As usual, draw a diagram. If the angle subtended by the arc walked around the lake is θ, then the distance traveled
on foot = 4θ
by boat = √(32(1+cosθ)) = 8cos(θ/2)
Use the law of cosines to get this.
So, the total time t is
t = 4θ/10 + 8/5 cos(θ/2)
dt/dθ = 2/5 - 4/5 sin(θ/2)
= 2/5 (1-2sin(θ/2))
dt/tθ = 0 when θ = π/3 or 2π/3
I'll leave it to you to figure out which is the min or max. Better check my math while you're at it.
A woman at a point A on the shore of a circular lake with radius r=4 wants to arrive at the point C diametrically opposite A on the other side of the lake in the shortest possible time. She can walk at the rate of 10mph and row a boat at 5mph. What is the shortest amount of time it would take her to reach point C?
I'm not sure where to start... :(
2 answers
I drew a circle and placed the woman at point A.
She wants to go to point B, clearly where AB is a diameter. Label the centre as O.
Pick a point P on the circle so that she will row from A to P, and then walk along the circumference from P to B
I labeled AP = x and arc PB as a (a for arc)
let the central angle for arc a be Ø radians
so a = 4Ø (a formula you should know)
so the time to walk along the arc will be 4Ø/10 or 2Ø/5
now if angle POB = Ø, then angle POA = π - Ø
recall that cos(π-Ø) = cosπcosØ + sinπsinØ
= (-1)cosØ + (0)sinØ
= -cosØ
Now in triangle PAO, using the cosine law:
x^2 = 4^2 + 4^2 - 2(4)(4)cos(π-Ø)
= 32 - 32(-cosØ) , from above
= 32 + 32cosØ
x = √(32 + 32cosØ) = (32+32cosØ)^(1/2
and the time rowing = (32+32cosØ)^(1/2)
total time = T
= (32+32cosØ)^(1/2) + 2Ø/5
d(T)/dØ = (1/2)(32+32cosØ)(-1/2) sinØ + 2/5
= sinØ/√2(32+32cosØ) + 2/5
=0 for a min of T
sinØ/2√(32+32cosØ) = -2/5
square both sides
sin^2 Ø/(4(32+32cosØ)) = 4/25
(1- cos^2 Ø)/(32+32cosØ) = 16/625
512+515cosØ = 625 - 625cos^2 Ø
625cos^2 Ø + 512cosØ - 113 = 0
wow, this factors!
(625cosØ - 113)(cosØ + 1) = 0
cosØ = 113/625 or cosØ = -1
Ø = 1.3889965 or Ø = π or 180° , which is not feasible
plug that into T
min time = √(32 + 32(113/625) + 2(1.3889965)/5
= appr 6.7 hrs
WOW#2, nice question, but you better check all that messy arithmetic.
She wants to go to point B, clearly where AB is a diameter. Label the centre as O.
Pick a point P on the circle so that she will row from A to P, and then walk along the circumference from P to B
I labeled AP = x and arc PB as a (a for arc)
let the central angle for arc a be Ø radians
so a = 4Ø (a formula you should know)
so the time to walk along the arc will be 4Ø/10 or 2Ø/5
now if angle POB = Ø, then angle POA = π - Ø
recall that cos(π-Ø) = cosπcosØ + sinπsinØ
= (-1)cosØ + (0)sinØ
= -cosØ
Now in triangle PAO, using the cosine law:
x^2 = 4^2 + 4^2 - 2(4)(4)cos(π-Ø)
= 32 - 32(-cosØ) , from above
= 32 + 32cosØ
x = √(32 + 32cosØ) = (32+32cosØ)^(1/2
and the time rowing = (32+32cosØ)^(1/2)
total time = T
= (32+32cosØ)^(1/2) + 2Ø/5
d(T)/dØ = (1/2)(32+32cosØ)(-1/2) sinØ + 2/5
= sinØ/√2(32+32cosØ) + 2/5
=0 for a min of T
sinØ/2√(32+32cosØ) = -2/5
square both sides
sin^2 Ø/(4(32+32cosØ)) = 4/25
(1- cos^2 Ø)/(32+32cosØ) = 16/625
512+515cosØ = 625 - 625cos^2 Ø
625cos^2 Ø + 512cosØ - 113 = 0
wow, this factors!
(625cosØ - 113)(cosØ + 1) = 0
cosØ = 113/625 or cosØ = -1
Ø = 1.3889965 or Ø = π or 180° , which is not feasible
plug that into T
min time = √(32 + 32(113/625) + 2(1.3889965)/5
= appr 6.7 hrs
WOW#2, nice question, but you better check all that messy arithmetic.