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A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is propotional to the surface. Prove that the radius is decreasing at a constan - Mathematics

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Question

A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate

Sum

Solution

Ball of salt is spherical

∴ Volume of ball, V = `4/3  pi"r"^3`

Where r = radius of the ball

As per the question, `"dV"/"dt" oo  "S"`

Where S = surface area of the ball

⇒ `"d"/"dt" (4/3 pi"r"^3) oo  4pi"r"^2`   .....[∵ S = 4πr2]

⇒ `4/3 pi * 3"r"^2 * "dr"/"dt" oo  4pi"r"^2`

⇒ `4pi"r"^2 * "dr"/"dt" = "K" * 4pi"r"^2`  ......(K = Constant of proportionality)

⇒ `"dr"/"dt" = "K" * 4pi"r"^2`

∴ `"dr"/"dt" = "K" * 1` = K

Hence, the radius of the ball is decreasing at constant rate.

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Chapter 6: Application Of Derivatives - Exercise [Page 135]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 6 Application Of Derivatives
Exercise | Q 1 | Page 135

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