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Derivative as a Rate Measurer — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDerivative as a Rate Measurer1 Mark
Question
A cone whose height is always equal to its diameter is increasing in volume at the rate of 40cm3/ sec. At what rate is the radius increasing when its circular base area is 1m2?
  1. 1mm/ sec.
  2. 0.001cm/ sec.
  3. 2mm/ sec.
  4. 0.002cm/ sec.
✓

Answer

  1. 0.002cm/ sec.

Solution:

$\text{V} = \frac{1}{3} \pi \text{r}^{2}\text{h}$

Given that height is equals diamiter.

$\Rightarrow \text{h} = 2\text{r}$

$\text{V} = \frac{1}{3} \pi\text{r}^{2}\text{2r}$

$\text{V} = \frac{2}{3} \pi\text{r}^{3}$

$\Rightarrow \frac{\text{dV}}{\text{dt}} = 2\pi\text{r}^{2} \frac{\text{dr}}{\text{dt}}$

$\Rightarrow \frac{\text{dr}}{\text{dt}} = \frac{1}{2\times10^{4}} \times40 \ \ \begin{pmatrix}\because \pi\text{r}^{2} = 1\text{m}^{2}\\\Rightarrow1\text{m}^{2} = 10^{4} \text{cm}^{2} \end{pmatrix}$

$\Rightarrow \frac{\text{dr}}{\text{dt}} = \frac{1}{2\times10^{4}} \times40 = 0.002\text{cm}/\sec.$

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