I'm not sure how to do this problem

Question

A women 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 a 5mph .

What is the shortest amount of time it would take her to reach point C? What is the longest amount of time it would take her to reach point C following a path like the path in the figure ?

Answer 1

If you sketch the circle, the woman could row all the way across for a value of 2r/5 = 1.6 or she could walk the semicircle for a value of πr/10 ≈ 1.26.

Or, she could row to a point C on the semicircle and walk the rest of the semicircle segment.

If you add a chord for the segment, you get an inscribed triangle with the diameter as one side of the triangle.

By a geometry theorem, this makes the triangle angle opposite the diameter a right triangle.

By another geometry theorem, the remaining segment length will be 2rθ, where θ is the angle formed between the chord AC and the radius. So the total time function using time = distance/rate is:

T(θ)=2rcos(θ)/5 + 2rθ/10 = (8/5)cos(θ) + (8/10)θ

T'(θ) = -(8/5)sin(θ)+8/10

Setting T'(θ) to zero and solving for θ.

sin(θ) = 1/2, θ=π/6 or 5π/6 to be within the domain of [0, π/2]. Use only π/6 since ≤ π/2

T(π/6) ≈ 1.8 <--- Maximum

Minimum = 1.26

Answer 2

Sketch a diagram of the situation (we only need a semicircle). We could place this in the coordinate plane but there is no need to do so. We need to find a way to write the time her trip will take as a function of which direction she heads at the outset. To do this, we recall that rate·time = distance so time = distance/rate.

There are a couple of different ways to choose an independent variable here, but one that makes good sense is to let Θ: the acute angle her original rowing direction makes with the diameter of the circle. (So, if she aims directly at point C, planning to row the entire time, Θ = 0. Whereas, if she will walk the entire semicircle (which ends up minimizing the time), then Θ = π/2.)

When we do this, both her rowing distance and her walking distance can be written as a function of Θ:

Sketch an arbitrary Θ with vertex at point A with initial ray thru point C and terminal ray thru an arbitrary point on the top half of the circle. Label her "aim-to" point on the semicircle P and label the center of the circle O. Notice that ΔAOP is isosceles, since two of its sides are radii, with r = 4. We have called ∠OAP Θ, and note that the other base angle, ∠APO, is also Θ. Thus, the central angle, ∠AOP, is (π - 2Θ).

By law of cosines, her rowing distance is c = √(16 + 16 - 32cos(π - 2Θ)).

The other central angle adjacent to ∠AOP is ∠COP and its measure is 2Θ, since the 2 are supplementary.

We can use the arc length formula s = rΘ to write her walking distance (which is an arc) as s = 4(2Θ) = 8Θ.

Putting this all together we have t: total time for trip (in hrs)

t = √(16 + 16 - 32cos(π - 2Θ))/5 + 8Θ/10

If you are not required to find the extrema (max and min times) analytically, graph this function over the domain 0 ≤ Θ ≤ π/2 to see that we have relative mins at Θ = 0 and Θ = π/2 and a relative max at Θ = .523 (π/6)

The absolute minimum occurs when Θ = π/2, in other words she minimizes her time by walking the entire way. The semicircular path she walks is 4π miles so walking at a rate of 10 mph gives t_min = 4π/10 ∼ 1.2566 hrs.

Maximum is when Θ = π/6 and t_max ∼ 1.8045 hrs.