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Provide solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise  12.2 question 14

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Answer: \frac{dM}{dt}=(48)

Hint: Here we use the equation of given curve.

Given: Equation of the curve y=7x-x^{3}  and x  increases at the rate of 4 units per second

Solution: As the differentiate, the above equation with respect to x , we get the slope of the curve

\frac{d y}{d x}=\frac{d\left(7 x-x^{3}\right)}{d x}

\frac{d y}{d x}=\frac{d(7 x)}{d x}-\frac{d\left(x^{3}\right)}{d x}=7-3 x^{2}     …….(i)                                                                                    


Suppose M  be the slope of the given curve then the above equation M=7-3x^{2}


Given x  increases at the rate of 4 units per second therefore,

\frac{d x}{d t}=4 \text { units } / \mathrm{sec} ……… eq (ii)                                                                                                                                                  


Differentiate the equation of the slope that is equation (ii)

 \begin{aligned} &\frac{d M}{d t}=\frac{d\left(7-3 x^{2}\right)}{d t} \\\\ &\frac{d M}{d t}=\frac{d(7)}{d t}-\frac{d\left(3 x^{2}\right)}{d t} \end{aligned}

\begin{aligned} &\frac{d M}{d t}=0-(3 \times 2 x) \frac{d x}{d t} \\\\ &\frac{d M}{d t}=-6 x \times \frac{d x}{d t} \end{aligned}


 Let’s put value of   \frac{d x}{d t}=4 \text { units } / \sec

\frac{d M}{d t}=-6 x \times 4


Let’s put value of x = 2(given)

\frac{d M}{d t}=-6 x \times 4=-6 \times 2 \times 4=-48


The slope cannot be negative,

Thus, the slope of the curve is changing at the rate of 48 units/sec when x = 2

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