Answer :
Sure! To determine the fractional error in [tex]\(\mu\)[/tex] given by:
[tex]\[ \mu = \frac{\sin \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)}, \][/tex]
we need to follow these steps:
1. Calculate partial derivatives [tex]\(\frac{\partial \mu}{\partial A}\)[/tex] and [tex]\(\frac{\partial \mu}{\partial B}\)[/tex]:
First, let's compute the partial derivative of [tex]\(\mu\)[/tex] with respect to [tex]\(A\)[/tex]:
[tex]\[ \frac{\partial \mu}{\partial A} = \frac{\cos \left(A + \frac{B}{2}\right) \cdot \sin \left(\frac{A}{2}\right) - \sin \left(A + \frac{B}{2}\right) \cdot \frac{1}{2} \cos \left(\frac{A}{2}\right)}{\sin^2 \left(\frac{A}{2}\right)} = \frac{\cos \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)} - \frac{\sin \left(A + \frac{B}{2}\right) \cdot \cos \left(\frac{A}{2}\right)}{2 \sin^2 \left(\frac{A}{2}\right)} \][/tex]
Now, let's compute the partial derivative of [tex]\(\mu\)[/tex] with respect to [tex]\(B\)[/tex]:
[tex]\[ \frac{\partial \mu}{\partial B} = \frac{\frac{1}{2} \cos \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)} \][/tex]
2. Using error propagation to find the fractional error in [tex]\(\mu\)[/tex]:
The propagation of errors formula for a function [tex]\(z = f(x, y)\)[/tex] due to the errors in [tex]\(x\)[/tex] and [tex]\(y\)[/tex] is given by:
[tex]\[ \left( \frac{\Delta z}{z} \right)^2 = \left( \frac{\partial z}{\partial x} \cdot \frac{\Delta x}{x} \right)^2 + \left( \frac{\partial z}{\partial y} \cdot \frac{\Delta y}{y} \right)^2 \][/tex]
In our case, [tex]\(z = \mu\)[/tex], [tex]\(x = A\)[/tex], [tex]\(y = B\)[/tex], and the fractional errors are [tex]\(\frac{\Delta A}{A}\)[/tex] (denoted as [tex]\(\delta_A\)[/tex]) and [tex]\(\frac{\Delta B}{B}\)[/tex] (denoted as [tex]\(\delta_B\)[/tex]). Hence, the fractional error in [tex]\(\mu\)[/tex] is given by:
[tex]\[ \delta_{\mu} = \sqrt{\left( \frac{\partial \mu}{\partial A} \cdot \frac{\Delta A}{A} \right)^2 + \left( \frac{\partial \mu}{\partial B} \cdot \frac{\Delta B}{B} \right)^2} \][/tex]
Replace the partial derivatives and simplify:
[tex]\[ \delta_{\mu} = \left| \frac{\partial \mu}{\partial A} \cdot \frac{A}{\mu} \right| \cdot \delta_A + \left| \frac{\partial \mu}{\partial B} \cdot \frac{B}{\mu} \right| \cdot \delta_B \][/tex]
Plugging in the partial derivatives:
[tex]\[ \delta_{\mu} = \left|\left(\frac{\cos \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)} - \frac{\sin \left(A + \frac{B}{2}\right) \cdot \cos \left(\frac{A}{2}\right)}{2 \sin^2 \left(\frac{A}{2}\right)}\right) \cdot \frac{A}{\frac{\sin \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)}}\right| \cdot \delta_A + \left|\frac{\frac{1}{2} \cos \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)} \cdot \frac{B}{\frac{\sin \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)}}\right| \cdot \delta_B \][/tex]
Simplify further:
[tex]\[ \delta_{\mu} = \left|\left(\frac{\cos \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)} - \frac{\sin \left(A + \frac{B}{2}\right) \cdot \cos \left(\frac{A}{2}\right)}{2 \sin^2 \left(\frac{A}{2}\right)}\right) \cdot \frac{A \cdot \sin \left(\frac{A}{2}\right)}{\sin \left(A + \frac{B}{2}\right)}\right| \cdot \delta_A + \left|\frac{\frac{1}{2} \cos \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)} \cdot \frac{B \cdot \sin \left(\frac{A}{2}\right)}{\sin \left(A + \frac{B}{2}\right)}\right| \cdot \delta_B \][/tex]
[tex]\[ \delta_{\mu} = \left| \frac{\left(\cos \left(A + \frac{B}{2}\right) \cdot \sin \left(\frac{A}{2}\right) - \frac{1}{2} \sin \left(A + \frac{B}{2}\right) \cdot \cos \left(\frac{A}{2}\right)\right) \cdot A}{\sin \left(A + \frac{B}{2}\right)} \right| \cdot \delta_A + \left| \frac{\frac{1}{2} \cos \left(A + \frac{B}{2}\right) \cdot B}{\sin \left(A + \frac{B}{2}\right)} \right| \cdot \delta_B \][/tex]
After the absolute values and correct terms simplifications, the expression for the fractional error in [tex]\(\mu\)[/tex] is obtained.
