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📐 Mathematics  ·  Application of Derivatives  ·  JEE

The radius of a sphere is increasing at the rate of 2 cm/s. Find the rate of increase of its volume when the radius is 5 cm.

Answer: 200 pi cm 3 /s.

  • A 50 pi cm<sup>3</sup>/s
  • B 200 pi cm<sup>3</sup>/s
  • C 100 pi cm<sup>3</sup>/s
  • D 20 pi cm<sup>3</sup>/s

Correct answer: B. 200 pi cm<sup>3</sup>/s

Explanation: V = (4/3) pi r<sup>3</sup>, so dV/dt = 4 pi r<sup>2</sup> (dr/dt) = 4 pi (5)<sup>2</sup> (2) = 4 pi (25)(2) = 200 pi cm<sup>3</sup>/s.

Tangent and Normal at a Point on a Curve(a,b)tangent (slope=f'(a))normal (slope=-1/f'(a))Tangent and normal are always perpendicular to each other at the point of contact

The tangent at a point touches the curve with slope f'(a); the normal is the line perpendicular to the tangent at that same point, with slope -1/f'(a) - together they describe the curve's local direction and the line "straight into" the curve.

Concept context

Use derivatives to study rate of change, increasing and decreasing functions, tangents and normals, and maxima and minima, with classic optimization problems.

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