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Manipal MET2012MathematicsApplication of Derivatives

If an edge of a cube measure 2 m with a possible error of 0.5 cm . Find the corresponding error in the calculated volume of the cube

Options

  1. A0.6 ~m ^3
  2. B0.06 ~m ^3
  3. C0.006 ~m ^3
  4. D0.0006 ~m ^3

Correct answer

B. 0.06 ~m ^3

Step-by-step solution

Let x be the length of an edge of a cube and V be the volume of that cube. V=x^3 On differentiating w.r.t. x , we get d V d x =3 x^2 Let V be error in V and corresponding error x in x . V= d V d x x=3 x^2 x Given that, x=2 ~m and x=0.5 ~cm = 0.5 100 ~m aligned V & =3(2)^2 ( 0.5 100 ) & = 12 0.5 100 = 6 100 =0.06 ~m ^3 aligned

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