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2 votes
At a certain time, the length of a rectangle is 5 feet and its width is 3 feet. At that same moment, the length is decreasing at 0.5 feet per second and the widthis increasing at 0.4 feet per second.

What is the length of the diagonal at that time?
How fast is the length of the diagonal changing? Is this length increasing or decreasing?

User Pearman
by
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2 Answers

4 votes
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.
User Lincoln Mullen
by
6.5k points
2 votes
check the picture below


\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(dx)/(dt)+y(dt)/(dt)}{r} \\\\\\ \cfrac{dr}{dt}=\cfrac{(5\cdot -0.5)+(3\cdot 0.4)}{√(34)}

if it's a negative value, thus a negative rate, thus is decreasing, if it is a positive value, then increasing.
At a certain time, the length of a rectangle is 5 feet and its width is 3 feet. At-example-1
User Riley
by
6.1k points
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