A swimmer S is in the sea at a distance of 4km from the closest point A on a straight shore.His house B is on the shore at a distance 5km from A.He can swim at a speed of 3km/hr.The distance of the point on the shore from B so that he reaches his home in the minimum time is=?
Being familiar with this type of optimization problem, I was expecting to be given the rate at which he can run or walk along the beach.
If the rate along the beach is faster than his swimming rate, then the hypotenuse that Henry found will not give him the minimum time.
he would aim for some point between A and B
I will solve it with some arbitrary speed along the beach.
Suppose the speed running along the beach is 6 km/h
Pick a point P somewher between A and B
Let AP = x , then BP = 1-x
ST^2 = x^2 + 4^2
ST = √(x^2 + 16) or (x^2+16)^(1/2)
time swimming = (x^2+16)^(1/2) /3
time running along beach = (5-x)/6 = 5/6 - x/6
Total time = T = (x^2+16)^(1/2) /3 + 5/6 - x/6
dT/dx = (1/2)(1/3)(x^2+16)^(-1/2) (2x) - 1/6
= 0 for a min of T
2x/(6√(x^2+16) = 1/6
2x/√(x^2+16) = 1
2x = √(x^2+16)
square both sides
4x^2 = x^2 + 16
3x^2 = 16
x^2 = 16/3
x = appr 2.3
so distance from point B is 5-x = 2.7 km
Adjust the above solution using your given rate along the beach.
To find the point on the shore where the swimmer should reach in minimum time, we need to minimize the time taken to swim from point A to the shore and then to his house B.
Let's consider the following scenario:
1. Let x be the distance from point A to the point on the shore.
2. The swimmer needs to swim a distance of 4 km from the sea to Point A.
3. Since the swimmer swims at a speed of 3 km/hr, the time taken to swim from the sea to Point A is (4/3) hours.
Now, let's consider the time taken to swim from the shore to his house:
1. The distance from the point on the shore to his house B is (5 - x) km.
2. The swimmer swims at a speed of 3 km/hr, so the time taken to swim from the shore to his house is (5 - x)/3 hours.
To find the total time taken, we need to sum up the time taken to swim from the sea to Point A and the time taken to swim from the shore to his house:
Total Time = (4/3) + (5 - x)/3
To minimize the total time, we differentiate it with respect to x and set it to zero:
d(Total Time)/dx = 0
Differentiating the equation with respect to x gives us:
(-1/3) = 0
Since (-1/3) is a constant, it does not depend on x. Therefore, it cannot be zero.
This implies that there is no minimum time point on the shore. The swimmer can swim directly to his house from the sea, along the shore.
So, the distance of the point on the shore where he reaches his home in the minimum time is 0 km.
d^2 = X^2 + Y^2 = 4^2 + 5^2 = 41
d = 6.4 km. = Shortest distance to his
home.