A woman at a point A on the shore of a circular lake with radius 2 mi 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 4 mi/h and row a boat at 2 mi/h. How should she proceed

Ratio of the arc length to chord length between two points on the lake is

always lesser than the ratio of the walking and rowing speeds.

She should walk around the half the perimeter of the circular lake from

point A to point C.

Reasons:

Location of woman = Point A

Radius of the lake = 2 miles

Location of the destination = Point B

Distance between point A and point B = Diameter of the lake

Rate of walking = 4 mi/h

Rate at which she rows the boat = 2 mi/h

Required:

How she should proceed

Solution:

If she Journey's by boat;

Distance she would have to row the boat = The diameter of the lake

Distance = Diameter of lake = 2 × Radius

Distance by boat = 2 × 2 miles = 4 miles

Therefore;

If she journey's by walking

Distance she would have to walk, D = Half the circumference of the circle

∴ D = π × Radius

Distance she would have to walk, D = π × 2 miles = 2·π miles

Arc length = Radius × θ

Chord length = Radius × 2 × sin(θ/2)

For 0 ≤ θ ≤ π, we have; 0 ≤ sin(θ/2) ≤ 1

From the attached graph, as θ increases, the ratio of chord length to arc

length decreases, therefore, the maximum gain by boat is the direct route,

which takes more time.

Therefore, given that it takes less time walking around the lake, than to

row the boat across the lake, the she should walk around the lake.

Learn more here:

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She should only walk, and not take the boat ride

Step-by-step explanation:

please see attached a drawing for the problem

we want to minimize the travel time from A to C

given data

Radius of lake= 2mi

walk rate= 4mi/h

boat speed= 2mi/h

The distance from A to B from the drawing shows that

cosθ=adj/hyp=d1/4

d1=4cosθ

The arc length from A to C

S=rθ

but arc length is expressed as 2θ

S=2(2θ)

S=4θ=d2

d2=4θ

we aim to minimize the time of travel

time= distance/speed

t1=time taken from A to B

t1= time taken from B to C

t1=d1/2= 4cosθ/2=2coθ (note speed of boat is 2mi/h)

t2=d2/4=4θ/4=4 (note speed of walk is 4mi/h)

total time =t1+t2

T=2cosθ+θ

the time is a function of θ

T(θ)=2cosθ+θ

since we aim to minimize the travel time, let us find the derivative of the time function and equate it to zero(0)

T'(θ)= -2sinθ+1

0=-2sinθ+1

2sinθ=1

sinθ=1/2

there are two solutions 30° and 150°

but 150° is too much for the angel.

let us try som critical angle between 30° and 90°

like

0= 0 radians

30°= π/6

90°= π/62

for 0 we have

T(0)=2cos0+0

T(0)= 2hours

for 30 we have

T(π/6)=2.25hours

for 90 we have

T(π/2)=1.57hours

since we aim to minimize the travel time, the angle she should follow is 90° from the start point, this clearly shows that she should only walk along the circumference from A to C

plz refer to the attachement..