Asked by Rainbows
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... :(
I'm not sure where to start... :(
Answers
Answered by
Steve
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.
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.
Answered by
Reiny
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.