Practical lesson 2. The concept of the first- and second-order differential
Our practical lesson is devoted to the concept of the first- and second-order differential, and this lesson will also cover the tangent plane and the normal.
The function of two variables z=e^(2x+5y) is given, the problem is to find the differential of the first and second order, using the properties of the differential but not the formula. Then, having obtained this result, we must say what the partial derivatives of the first and second order are equal to.
Let’s try to solve it. So, z=e^(2x+5y). We find the first-order differential: dz=de^(2x+5y). How do we find the differential of a function of one variable? Note that dU=U’dt if the function U is the function of the variable t. Here, the variable is 2x+5y, so it is e to the same power multiplied by the differential of the exponent: dz=de^(2x+5y)=e^(2x+5y)d(2x+5y).
Then, using the linearity property of the differential, we have: e^(2x+5y) (2dx+5dy).
After opening the brackets, we conclude that the first-order differential is found: 2e^(2x+5y)dx+5e^(2x+5y)dy.
Let’s find the second-order differential, this is the differential from the first-order differential: d^2z=d(2e^(2x+5y)dx+5e^(2x+5y)dy).
Let’s use the properties of the differential. In addition, when calculating the second-order differential, we assume that dx and dy are constants, as well as the numbers 2 and 5.
We rewrite the expression as follows: 2dxe^(2x+5y)+5dye^(2x+5y)= (2dx+5dy)de^(2x+5y). We have already found the differential of the function, we use this result: (2dx+5dy)(2e^(2x+5y)dx+5e^(2x+5y)dy).
We open the brackets: 4e^(2x+5y)dx^2+20e^(2x+5y)dxdy+25e^(2x+5y) dy^2. The result is obtained, but we were also asked to find partial derivatives. Let’s write down the differential formula.
The first-order differential is dz=zx’dx+zy’dy, which means that from the reasoning earlier, zx’=2e^(2x+5y), and zy’=5e^(2x+5y).
The second-order differentials d^2z=zxx’’dx^2+2zyx’’dydx+ zyy’’dy^2. Look, it is not difficult to write out zxx’’=4e^(2x+5y), zyx’’=10e^(2x+5y), zyy’’=25e^(2x+5y). The problem is solved.
In reality, when we solve a problem in which we need to find a differential, we, of course, use the differential formulas, find partial derivatives and substitute them into the desired formula.
The task is to make the equation of the tangent plane and the normal to the surface z=sinx*cosx at the point (π/4; π/4; 1/2). Let’s check that the point belongs to the surface: sin(π/4)*cos(π/4)=2/4=1/2. So, this point belongs to the surface, so you should draw tangent planes to the surface point.
So, first, let’s remember the equation of the tangent to the plane z-z0=zx’(x0, y0) (x-x0)+ zy’(x0, y0) (y-y0). Let’s consider what data we have: x0=π/4, y0=π/4, z0=1/2. So, the three numbers are already known to us, we just need to find out the numbers – the values of partial derivatives. Let’s find them: zx’=sinx*cosy, zx’(π/4; π/4)=1/2; zy’= -sinx*siny, zy’(π/4; π/4)=-1/2.
Let’s go to the plane equation: z-1/2=1/2(x-π/4)-1/2(y-π/4). It is most rational to multiply both parts of this equation by 2, we get: 2z-1=x-π/4-y+π/4. We obtain the equation of the tangent plane x-y-2z+1=0.
Let’s turn to the normal. You can use the values of partial derivatives, or you can use the fact that the normal vector for the plane is made up of coefficients for x, y, and z: n=(1;-1;-2), so you don’t need to use the normal formula. We write: (x-π/4)/1=(y-π/4)/(-1)=(z-1/2)/(-2). So, the normal equation is also obtained.
Another, more interesting task. The paraboloid of rotation is the graph of the function z=x^2+y^2. The tangent plane is drawn so that it passes through the points (1,0,0), (0,1,0) located on the x and y coordinate axes. The tangent plane contains a straight line connecting these two points and passes through a point on the surface of the paraboloid. How to find this point and how to make the equation of the tangent plane? Let’s solve this problem.
We will assume that we found this point: (x0, y0, x0^2+y0^2). We also need the values of partial derivatives: zx’=2x, zy’=2y. As a result, the equation of the tangent to the plane looks like this: z-x0^2+y0^2=2x0(x-x0)+2y0(y-y0). After pre-formation, we get: z+x0^2+y0^2=2x0x+2y0y. We got this equation for the tangent plane, but we don’t know the numbers x0 and y0.
The points with coordinates (1, 0, 0) and (0, 1, 0) belong to this plane, and the tangent plane is drawn through these points. Substituting the coordinates of these points in the tangent equation, we get x0^2+y0^2=2x0, x0^2+y0^2=2у0. As a result, we have two cases: x0=0, y0=0 and x0=1, y0=1. For the first case, z=0, for the second – 2x+2y-z=2. So what do we see? The problem does not have a unique solution, and two tangent planes pass through two points to this surface. The problem is solved.