0+0+\frac{1}{2}\frac{\delta L}{L}-\frac{1}{2}\frac{\delta g}{g}$$ $$\frac{\delta T}{T}=\frac{1}{2}\frac{\delta L}{L}-\frac{1}{2}\frac{\delta g}{g}$$ $$\frac{\delta T}{T}\times 100=\frac{1}{2}\frac{\delta L}{L}\times 100-\frac{1}{2}\frac{\delta g}{g}\times 100$$ $$\frac{\delta T}{T}\times 100=\frac{1}{2}-\frac{2}{2}$$ $$\boxed{{\displaystyle \frac{\delta T}{T}\times 100=-0.5\mathrm{\%}}}$$