answer:To find the equation of the tangent plane to the surface at the point (3, 4, -2), we first need to find the gradient of the surface at that point. The gradient is given by the partial derivatives of the equation with respect to x, y, and z. The equation of the surface is x² + y² + z² = 25. Let's find the partial derivatives: ∂f/∂x = 2x ∂f/∂y = 2y ∂f/∂z = 2z Now, we need to evaluate these partial derivatives at the point (3, 4, -2): ∂f/∂x(3, 4, -2) = 2(3) = 6 ∂f/∂y(3, 4, -2) = 2(4) = 8 ∂f/∂z(3, 4, -2) = 2(-2) = -4 The gradient of the surface at the point (3, 4, -2) is given by the vector (6, 8, -4). This vector is also the normal vector to the tangent plane at that point. Now, we can use the point-normal form of the equation of a plane, which is given by: A(x - x₀) + B(y - y₀) + C(z - z₀) = 0 where (A, B, C) is the normal vector and (x₀, y₀, z₀) is the point on the plane. In our case, (A, B, C) = (6, 8, -4) and (x₀, y₀, z₀) = (3, 4, -2). Plugging in these values, we get: 6(x - 3) + 8(y - 4) - 4(z + 2) = 0 Simplifying the equation, we get: 6x + 8y - 4z = 50 This is the equation of the plane tangent to the surface at the point (3, 4, -2).,To find the equation of the tangent plane to the given surface at the point (3, 4, -2), we first need to find the gradient of the surface at that point. The gradient is given by the partial derivatives of the surface equation with respect to x, y, and z. Let F(x, y, z) = x² + y² + z² - 25 Now, we compute the partial derivatives: ∂F/∂x = 2x ∂F/∂y = 2y ∂F/∂z = 2z At the point (3, 4, -2), the gradient is: ∇F(3, 4, -2) = (6, 8, -4) Now, we can use the gradient as the normal vector to the tangent plane. The equation of the tangent plane can be written as: A(x - x0) + B(y - y0) + C(z - z0) = 0 where (A, B, C) is the normal vector and (x0, y0, z0) is the point on the plane. In our case, (A, B, C) = (6, 8, -4) and (x0, y0, z0) = (3, 4, -2). Plugging these values into the equation, we get: 6(x - 3) + 8(y - 4) - 4(z - (-2)) = 0 Simplifying this equation, we get the equation of the tangent plane: 6x + 8y + 4z = 68 So, the equation of the tangent plane to the surface at the point (3, 4, -2) is 6x + 8y + 4z = 68., To find the equation of the plane tangent to the algebraic surface at a given point, we need to find the gradient of the function at that point. The general form of the surface is F(x, y, z) = x² + y² + z² - 25 First, let's find the partial derivatives of F with respect to x, y, and z: F_x(x, y, z) = 2x F_y(x, y, z) = 2y F_z(x, y, z) = 2z Now, let's evaluate these partial derivatives at the point (3, 4, -2): F_x(3, 4, -2) = 2 * 3 = 6 F_y(3, 4, -2) = 2 * 4 = 8 F_z(3, 4, -2) = 2 * (-2) = -4 So, the normal vector to the surface at the point (3, 4, -2) is given by N = (6, 8, -4) Since the tangent plane is perpendicular to the normal vector, its equation can be found using the dot product and the point-normal form of the plane equation: (X - X0) . N = 0 Where X = (x, y, z) is a general point on the plane, X0 = (3, 4, -2) is the given point, and N is the normal vector. Expanding this equation, we get: (x - 3) * 6 + (y - 4) * 8 + (z + 2) * (-4) = 0 Simplifying and rearranging, we finally obtain the equation of the tangent plane: 6x + 8y - 4z - 58 = 0

