Help:Maths

Some students on an expedition reach the cornor of a very muddy field. They need to reach the opposite cornor as quickly as possible as they are behind schedule. They estimate that they could walk along the edge of the field at 5kmh^-1 and across the field at 3kmh^-1.They know from their map that the field is a square of side 0.5 km

How far should they walk along the edge of the field before cutting across?

Thank you.

Reply 1

theone

superkillball

Shouldn't they just go all the way round?

Reply 2

thefish_uk

huh?

do you have a picture? I don't get the "how far should they walk along before cutting across"

Anyway, you shouldn't walk straight across a field you should follow the Public Rights of Way.

Reply 3

meegan

think there's more to it than that. imagine you went (0.5 - x) kilometers around one side, then cut across the field to the opposite corner [i have choosen to go 0.5 - x first rather than x first to make the algebra easier]). i may well have made algebraic mistake so you should work through this, btu reckon this what they want you to do...

when you are going along the side you are travel (0.5 - x) kilometers.

when you are cutting across the field you are travel
sqrt(0.5 * 0.5 + x * x), the distance across the field is the hypotenuse of a right angled triangle with sides 0.5 and x. you can rewrite this as
(1/2) * sqrt(4 x^2 + 1).

using speed = distance / time => time = distance / speed the total time taken is

T = ((1/2) - x) / 5 + ((1/2) * sqrt(4 x^2 + 1))/3

now differentiate this wrt x

dT/dx = -1/5 + (1/12) * ( 8 x / sqrt(4 x^2 + 1))

setting equal to zero

1/5 = 2x/(3 * sqrt(4x^2 + 1))

squaring both sides

1/25 = (4x^2)/(9 * (4x^2 + 1))

36 x^2 + 9 = 100 x^2

x^2 = (9/64) => x = 3/8 ignoring minus root as makes no sense

i cant be bothered to do it, but if you want to be fully correct you should differentiate dT/dx wrt x to find d^2 T/dx^2 to check value is a minimum.