If the surface area of a cube is increasing at a rate of 3.6 cm2/sec, retaining its shape; then the rate of change of its volume (in cm3/sec), when the length of a side of the cube is 10 cm, is :
Solution
For cube of side 'a'<br><br>A = 6a<sup>2</sup> and V = a<sup>3</sup><br><br>Given ${{dA} \over {dt}} = 3.6$<br><br>$\Rightarrow$$12a{{da} \over {dt}}$ = 3.6<br><br>$${{dV} \over {dt}} = 3{a^2}.{{da} \over {dt}} = 3{a^2}\left( {{{3.6} \over {12a}}} \right)$$<br><br>at a = 10<br><br>${{dV} \over {dt}} = 9$
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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