So far I haven't yet seen anyone comment on the general solution to this
problem for arbitrary values of the problem's parameter's. Therefore I
thought I would make such comments.
Let v_s == J's sprinting speed.
Let v_r == J's rowing speed relative to the a frame for which the water is
at rest.
Let v_w == the water's speed of the river.
Let W == the width of the river.
Let T_0 == W/v_s = a characteristic "time" for the problem.
Let r == v_w/v_r.
Let s == v_s/v_r.
Let u == sine of the heading angle upstream of straight across the river.
Let t_1 == time interval for J to row across the river.
Let t_2 == time interval for J to sprint along the far shore of the river
to the apparition.
Let t_t == t_1 + t_2 = total time interval for J's trip starting from the
moment of J's entry into the boat.
Using these definitions we can get expressions for t_1 & t_2 as:
This expression is to be minimized wrt u for -1 <= u <= 1. The solution
formula for u which minimizes this expression depends crucially on the
value of the rowing speed v_r. For sufficiently low values of v_r, i.e.
so low that r*(r + s) >= 1 or equivalently, r >= sqrt(1 + (s/2)^2) - s/2),
then the (sine of the) heading for minimal t_t is given by:
u = 1/(r + s),
and the minimal t_t value is:
t_min = T_0*sqrt((r + s)^2 - 1).
OTOH, if the rowing speed v_r is so large that r*(r + s) <= 1 or
equivalently r <= sqrt(1 + (s/2)^2) - s/2, then the (sine of the) heading
for minimal t_t is given by:
u = r,
and the minimal t_t value is:
t_min = T_0*s/sqrt(1 - r^2)
It should be noted that the minimal value in the first case occurs where
the derivative of t_t wrt u vanishs, but in the second case the minimal
value occurs at the cusp singularity of an absolute value function where
the derivative has a jump discontinuity (from negative to positive) and
the second derivative effectively has a positive infinite value. Also,
in this second case the rowing speed is so great that J only needs to
head upstream at whatever angle is necessary so that he/she goes straight
across the river with no sprinting required.
For the special case of the parameters given by James Harris we have:
r = (4 km/h)/(6 km/h) = 2/3
s = (10 km/h)/(6 km/h) = 5/3
T_0 = (1 km)/(10 km/h) = (1/10) h.
Since r*(r + s) = (2/3)*(2/3 + 5/3) = 14/9 > 1 we see that the parameters
are in the regime of the first case solution. This means that the heading
angle is: