Question

1+(f'(x))²-f''(x)=0

1) make substitution u=f' in the above equation. Solve this first order for u(x)
2) Integrate your answer from problem 1 to find f(x).
3) What is the cheapest path from(-1,0) to (1,0)?
4) what happens when you try to use the formula you found in problem to to write the cheapest path from (-3,0) to (3,0)
5) Expalin why the formula doesn't work in problem 4 in terms of the orginal problem of finding a cheapest path.

Answer 1

The question pertains to solving a second-order differential equation by substitution, integrating to find a function, and determining the cheapest path between two points. It explores the application of this solution to a specific path and the explanation of the limitations when applying the formula to a different path.

Step-by-step explanation:

The problem involves solving a differential equation with a substitution, integrating to find a function, and then applying this solution to find the cheapest path between two points, likely in the context of a calculus of variations problem or a related physical problem such as finding the curve of least time or brachistochrone.

First step: Make the substitution u = f'(x) in the equation 1 + (f'(x))^2 - f''(x) = 0 to reduce it to a first-order equation involving u and its derivative u' with respect to x. Solve for u(x).

Second step: Integrate u(x) to find f(x).

Third step: Using the function f(x), determine the cheapest path from (-1,0) to (1,0), presumably by applying the solution in the context of the given problem constraints.

Fourth step: Attempt to use the f(x) function to determine the cheapest path from (-3,0) to (3,0), and then analyze why it might not be possible or appropriate to use the formula in this way.

Fifth step: Explain why the formula does not work for the path from (-3,0) to (3,0), potentially by discussing the limitations of the original problem's conditions or constraints on the solution for a cheapest path.