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vish97

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tz2, how was it that you guys calculated the percentage area of the field the goat could graze? 

You had to calculate two areas. One was the area of the sector that is created when the goat pivots across the field, and one was the triangle below that area. I got 44% as well.

 

How did you guys do on paper 1 with the geometric sequence created by the products of the roots or whatever it was?

 

 

the sector created by the goats pivot went outside of the area of the field, is the "triangle below that area" an approximation of the extra area the sector created by the goat and your taking that away from the sector? 

 

 

Not sure what you mean by "triangle below that area", but you're right that if you draw a sector created by the goat's pivot, you'd eventually go outside of the field. That's why when you draw the sector, you stop once you reach the other side of the field. You'll find that you now have a sector of radius 5m, and a right-angled triangle with height 4m and hypotenuse 5m. Calculate the area of the sector, and the right-angled triangle, and you're good.

 

 

The triangle is a very close appromixation to the area of the land that Gruff cannot graze on. You end up with a sector that's 5 m radius, and you take out the area of the whole sector (pi*r*r/4).

 

From this you subtract the area of the triangle with a base of (the extension - i.e. 1 m) and the height (3 m) using 0.5*b*h. The right angle triangle has neither height of 4 m or hypotenuse of 5 m, according to my calculations at least :/

 

If you want to be really precise, you're correct, the area that gruff cannot graze on is not exactly a triangle but it isn't a sector either. Its more of a quarter of an elliptical orbit, the calculations of which are beyond HL math. Hence, you have to use the triangle method. I got 44% too. 

 

 

Thanks for the input, Megamind. I'm not sure but according to your calculations, perhaps we used different approaches? I didn't focus on finding the "extra area" Gruff cannot graze, but seeing both of us got correct answers we both should be fine :)

 

 

Oh alright, sorry, I thought we had the same approaches that's why I found the dimensions different from mine. But same answer in Math HL, Yay! 

 

 

okay i understand your method, i used an alternative method of using the line of the 5m radius as part of a circle with equation y^2 + x^2 = 5^2, rearranged for y, input that into calculator and integrated between 0 and 4, gave me 43.98%, no approximations needed.

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