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  • Provide solution RD Sharma maths class 12 chapter 12 derivative as a Rate Measurer exercise case study base question coma question 2 sub question 5 maths textbook

Provide solution RD Sharma maths class 12 chapter 12 derivative as a Rate Measurer exercise case study base question, question 2 sub question 5

Answers (1)

Answer: 

4 cm

Hint:
V=a^{3}

Given:

Solution:

Volume of the cube,

\begin{aligned} &V=a^{3} \\ &\frac{d v}{d t}=3 a^{2} \cdot \frac{d a}{d t} \\ &\text { at } \frac{d a}{d t}=0.2 \mathrm{~cm} / \mathrm{sec} \\ &\frac{d v}{d t}=3 a^{2} \times 0.2-(1) \end{aligned}

Surface Area of cube,

\begin{aligned} &s=6 a^{2} \\ &\frac{d s}{d t}=12 a \cdot \frac{d a}{d t} \\ &\text { At } \frac{d a}{d t}=0.2 \mathrm{~cm} / \mathrm{sec} \\ &\frac{d s}{d t}=12 a \times 0.2-(2) \end{aligned}

From equation (1) & (2),

\frac{d v}{d t}= \frac{d s}{d t}

\begin{aligned} &\text { at } \frac{d a}{d t}=0.2 \mathrm{~cm} / \mathrm{sec} \\ &3 a^{2} \times 0.2=12 a \times 0.2 \\ &3 a^{2}-12 a=0 \\ &3 a(a-4)=0 \\ &a=0, \\ &a=4 \end{aligned}

Side can’t be zero

\therefore a=4cm

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