I also made a "careful" diagram, and labeled the position of the canoe as T
Let TR = d
let PR = x, then RQ = 10-x
d^2 = 1^2 + x^2
d = (x^2 + 1)^(1/2)
Total time = (1/3)(x^2 + 1)^(1/2) + (10-x)/5
d(total time)/dx = (1/6)(x^2 + 1)(-1/2) (2x) - 1/5
= 0 for a minimum of total time
x/(3√(x^2 + 1) - 1/5 = 0
x/(3√(x^2 + 1) = 1/5
5x = 3 √(x^2 + 1)
square both sides
25x^2 = 9x + 9
16x^2 = 9
4x=3
x = 4/3
take it from there, but check my arithmetic
A person on a lake in a canoe 1 mile from the nearest point "P" on a straight shore line; the person wishes to get to a point "Q" , 10 miles along the shore from "P". To do so, the canoe moves to a point "R" between P and Q and then walks the remaining distance to"Q". The canoe can move at 3 mph and the person can walk at 5 mph. Where should "R" be selected such that the person gets to "Q" in the least amount of time?
Draw a careful diagram.
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