Presentation on theme: "This is problem 49 page 219: A man is in a boat 2 miles from the nearest point on the coast. He is to go to a point Q, located 3 miles down the coast and."— Presentation transcript:
This is problem 49 page 219: A man is in a boat 2 miles from the nearest point on the coast. He is to go to a point Q, located 3 miles down the coast and 1 mile inland. If he can row at 2 miles per hour and walk at 4 miles per hour, toward what point on the coast should he row in order to reach point Q in the least amount of time.
Need a picture: Q 1 mile 2 miles 3 miles x 3 - x Find x to minimize time
Use the Pythagorean theorem twice to find the distance to row and walk To row: To walk:
The time to row is: Notice I used unit analysis to find I had to divide by 2 and not multiply by 2 The time to walk is:
The primary equation is the total time: We don’t need a “secondary equation”. Why? The primary equation only has one variable, there is no need to substitute for a second variable.
To minimize, find the derivative = 0 This will be good chain rule practice OutsideInside OutsideInside Do you see another chain rule here?
We need to solve this mess = 0 First, let’s clean it up a bit by multiplying constants Wow, let’s cheat and use MathPert to graph this and find the solution
Use the Point and Slope Tool click it and center the crosshairs on the intercept
y = 0 and x = 1 this is where the derivative is equal to 0! We just use a ‘numerical” technique to solve the equation
Just to make sure, evaluate at x = 1: You should get: The man should aim for a point 1 mile down the coast
To review what I typed into MathPert: 1/2((x^2+4)^(-1/2)(x))-1/4((3-x)^2+1)(-1/2)(3-x)) “Count” the number of opening (( versus the number of closing )), they must match in any expression. I could also type, which is easier: x/2sqrt(x^2+4)-(3-x)/4sqrt((3-x)^2+1)