KCET · Maths · Application of Derivatives
Approximate change in the volume \( V \) of a cube of side \( x \) metres caused by increasing the side
by \( 3 \% \) is
- A \( 0.09 x^{3} m^{3} \)
- B \( 0.03 x^{3} m^{3} \)
- C \( 0.06 x^{3} m^{3} \)
- D \( 0.04 x^{3} m^{3} \)
Answer & Solution
Correct Answer
(A) \( 0.09 x^{3} m^{3} \)
Step-by-step Solution
Detailed explanation
Given sides of a cube \( x \mathrm{~m} \).
We know that, volume of cube is given by,
\( V=x^{3} \)
and \( \frac{d x}{x} \times 100=3 \)
Now, \( d V=3 x^{2} d x \)
\( d V=3 x^{2} d x \)
\( =3 x^{3} \frac{d x}{x}=3 x^{3} \times \frac{3}{100} \)
\( =0.09 x^{3} \)
We know that, volume of cube is given by,
\( V=x^{3} \)
and \( \frac{d x}{x} \times 100=3 \)
Now, \( d V=3 x^{2} d x \)
\( d V=3 x^{2} d x \)
\( =3 x^{3} \frac{d x}{x}=3 x^{3} \times \frac{3}{100} \)
\( =0.09 x^{3} \)
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