Respuesta :
check the picture below
[tex]\bf r^2=x^2+y^2\implies 2r\cfrac{dr}{dt}=2x\cfrac{dx}{dt}+2y\cfrac{dy}{dt}\implies \cfrac{dr}{dt}=\cfrac{x\frac{dx}{dt}+y\frac{dt}{dt}}{r} \\\\\\ \cfrac{dr}{dt}=\cfrac{(5\cdot -0.5)+(3\cdot 0.4)}{\sqrt{34}}[/tex]
if it's a negative value, thus a negative rate, thus is decreasing, if it is a positive value, then increasing.
[tex]\bf r^2=x^2+y^2\implies 2r\cfrac{dr}{dt}=2x\cfrac{dx}{dt}+2y\cfrac{dy}{dt}\implies \cfrac{dr}{dt}=\cfrac{x\frac{dx}{dt}+y\frac{dt}{dt}}{r} \\\\\\ \cfrac{dr}{dt}=\cfrac{(5\cdot -0.5)+(3\cdot 0.4)}{\sqrt{34}}[/tex]
if it's a negative value, thus a negative rate, thus is decreasing, if it is a positive value, then increasing.

The diagonal is the hypotenuse of a 5 by 3 triangle.
d = (L^2 + W^2)^.5 = SQRT(34) or 34^.5
Taking the derivative of d:
d' = (1/2)(2LL' + 2WW')(L^2 + W^2)^(-.5)
Solving for d' given the L=5, L'=-.5, W=3, W'=+.4
yields d is decreasing at a rate of -2229 feet/sec.
d = (L^2 + W^2)^.5 = SQRT(34) or 34^.5
Taking the derivative of d:
d' = (1/2)(2LL' + 2WW')(L^2 + W^2)^(-.5)
Solving for d' given the L=5, L'=-.5, W=3, W'=+.4
yields d is decreasing at a rate of -2229 feet/sec.