If P = frac{A^3}{B^{5/2}} and Delta A is the | vedclass

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(A) Given the relation $P = \frac{A^3}{B^{5/2}}$.Taking the natural logarithm on both sides,we get $\ln P = 3 \ln A - \frac{5}{2} \ln B$.Differentiati

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$\Delta P = \pm \left( 3 \frac{\Delta A}{A} + \frac{5}{2} \frac{\Delta B}{B} \right) P$

Vedclass · Answer

(A) Given the relation $P = \frac{A^3}{B^{5/2}}$.Taking the natural logarithm on both sides,we get $\ln P = 3 \ln A - \frac{5}{2} \ln B$.Differentiating both sides,we obtain $\frac{dP}{P} = 3 \frac{dA}{A} - \frac{5}{2} \frac{dB}{B}$.For maximum relative error,we consider the absolute values of the terms,so $\frac{\Delta P}{P} = \pm \left( 3 \frac{\Delta A}{A} + \frac{5}{2} \frac{\Delta B}{B} \right)$.Multiplying both sides by $P$,we get the absolute error $\Delta P = \pm \left( 3 \frac{\Delta A}{A} + \frac{5}{2} \frac{\Delta B}{B} \right) P$.

If $P = \frac{A^3}{B^{5/2}}$ and $\Delta A$ is the absolute error in $A$ and $\Delta B$ is the absolute error in $B$,then the absolute error $\Delta P$ in $P$ is:

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