Lens Formula and Derivative for Error Analysis:
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Taking the derivative of both sides with respect to \(u\) and \(v\), we get:
\[ -\frac{df}{f^2} = -\frac{dv}{v^2} + \frac{du}{u^2} \]
Rearranging for \(df\):
\[ df = f^2 \left( \frac{dv}{v^2} + \frac{du}{u^2} \right) \]
Error in Measurement of Focal Length:
Since \(dv\) and \(du\) represent the measurement errors in \(v\) and \(u\),
respectively, we can substitute \(dv = \Delta v\) and \(du = \Delta u\):
\[ \Delta f = f^2 \left[ \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} \right] \]
Conclusion:
The error in the measurement of the focal length \(f\) is:
\[ \Delta f = f^2 \left[ \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2} \right] \]
If \( T = 2\pi \sqrt{\frac{L}{g}} \), \( g \) is a constant and the relative error in \( T \) is \( k \) times to the percentage error in \( L \), then \( \frac{1}{k} = \) ?
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32