Answer
$-0.01648$ mg and tension decreases
Work Step by Step
Write the partial derivatives.
$dT=\dfrac{\partial T}{\partial r} dr + \dfrac{\partial T}{\partial R} dR=[\dfrac{-4mgRr}{(2r^2+R^2)^2} ]dr + [\dfrac{mg(2R^2-r^2)}{(2r^2+R^2)^2} ] dR$
The above equation can be re-write as:
$\triangle T \approx [\dfrac{-4mgRr}{(2r^2+R^2)^2} ] \triangle r + [\dfrac{mg(2R^2-r^2)}{(2r^2+R^2)^2} ] \triangle R$
or, $=[\dfrac{-4mg(3) \times (0.7)}{(2(0.7)^2+(3)^2)^2} ] \times (0.1) + [\dfrac{mg(2(0.7)^2-(3)^2)}{(2(0.7)^2+(3)^2)^2} ] \times (0.1)$
or, $=mg [\dfrac{-12 \times (0.7)}{(2(0.7)^2+(3)^2)^2} ] (0.1) + mg [\dfrac{[2(0.7)^2-(3)^2]}{[2(0.7)^2+(3)^2} ] \times (0.1)$
or, $ \approx -0.01648$ mg
$\triangle T$ is negative, so it decreased.
