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The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side

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प्रश्न

The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side

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उत्तर

Let x be the length of the cube

∴ Volume of the cube V = x3   ......(1)

Given that `"dV"/"dt"` = K

Differentiating Equation (1) w.r.t. t, we get

`"dV"/"dt" = 3x^2 * "dx"/"dt"` = K  .....(constant)

∴ `"dx"/"dt" = "K"/(3x^2)`

Now surface area of the cube, S = 6x2

Differentiating both sides w.r.t. t, we get

`"ds"/"dt" = 6 * 2 * x * "dx"/"dt"`

= `12x * "K"/(3x^2)`

⇒ `"ds"/"dt" = (4"K")/x`

⇒ `"ds"/"dt" oo 1/x`   .....(4K = constant)

Hence, the surface area of the cube varies inversely as the length of the side.

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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 10 | पृष्ठ १३६

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