In lecture you are posed with a question of why is the minimum velocity curve also the shortest distance curve? In this segment, we'll discuss the answer to this question. Recall that to find the minimum velocity curve, we solved for the trajectory x star of t that minimizes the interval of the cost function of x dot squared. We were able to use the Euler-Lagrange equation to find that the general form of the equation is x of t equals c1 times t plus c0. Now, let's find the minimum distance trajectory. Given two points, a and b, we wanna find the trajectory that is the shortest in total length. We can find the length of a trajectory between a and b by integrating infinitesimal segments ds along the curve. Each infinitesimal segments ds has corresponding change in dt and a change in dx. We can find the length of the segment ds using the distance function. We can rewrite this function by factoring out a factor of dt from under the square root. And the second step, we use the fact that by definition x dot is equal to dxdt. To find the total length of the curve we wanna integrate dx along the entire curve. Substitute in the expression for ds we just found, you'll find that the length of the curve can be found by taking the integral from zero to t of the square root of one plus x dot squared dt. We can now mathematically represent the problem of finding the minimum distance trajectory in the familiar form. Finding the function, x star of t, that minimizes the integral of a cause function with respect to t. In this case, the cause function L is the square root of 1 plus x star squared. Again, the necessary condition for the optimal trajectory is given by the Euler-Lagrange equation. To find this condition, we need to evaluate the Euler-Lagrange Equation for our cost-function. We can start by evaluating each term in the equation. The partial derivative of L with respect to x is 0. This is because x does not appear in the cost-functional. The partial derivative of L with respect to x dot, is x dot over the square root of 1 plus x dot squared. Know that this is the partial derivative of L with respect to x dot, that is we are not taking any derivatives with respect to time yet. To use the Euler-Lagrange equation we need to find the time derivative of partial o, partial x dot. However, it turns out that we don't need to explicitly calculate this. Substituting the terms we found into the Euler-Langrange Equation. We get the following expression for the necessary condition for the minimum distance trajectory. We can directly integrate this expression with respect to time. To get an expression for the velocity of the trajectory. Here k is an arbitrary constant. We can solve this equation for x dot. We see that x dot is a function of only the constant K. Therefore, x dot is a constant and does not vary with time. We can relabel this constant as c1. We can integrate the expression, x dot equals c1 to get the position function of the minimum distance trajectory. This gives us the expression x of t equals c1 times t plus c0 or c1 and c0 are obituary constants. We see that this is the equation for the minimum velocity curve. Thus, given the same set of boundary conditions, the minimum velocity trajectory will be the same as the minimum distance trajectory.