Question

1. The radius and height of a cone are measured as 6cms each by scale in which there is an error of 0.01cm in each cm. Then the

approximate error in its volume is​

Answer 1

Explanation:

FORMULA

$= \pi {r}^{2} \frac{h}{3} \\ = \pi \times {6}^{2} \frac{6}{3} = 226.19 \\ </p><p> \: \frac{ triangle \: v\: }{v} \: \: = \frac{\pi}{3} 6 \times 0.01 \times 0.01 = 0.000628$

The Change is volume is,

V=0.000628×226.19 = 0.14%

Answer 2

The radius and height of a cone are measured as 6cm each by scale in which there is an error of 0.01cm in each cm.

The approximate error in its volume is 2.16π.

We have to find the approximate error in its volume.

We know the formula of the volume of cone is given by,

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

but here r = h = x = 6cm

$\implies V=\frac{1}{3}\pi x^3\:\:...(1)$

differentiating both sides we get,

$\frac{dV}{dx}=\pi x^2\\\\\implies\frac{\Delta V}{\Delta x}=\pi x^2\:\: ...(2)$

From equations (1) and (2) we get,

$\Delta V=\left(\frac{3\Delta x}{x}\right)V$

Here Δx = 0.01 × 6 = 0.06 cm [∵ 0.01 cm error in each cm] , x = 6 cm,

$V=\frac{1}{3}\pi x^3=\frac{1}{3}\pi(6)^3=72\pi$

∴ the approximate error in its volume,

$\Delta V=\left(\frac{3\times0.06}{6}\right)72\pi=2.16\pi$

Therefore the approximate error in its volume is 2.16π.

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