[MUSIC] I would like to remind you that when we considered partial derivatives and gradients, we were interested in finding the direction of the gradient vector, given a production function. So we dealt with the production function, which has the name of Cobb-Douglas, a square root of the product of labor and capital. Let me draw one of the isoquants, that'd be labor and capital, and we choose some point here. So when the firm with this production function employs four units of labor and one unit of capital, the question was, in what proportion should we employ extra units of labor and capital in order to increase the production in the most fastest way possible? And it said that first of all, we need to draw the gradient of this function at a given point. So how to draw this gradient? This gradient and this title tells us exactly what we to do to draw the vector, which is perpendicular to the isoquant, which is a particular case of a production function, And look like that. So this is our gradient, I'm using nabla notation, nabla q. And let us prove the proposition which states the fact that whenever we draw a level curve of a function, a gradient of this function at a given point is perpendicular to a level curve of this function. Now let's consider a formal theoretical statement. Let us suppose a function of two variables is given, And it is continuously differentiable on some ball. And there is a point, x0, y0, which belongs to this ball, B. And moreover, when we substitute the coordinates of this point, they satisfy this equation. So let's suppose in x, y plane, I am not drawing the ball, there is some. We have this point. And this is a little curve which passes through this point. Now, we suggest that under some conditions, the gradient of this function will be perpendicular to the level curve, or perpendicular to the tangent line which touches this level curve at a given point. But first of all, we need to set a condition on derivatives, partial derivatives of a given function. So let's suppose, this is an assumption. Suppose, The gradient, Of this function at x0, y0 shouldn't be 0 vector. We split the proof into two parts and we start with the main part. In this main part, let's suppose that, This derivative, which we use in implicit function theorem, dfo(y), is a nonzero number at this point. So this is an assumption. It's clear how to find the tangent vector to the graph of an explicitly defined function. So if we have a function, f of x, and this function is differentiable at point x0, So then graph passes through x0, y0. How to find the tangent vector? The tangent vector, v, Can be found, If we use for its coordinates two differentials. The first being the differential of an independent variable and the second is a differential or this function f of x. So we can substitute, according to the definition or the differential, the product, f prime x0 times dx. The value of dx is unimportant because this is a common factor, so we can factor it out. And later on, we'll be dealing with a vector whose coordinate is 1, the first coordinate, and the second is simply f prime. Now, we need to calculate the value of the inner products of two vectors. The first vector is the tangent vector, the second vector is the gradient vector of this function. Let me remind you that a formula we use to calculate the inner product of two vectors from the Cartesian plane, We use the formula when we take the first coordinates, so both vectors, multiply them. After that, take the second coordinates of these vectors, also multiply them and add them up. In our case, we take tangent vector to the level curve, we take gradient, At point x0, y0, and we multiply them. But the first coordinate, or vector v, is simply 1. So 1 is multiplied by dF over dx taken at x0, y0, plus, This is the f prime, the value of the derivative. And here we multiply by dF over dy, Taken at this point. We can't finish here. At this point, we need to apply IFT, which provides the value for this derivative because f of x is the implicit function defined by equation. Equation is f of x and y equals 0. And all conditions of this theorem are met, And we can substitute using the formula, substitute for f prime using the formula 1 dF over dx plus, here I write negative and a quotient, dF over dx and dF over dy, Multiplied by, dF over dy. And easily to check that we get 0. And we have concluded the proof of the main part of this proposition. The remaining part, the remaining part is about 0 partial derivative. So what if, in our case, dF over dy at the point is 0? We have a condition, the gradient is not 0. Then that means that dF over dx taken at x0, y0 is a nonzero number. As I said earlier, we can find implicit function. This time, not the function y of x, it doesn't exist. But we can find x of y. And the derivative dx over dy, according to the theorem, is the quotient of Fx and Fy, but this is 0. That means that the tangent vector drawn to the level curve will be vertical. It can be expressed as the first coordinate will be 0. And for the second we can chose any non-negative, sorry, nonzero number. So when we multiply again, we need to calculate inner product. Again, we get 0 as well. So once again, in this case, v times grad f is 0. And that concludes the proof of this proposition. [MUSIC]
