The rate of change of the diagonal R of a squ | vedclass

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(B) Let the side length of the square be $x$.The diagonal $R$ is given by $R = \sqrt{x^2 + x^2} = x\sqrt{2}$.The area $A$ is given by $A = x^2$.From t

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$1/R$

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(B) Let the side length of the square be $x$.The diagonal $R$ is given by $R = \sqrt{x^2 + x^2} = x\sqrt{2}$.The area $A$ is given by $A = x^2$.From the area equation,we have $x = \sqrt{A}$.Substituting this into the diagonal equation,we get $R = \sqrt{2} \cdot \sqrt{A} = \sqrt{2A}$.To find the rate of change of $R$ with respect to $A$,we differentiate $R$ with respect to $A$:$\frac{dR}{dA} = \frac{d}{dA}(\sqrt{2} \cdot A^{1/2}) = \sqrt{2} \cdot \frac{1}{2} A^{-1/2} = \frac{\sqrt{2}}{2\sqrt{A}} = \frac{1}{\sqrt{2A}}$.Since $R = \sqrt{2A}$,we can substitute this back into the expression:$\frac{dR}{dA} = \frac{1}{R}$.

The rate of change of the diagonal $R$ of a square with respect to its area $A$ is:

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