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Provide solution RD Sharma maths class 12 chapter 12 derivative as a Rate Measurer exercise case study base question, question 2 sub question 2

Answers (1)

Answer:

60cm^{3}/\sec

Hint: 

Here, we use the basic concept

Given:

Theedge Increases at rate of \frac{da}{dt}=0.2cm/\sec

Length of edge of cube =10cm

Solution: 

we have \frac{da}{dt}=0.2cm/\sec

Volume of cube =a^{3}

Differentiate it we get

\frac{d v}{d t}=3 a^{2} \frac{d a}{d t}=3 \times 10 \times 10 \times 0.2=60 \mathrm{~cm}^{3} / \mathrm{sec}

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