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# Related Rates and Areas

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• Related Rates and Areas
• ### Related Rates and Areas

alyssaac11 > 01-17-2020, 03:05 AM

Suppose that a square sheet of aluminum is placed in the hot sunIt begins to expand very slowly so that its diagonal is increasing at the rate of 1 millimeter per minuteAt the moment that the diagonal is 100 millimeters, at what rate is the area increasing?Then a similar question I\'m stuck on, it\'s pretty much the same, but with a triangleIn the heat of the sun, a sheet of aluminum in the shape of an equilateral triangle is expanding so that its side length increases by 1 millimeter per hourWhen the side length is 100 millimeters, how is the area increasing?The way I tried to solve them, seems totally wrongAny suggestions?Suppose that a square sheet of aluminum is placed in the hot sunIt begins to expand very slowly so that its diagonal is increasing at the rate of 1 millimeter per minuteAt the moment that the diagonal is 100 millimeters, at what rate is the area increasing? let x length of a side of a square h length of diagonal dh/dt 1 mm/min A x^2 dA/dx 2x h^2 x^2 + x^2 h^2 2x^2 h sqrt(2)x dh/dx sqrt(2) by the chain rule, the rate of change of the area with time dA/dt dA/dx dx/dh dh/dt dA/dt 2x (1/sqrt(2)) 1 dA/dt 2x/sqrt(2) dA/dt 2xsqrt(2)/2 dA/dt xsqrt(2) when x 100 mm the area increases by dA/dt 100sqrt(2) in mm^2/min In the heat of the sun, a sheet of aluminum in the shape of an equilateral triangle is expanding so that its side length increases by 1 millimeter per hourWhen the side length is 100 millimeters, how is the area increasing? let x side length dx/dt 1 mm/min h ht of the rectangle x^2 (x/2)^2 + h^2 h^2 x^2 - (x/2)^2 x^2 - x^2/4 x^2 (1 - 1/4) h^2 3x^2 / 4 h sqrt(3/4)x dh/dt sqrt(3/4) A xh/2 A x(sqrt(3/4)x/2 A (sqrt(3/4)x^2 / 2 dA/dx (sqrt(3/4)x dA/dt dA/dx dx/dt dA/dt sqrt(3/4) x 1 dA/dt sqrt(3/4) x when x 100 mm dA/dt sqrt(3/4)(100) dA/dt 100 sqrt(3/4) dA/dt 100 sqrt(3)/2 dA/dt 50sqrt(3)Other related question