The minute hand of a certain clock is 4 in long, Starting from the moment when the hand is pointing straight up, how fast is the area of the sector that is swept out by the hand increasing at any instant during the next revolution of the hand?

This article aims to find the area of a sector. This article uses the concept of the area of a sector. The reader should know how to find the area of the sector. Area of sector of a circle is the amount of space enclosed within the boundary of the sector of circle. The sector always starts from the center of the circle.

The area of sector can be calculated using the following formulas:

Area of a circular section = $(\dfrac{\theta}{360^{\circ}}) \times \pi r ^ {2}$ where $\theta$ is sector angle subtended by the arc at the center in degrees and $r$ is the radius of the circle.

Area of a circular section = $\dfrac {1} {2} \times r ^ {2} \theta$ where $\theta$ is sector angle subtended by the arc at center and $r$ is the radius of the circle.

Let $A$ represent the area swept out and $\theta$ the angle through which the minute hand has turned.

$A = \dfrac {1} {2} r ^ {2} \theta$

$\dfrac { dA }{ dt } = \dfrac {1}{2} r ^ {2} \dfrac{ d\theta }{ dt }$

We know that:

$\dfrac {the\:area\: of \:sector }{the\: area\: of\: circle } = \dfrac { A }{ \pi r ^ {2} }$

$= \dfrac{ \theta }{2 \pi }$

The minute hand lasts $60$ minutes per rotation. Then the angular velocity is one revolution per minute.

$\dfrac{d\theta }{dt} = \dfrac { 2\pi }{ 60 } = \dfrac { \pi }{ 30 } \dfrac { rad }{ min }$

Thus

$\dfrac{dA }{ dt } = \dfrac{1}{2} r^{2} \dfrac { d\theta }{ dt } = \dfrac { 1 }{ 2 } . (4)^{ 2 }. (\dfrac {\pi}{30})$

$= \dfrac{4\pi}{15} \dfrac{in^{2}}{min}$

Numerical Result

The area of sector that is swept out is $\dfrac{ 4\pi }{ 15 } \dfrac{ in ^ {2}}{min}$.

Example

The minute hand of a particular clock is $5\: inches$ long. Starting when the hand points straight up, how fast does the area of the sector swept by the hand increase at each instant during the next hand revolution?

Solution

The $A$ is given by:

$A = \dfrac{1} {2} r ^ {2} \theta$

$\dfrac { dA }{ dt } = \dfrac{ 1 }{ 2 } r ^ {2} \dfrac { d\theta}{ dt }$

We know that:

$\dfrac { the\:area\: of \:sector }{the\: area\: of\: circle } = \dfrac { A }{ \pi r ^ {2} }$

$= \dfrac{ \theta }{2 \pi }$

The minute hand lasts $60$ minutes per rotation. Then the angular velocity is one revolution per minute.

$\dfrac{ d\theta }{ dt } = \dfrac{ 2\pi }{ 60 } = \dfrac{ \pi }{ 30 } \dfrac{ rad }{ min }$

Thus

$\dfrac{dA}{dt} = \dfrac{1}{2} r^{2} \dfrac{d\theta}{dt} = \dfrac{1}{2} . (5)^{2}. (\dfrac{\pi}{30})$

$= \dfrac{5\pi}{12} \dfrac{in^{2}}{min}$

The area of sector that is swept out is $\dfrac{5\pi}{12} \dfrac{in^{2}}{min}$.