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  • provide solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise 12.2 question 22

provide solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise 12.2 question 22

Answers (1)

Answer:  33 \pi \mathrm{cm}^{3} / \mathrm{sec}

Hint:  we know the volume of the cylinder is V=\pi r^{2} h

Given:  The radius of a cylinder is increasing at the rate 2cm/sec and its altitude is increasing at rate of 3cm

Solution: The radius of a cylinder is increasing at the rate 2 cm/sec and its altitude is decreasing at the rate of 3 cm/sec

 To find the rate of change of volume when radius is 3 cm and altitude 5 cm

Let V be the volume of the cylinder, r be its radius and h be its altitude at any instant of time ‘t’.

We know volume of the cylinder is V=\pi r^{2} h

Differentiating this with respect to time we get

\begin{aligned} &\frac{d V}{d t}=\frac{d\left(\pi r^{2} h\right)}{d t} \\\\ &\frac{d v}{d t}=\pi\left[\frac{d\left(r^{2} h\right)}{d t}\right] \end{aligned}

Now will apply the product rule of differentiation

\frac{d(u v)}{d x}=v \frac{d(u)}{d x}+u \frac{d(v)}{d x}

So above equation becomes

\begin{aligned} &\frac{d V}{d t}=\pi\left[h \frac{d\left(r^{2}\right)}{d t}+r^{2} \frac{d(h)}{d t}\right] \\\\ &\frac{d V}{d t}=\pi\left[h \times 2 r \times \frac{d r}{d t}+r^{2} \frac{d h}{d t}\right] \end{aligned}

But given of a cylinder is increasing at the rate 2 cm/sec, i.e., \frac{d r}{d t}=2\; cm/secand its altitude is decreasing at the rate of 3 cm/sec, i.e.,  \frac{d h}{d t}=-3\: cm/sec

by substituting the above values in equation we get

 \begin{aligned} &\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=\pi\left[\mathrm{h} \times 2 \mathrm{r}(2)+\mathrm{r}^{2}(-3)\right] \\\\ &\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=\pi\left[4 \mathrm{hr}-3 \mathrm{r}^{2}\right] \end{aligned}

When radius of the cylinder, r = 3 cm and its altitude,h = 5 cm, the above equation becomes

 \begin{aligned} &\mathrm{d} \mathrm{V} / \mathrm{dt}=\pi\left[4 \times 5 \times 3-3 \times 3^{2}\right] \\\\ &\frac{d V}{d t}=\pi[60-27] \\\\ &\frac{d V}{d t}=33 \pi \mathrm{cm}^{3} / \mathrm{sec} \end{aligned}

 Hence the rate of change of volume when radius is 3 cm and altitude 5cm is 33\pi cm3/sec

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