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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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प्रश्न

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.

योग
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उत्तर

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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अध्याय 6: Application Of Derivatives - Exercise [पृष्ठ १३६]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 6 Application Of Derivatives
Exercise | Q 11 | पृष्ठ १३६

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