how can I solve this problem using total differential

Question

The deflection y at the center of a uniform loaded simply supported beam is given by: $$y=\frac{5}{384}*\frac{wl^2}{EI}$$ Where $w$ is the uniform load, $l$ is the beam length, $E$ is the young modulus and $I$ is the moment of inertia of beam cross-section. If $w$ increases by $0.02%$, $l$ increases by $0.03%$, $E$ decreases by $0.02$, and $I$ decreases by $0.01%$, using the total differential to find the percentage increase in $y$?

how can I solve this problem using the total differential?

Answer 1

$y = \frac{5}{384}\frac{wl^2}{EI}$

$\ln(y)= \ln(5) - \ln(384) + \ln(w) + 2\ln(l)-\ln(E)-\ln(I)$

Taking differentials on both sides,

$\frac{dy}{y} = 0 - 0 +\frac{dw}{w}+2\frac{dl}{l}-\frac{dE}{E}-\frac{dI}{I}$

Given: $$dw = 0.02 , dl = 0.03 , dE = -0.02 , dI = -0.01$$

So,

$$\frac{dy}{y} = \frac{0.02}{w} + \frac{0.06}{l} +\frac{0.02}{E} +\frac{0.01}{I} = \bigg[\frac{2}{w}+\frac{6}{l}+\frac{2}{E}+\frac{1}{I}\bigg]\frac{1}{100}$$