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Question
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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Solution
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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