Rate of Change Question

Question

During the Sydney to Hobart Yacht Race the yacht Adelaide starts taking on water. Two other yachts are in the vicinity and are called for help: the yacht Eloise is exactly 40 kilometres south of the Adelaide and the yacht Matilda is exactly east of the Eloise and 80 kilometers away from the Adelaide. Both yachts start travelling directly towards the Adelaide, the Eloise with a constant speed of 40 kilometres per hour and the Matilda, due to a strong easterly wind, much faster with 60 kilometres per hour.

What is the rate of change of the distance between the Eloise and the Matilda at the moment when the Eloise is 20 kilometres away from the Adelaide?

I've set up a triangle and labeled three different points on each corner (A, E and M respectively), but I'm not sure how I should properly implement the cosine rule here.

Answer 1

EM^2 = AE^2 + AM^2 - 2AMcosA

= 400 + 2500 - 2000(1/2) = 1900

EM = 10sqr19= about 43.59 km

EM/sinA = 10sqr19/sqr3/2 = 20sqr(19/3) = AM/sinE = 50/sinE=20/sinA

sinE =50/20sqr(19/3/) = 2.5/sqr(19/3) = .9933992678

90<E<120

Angle E = sin^-1(2.5/sqr(19/3) = about 180-83.41 =about 96.59

Angle M =180- 96.59-60 = 180-156.59 = 24.01

Triangle is about 20 by 50 by 44 km

with angles about 24, 97, and 60 degrees

EM = about 44 km

Let EM=a =44, AM =e=50, EA =m=20, e'=40, m'=60

a^2 = e^2 +m^2 -2emCosA A =a constant 60 degrees, CosA =1/2

take the derivative with respect to time

2aa' = 2ee' + 2mm' -em' -me'

2(44)a' = 2(50)(40) +2(20)(60) -2000-800

88a' =4000+2400-2800

a' = 3600/88 =450/11 = about 40.91 = about 41 km per hour

about 41 kph = rate distance between Matilda and Eloise are moving apart when Eloise is 20 km from Adelaide

it definitely helps to draw a diagram of the original right triangle and then the obtuse triangle formed inside the right triangle's upper part, when Elioise moves 20 km north

some rounding in above calculations. (no guarantees of arithmetic mistakes)