MY QUESTION IS VERY DETAIL PLEASE CHECK DISCRIBTION

Question

you are on a boat 400 m from the closest point Perpendicular on the beach. you want to visit your friend who is located another 1km along the beach. if you can row pedal boat at a constant speed of 2km/h and you can walk at a constant speed of 3km/h, where should now row your boat to the shore in order to get the friend fast as possible?

Answer 1

Let x represents the distance from point P where we would like to start walking. Then, we will walk distance d = 1000 -x meters.

 

The distance we will row is

d_row = √(400²+x²)

time row = 1/2√(160,000 +x²)

time walk = 1/3(1000 -x)

y = time walk = 1/2√(160,000 + x²) + 1/3(1000 -x)

y' = 1/2(1/2(160,000 + x²)^1/2-12x) -1/3

 

We set y' =0 and solve for x.

 

1/2(1/2(160,000 + x²)^-1/22x-1/3=0

 

1/2(160,000 + x²)^-1/2 x = 1/3

 

3x = 2(160,000 + x²)^1/2

 

9x²=4(160,000 + x²)

 

9x²=640,000 +4x²

 

5x² =640,000

 

x²= 128000

x =√(128000)

x=357.77

The person must row approximately toward 358 meters to get to his friend the fastest.

 

In the solution above, the equation for y'=0 does not have a solution. x is a positive value.

Then,

-x/2√(160000-x²) -1/3 =0

does not have a solution. Check!

If we add two negative quantities, we cannot obtain 0. The solution x =221.88 is a false solution for the radical equation.

 

Answer 2

Hopefully you have a good picture of how to set this up. The shortest path time wise will be to row the boat somewhere between the point directly perpendicular and the location a kilometer away. Call the distance from the perpendicular point to the point that you want to land x.

 

You should now have a right triangle and you are going to row along the hypotenuses path. The length of that hypotenuse, using the pythagorean theorem, can be represented by√(160000-x²) meters. The portion that he has to walk can be represented as 1000-x which needed to be converted to meters so that we are talking about the same units.

 

Now Rate*time = distance so time = d/R

 

So the time it takes to row along the hypotenuse can be represented by (160000-x²)^1/2 / 2 and the walking distance is represented by (1000-x)/3.

 

So his trip, in terms of time is y=(160000-x²)^1/2 / 2    + (1000-x)/3

 

You can now take the derivative to determine the minimum by setting the derivative to 0. The derivative is

 

y' = 1/2(160000-x²)^-1/2 (-2x)/2 + 1/3(-1) cleaning this up and setting it equal to zero you get

 

-x/2√(160000-x²) -1/3 =0   bring the 1/3 onto the other side and cross multiply to get

-3x = 2√(160000-x²)    square both sides to remove the radical to get

9x²= 4(160000-x²)

9x² = 640000-4x²

13x² = 640000

x²    = 49230.76923

x      = 221.88

 

So, if I haven't made any careless mistakes, the person should row to a point 221.88 meters from the perpendicular point and 778.12 meters from the friend's location.

 

Hope this helped.