Asked by Daniella
the two bases of an isosceles trapezoid are 12 and 20 feet long, respectively. find the rate at which the area is changing when the equal sides are 5 feet long and are increasing at the rate of 2 feet per minute.
Answers
Answered by
Reiny
Make a sketch
I dropped perpendiculars at the ends of the 12 ft side to the 20 ft side and let the height be h ft
It is easy to see that
Area = 16h
also we know x^2 = 4^2 + h^2
h = √(x^2 - 16)
when x = 5, h = 3, and dx/dt = 2
Area = 16h = 16(x^2 - 16)^(1/2)
d(Area)/dt = 8(x^2 - 16)^(-1/2) (2x) dx/dt
which for the given data
= 8(25-16)^(-1/2)(10)(2)
= 8/√9 (20)
= 160/3
At that moment, the area is increasing at 160/3 ft^2 / min
check my arithmetic
I dropped perpendiculars at the ends of the 12 ft side to the 20 ft side and let the height be h ft
It is easy to see that
Area = 16h
also we know x^2 = 4^2 + h^2
h = √(x^2 - 16)
when x = 5, h = 3, and dx/dt = 2
Area = 16h = 16(x^2 - 16)^(1/2)
d(Area)/dt = 8(x^2 - 16)^(-1/2) (2x) dx/dt
which for the given data
= 8(25-16)^(-1/2)(10)(2)
= 8/√9 (20)
= 160/3
At that moment, the area is increasing at 160/3 ft^2 / min
check my arithmetic
Answered by
Ark Shem Catada
Where did 16 come from?
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