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Question
x and y are the sides of two squares such that y = x – x2. Find the rate of change of the area of second square with respect to the area of first square.
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Solution
Let area of the first square A1 = x2
And area of the second square A2 = y2
Now A1= x2 and A2 = y2 = (x – x2)2
Differentiating both A1 and A2 w.r.t. t, we get
`("dA"_1)/"dt" = 2x * "dx"/"dt"` and `("dA"_2)/"dt" = 2(x - x^2)(1 - 2x) * "dx"/"dt"`
∴ `("dA"_2)/("dA"_1) = ("dA"_2/"dt")/("dA"_1/"dt")`
= `(2(x - x^2)(1 - 2x) * "dx"/"dt")/(2x * "dx"/"dt")`
= `(x(1 - x)(1 - 2x))/x`
= (1 – x)(1 – 2x)
= 1 – 2x – x + 2x2
= 2x2 – 3x + 1
Hence, the rate of change of area of the second square with respect to first is 2x2 – 3x + 1.
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