We are looking for the gradient of the surface S: grad(S) = (2x-4, 2y, 2z+6). At the point M0(2, 1, –1) we have: grad(S) = (0, 2, 4). Since the tangent plane to the surface S at point M0 is parallel to the gradient of the surface, the equation of the tangent plane has the form: 0(x-2) + 2(y-1) + 4(z+1) = 0, that is, y + 2z - 1 = 0. The equation of the normal to the surface S at point M0 has the form: 2(x-2) + 2(y-1) + 4(z+1) = 0, that is, x + y + 2z - 8 = 0.
We calculate the first partial derivatives: z'x=2xe^(x^2-y^2), z'y=-2ye^(x^2-y^2). Next we find the second partial derivatives: z''xy=2e^(x^2-y^2)-4x^2e^(x^2-y^2), z''yx=2e^(x^2-y ^2)-4y^2e^(x^2-y^2). Note that z''xy = z''yx, which means that the function z=ex2-y2 satisfies the condition of equality of mixed derivatives.
Let us calculate Laplace from the function u(x,y,z): Δu = u''xx + u''yy + u''zz = 4 + 6 + 2 = 12. Since Δu is not equal to zero, then the function u(x ,y,z) does not satisfy the Laplace equation.
We calculate the partial derivatives: z'x= y/(2√x) - 1, z'y= √x - 4y + 14. Find the stationary points: z'x=0 => y/(2√x) = 1 = > y=2√x, z'y=0 => √x - 4y + 14 = 0 => √x - 8√x + 14 = 0 => x = 4, y = 4. Check the sufficient conditions for the extremum: z ''xx= -y/(4x^(3/2)) 0. Since z''xx
We express y in terms of x in the equation y=4: y=4. Substitute y=x into the function z=3x+y-xy: z=4x-2x^2. We calculate the derivatives: z'x=4-4x, z''xx=-4. Find the critical point: z'x=0 => x=1, then y=1. We check the sufficient conditions for the extremum: z''xx=-4
IDZ 10.2 – Option 1. Solutions Ryabushko A.P. is a digital product, which is a collection of solutions to mathematics tasks developed by A.P. Ryabushko. It contains detailed and clear solutions to assignments that will help students better understand the material and prepare for exams.
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IDZ 10.2 – Option 1. Solutions Ryabushko A.P. is a set of problems in mathematical analysis, which covers the following tasks:
It is necessary to find the equations of the tangent plane and normal to a given surface S at the point M0(x0, y0, z0). The surface S is given by the equation x2 + y2 + z2 + 6z – 4x + 8 = 0, and the point M0 has coordinates (2, 1, – 1).
It is necessary to find the second partial derivatives of the indicated functions and check that z''xy = z''yx. The function z(x,y) is given by the equation z = ex2-y2.
It is necessary to check whether the function u satisfies the specified equation.
It is necessary to examine the function z(x,y) = y√x – 2y2 – x + 14y for its extremum.
It is necessary to find the largest and smallest values of the function z(x,y) = 3x + y – xy in the region D, limited by the given lines y = x, y = 4, x = 0.
The set of problems is designed in Microsoft Word 2003 using the formula editor. Solutions to problems are presented in detailed form.
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