Thus, the final result for the fractional error in [tex]\(\mu\)[/tex] is:
\[
\delta_{\mu} = \delta_A \cdot \left| \frac{\cos\left( A + \frac{B}{2} \right)}{\sin\left( A + \frac{B}{2} \right)} \right| + \delta_B \cdot \left| \frac{\cos\left( A + \frac{B}{2}/2 \right)}{2 \sin\left( A + \frac{B}{2} \right)} \right|
[tex]\[ \mu = \frac{\sin \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)}, \][/tex]
we need to follow these steps:
1. Calculate partial derivatives [tex]\(\frac{\partial \mu}{\partial A}\)[/tex] and [tex]\(\frac{\partial \mu}{\partial B}\)[/tex]:
First, let's compute the partial derivative of [tex]\(\mu\)[/tex] with respect to [tex]\(A\)[/tex]:
[tex]\[ \frac{\partial \mu}{\partial A} = \frac{\cos \left(A + \frac{B}{2}\right) \cdot \sin \left(\frac{A}{2}\right) - \sin \left(A + \frac{B}{2}\right) \cdot \frac{1}{2} \cos \left(\frac{A}{2}\right)}{\sin^2 \left(\frac{A}{2}\right)} = \frac{\cos \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)} - \frac{\sin \left(A + \frac{B}{2}\right) \cdot \cos \left(\frac{A}{2}\right)}{2 \sin^2 \left(\frac{A}{2}\right)} \][/tex]
Now, let's compute the partial derivative of [tex]\(\mu\)[/tex] with respect to [tex]\(B\)[/tex]:
[tex]\[ \frac{\partial \mu}{\partial B} = \frac{\frac{1}{2} \cos \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)} \][/tex]
2. Using error propagation to find the fractional error in [tex]\(\mu\)[/tex]:
The propagation of errors formula for a function [tex]\(z = f(x, y)\)[/tex] due to the errors in [tex]\(x\)[/tex] and [tex]\(y\)[/tex] is given by:
[tex]\[ \left( \frac{\Delta z}{z} \right)^2 = \left( \frac{\partial z}{\partial x} \cdot \frac{\Delta x}{x} \right)^2 + \left( \frac{\partial z}{\partial y} \cdot \frac{\Delta y}{y} \right)^2 \][/tex]
In our case, [tex]\(z = \mu\)[/tex], [tex]\(x = A\)[/tex], [tex]\(y = B\)[/tex], and the fractional errors are [tex]\(\frac{\Delta A}{A}\)[/tex] (denoted as [tex]\(\delta_A\)[/tex]) and [tex]\(\frac{\Delta B}{B}\)[/tex] (denoted as [tex]\(\delta_B\)[/tex]). Hence, the fractional error in [tex]\(\mu\)[/tex] is given by:
[tex]\[ \delta_{\mu} = \sqrt{\left( \frac{\partial \mu}{\partial A} \cdot \frac{\Delta A}{A} \right)^2 + \left( \frac{\partial \mu}{\partial B} \cdot \frac{\Delta B}{B} \right)^2} \][/tex]
Replace the partial derivatives and simplify:
[tex]\[ \delta_{\mu} = \left| \frac{\partial \mu}{\partial A} \cdot \frac{A}{\mu} \right| \cdot \delta_A + \left| \frac{\partial \mu}{\partial B} \cdot \frac{B}{\mu} \right| \cdot \delta_B \][/tex]
Plugging in the partial derivatives:
[tex]\[ \delta_{\mu} = \left|\left(\frac{\cos \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)} - \frac{\sin \left(A + \frac{B}{2}\right) \cdot \cos \left(\frac{A}{2}\right)}{2 \sin^2 \left(\frac{A}{2}\right)}\right) \cdot \frac{A}{\frac{\sin \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)}}\right| \cdot \delta_A + \left|\frac{\frac{1}{2} \cos \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)} \cdot \frac{B}{\frac{\sin \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)}}\right| \cdot \delta_B \][/tex]
Simplify further:
[tex]\[ \delta_{\mu} = \left|\left(\frac{\cos \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)} - \frac{\sin \left(A + \frac{B}{2}\right) \cdot \cos \left(\frac{A}{2}\right)}{2 \sin^2 \left(\frac{A}{2}\right)}\right) \cdot \frac{A \cdot \sin \left(\frac{A}{2}\right)}{\sin \left(A + \frac{B}{2}\right)}\right| \cdot \delta_A + \left|\frac{\frac{1}{2} \cos \left(A + \frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)} \cdot \frac{B \cdot \sin \left(\frac{A}{2}\right)}{\sin \left(A + \frac{B}{2}\right)}\right| \cdot \delta_B \][/tex]
[tex]\[ \delta_{\mu} = \left| \frac{\left(\cos \left(A + \frac{B}{2}\right) \cdot \sin \left(\frac{A}{2}\right) - \frac{1}{2} \sin \left(A + \frac{B}{2}\right) \cdot \cos \left(\frac{A}{2}\right)\right) \cdot A}{\sin \left(A + \frac{B}{2}\right)} \right| \cdot \delta_A + \left| \frac{\frac{1}{2} \cos \left(A + \frac{B}{2}\right) \cdot B}{\sin \left(A + \frac{B}{2}\right)} \right| \cdot \delta_B \][/tex]
After the absolute values and correct terms simplifications, the expression for the fractional error in [tex]\(\mu\)[/tex] is obtained.
Thus, the final result for the fractional error in [tex]\(\mu\)[/tex] is:
\[
\delta_{\mu} = \delta_A \cdot \left| \frac{\cos\left( A + \frac{B}{2} \right)}{\sin\left( A + \frac{B}{2} \right)} \right| + \delta_B \cdot \left| \frac{\cos\left( A + \frac{B}{2}/2 \right)}{2 \sin\left( A + \frac{B}{2} \right)} \right|
