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!

1 Answer
Oct 5, 2017

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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~~11.7m.

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)

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LINK

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