Question

A goat grazes a rectangular paddock 10 metres by 20 metres. It is tethered to the fence at one corner of the paddock by an inextensible rope of length `x` metres, where `10<x<20`? (More in 'details' section).

A goat grazes a rectangular paddock 10 metres by 20 metres. It is tethered to the fence at one corner of the paddock by an inextensible rope of length `x` metres, where `10<x<20`.

(a) Show that the goat can graze an area of `A` `m^2`, where `A=1/2x^2sin^(-1)(10/x)+5sqrt(x^2-100)`

(b) If the goat can graze an area equal to half the area of the paddock, find the length of the rope to one decimal place using one application of Newton's method.

Thanks!

Answer 1

Let the rope of length `x` m subtends angles `theta` radian with the lengths `AD and BC` of the rectangular paddock ABCD in its particular position as shown in the above figure.

It is given that `10m < x < 20m`

The maximum circular area that the goat can access when it is tethered by an in-extensible rope of length x metre is `=pix^2` `m^2`

(a) The total possible area `A` of grazing within the rectangular paddock will be

`A="area of sector" APQ+ DeltaAQB`

`=>A=(pix^2)/(2pi)xxtheta+1/2xxBPxxAB`

`=>A=1/2x^2xxsin^-1(10/x)+1/2xxsqrt(x^2-100)xx10 m^2`

`=>A=1/2x^2sin^-1(10/x)+5sqrt(x^2-100) m^2`

(b) If the goat can graze an area equal to half the area of the paddock `(100m^2)`, then we get the following equation putting `A=100m^2` .

`=>100=1/2x^2sin^-1(10/x)+5sqrt(x^2-100) m^2`

`f(x)=1/2x^2sin^-1(10/x)+5sqrt(x^2-100) -100=0`

The length of the rope to one decimal place calculated using one application of Newton's method will be `x~~11.7`m.

This calculation can be done by online calculator .(Link given below)

Taking

`f(x)=1/2x^2sin^-1(10/x)+5sqrt(x^2-100) -100`

and

`f'(x)=xsin^-1(10/x)-5/sqrt(1-100/x^2)+(5x)/sqrt(x^2-100)`

LINK

http://keisan.casio.com/exec/system/1244946907