answer:To find the equation of the tangent plane to the given algebraic surface at the point (-1, 1, 2), we first need to find the gradient of the function f(x, y, z) = x^4 + y^4 + z^4 - 4xyz at this point. The gradient is a vector of partial derivatives: ∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z) Let's compute the partial derivatives: ∂f/∂x = 4x^3 - 4yz ∂f/∂y = 4y^3 - 4xz ∂f/∂z = 4z^3 - 4xy Now, we evaluate the partial derivatives at the point (-1, 1, 2): ∂f/∂x|_(-1,1,2) = 4(-1)^3 - 4(1)(2) = 4 - 8 = -4 ∂f/∂y|_(-1,1,2) = 4(1)^3 - 4(-1)(2) = 4 + 8 = 12 ∂f/∂z|_(-1,1,2) = 4(2)^3 - 4(-1)(1) = 32 + 4 = 36 So, the gradient of f at the point (-1, 1, 2) is ∇f = (-4, 12, 36). The equation of the tangent plane in vector form is given by: (x - x_0, y - y_0, z - z_0) · ∇f(x_0, y_0, z_0) = 0 Substitute the point (-1, 1, 2) and the gradient (-4, 12, 36) into the equation: (x - (-1), y - 1, z - 2) · (-4, 12, 36) = 0 Simplify the equation: (x + 1, y - 1, z - 2) · (-4, 12, 36) = 0 Multiply both sides by -1: (x + 1, y - 1, z - 2) · (4, -12, -36) = 0 Expand the dot product: 4(x + 1) - 12(y - 1) - 36(z - 2) = 0 Distribute the coefficients: 4x + 4 - 12y + 12 - 36z + 72 = 0 Simplify the equation: 4x - 12y - 36z + 88 = 0 This is the equation of the tangent plane to the given algebraic surface at the point (-1, 1, 2).,To find the equation of the tangent plane to the given algebraic surface at the point (-1,1,2), we first need to find the partial derivatives of the surface equation with respect to x, y, and z. Let F(x, y, z) = x^4 + y^4 + z^4 - 4xyz. Compute the partial derivatives: ∂F/∂x = 4x^3 - 4yz ∂F/∂y = 4y^3 - 4xz ∂F/∂z = 4z^3 - 4xy Now, evaluate these partial derivatives at the point (-1,1,2): ∂F/∂x(-1,1,2) = 4(-1)^3 - 4(1)(2) = -4 - 8 = -12 ∂F/∂y(-1,1,2) = 4(1)^3 - 4(-1)(2) = 4 + 8 = 12 ∂F/∂z(-1,1,2) = 4(2)^3 - 4(-1)(1) = 32 + 4 = 36 Now, we can find the normal vector to the tangent plane using these partial derivatives: N = <-12, 12, 36> Finally, we can find the equation of the tangent plane using the point-normal form: (-1 - (-1))^2 + (1 - 1)^2 + (2 - 2)^2 - 12(-1) - 12(1) - 36(0) = 0 4 - 0 + 0 = 4 So, the equation of the tangent plane at the point (-1,1,2) is: 12(x - (-1)) + 12(y - 1) + 36(z - 2) = 0,To find the equation of the tangent plane to the surface at the point (-1,1,2), we first need to find the gradient of the surface at that point. The gradient is given by the vector of partial derivatives with respect to x, y, and z. Let F(x, y, z) = x^4 + y^4 + z^4 - 4xyz. We compute the partial derivatives: ∂F/∂x = 4x^3 - 4yz ∂F/∂y = 4y^3 - 4xz ∂F/∂z = 4z^3 - 4xy Now, we evaluate these partial derivatives at the point (-1,1,2): ∂F/∂x(-1,1,2) = 4(-1)^3 - 4(1)(2) = -4 - 8 = -12 ∂F/∂y(-1,1,2) = 4(1)^3 - 4(-1)(2) = 4 + 8 = 12 ∂F/∂z(-1,1,2) = 4(2)^3 - 4(-1)(1) = 32 + 4 = 36 The gradient of F at the point (-1,1,2) is given by the vector <-12, 12, 36>. Now, we can find the equation of the tangent plane using the point-normal form: A(x - x₀) + B(y - y₀) + C(z - z₀) = 0 where (x₀, y₀, z₀) is the point (-1,1,2) and (A, B, C) is the gradient <-12, 12, 36>. -12(x - (-1)) + 12(y - 1) + 36(z - 2) = 0 Simplifying the equation, we get: -12x - 12 + 12y - 12 + 36z - 72 = 0 Combining the constants: -12x + 12y + 36z - 96 = 0 Divide by -12 to simplify further: x - y - 3z + 8 = 0 So, the equation of the tangent plane to the surface at the point (-1,1,2) is: x - y - 3z + 8 = 0

answer:To find the equation of the plane curve that is the intersection of the two surfaces, we can solve the system of equations by eliminating one variable. Let's eliminate x^2 by adding the two equations: (x^2 + y^2 + z^2) + (x^2 - y^2 - z^2) = 81 + 0 Simplifying, we get: 2x^2 = 81 Now, divide by 2: x^2 = frac{81}{2} Now, substitute the value of x^2 back into either of the original equations. Let's use the first equation: frac{81}{2} + y^2 + z^2 = 81 Now, subtract frac{81}{2} from both sides: y^2 + z^2 = 81 - frac{81}{2} y^2 + z^2 = frac{81}{2} This is the equation of the plane curve that is the intersection of the two surfaces. It represents a circle in the yz-plane with a radius of sqrt{frac{81}{2}}.,To find the equation of the plane curve that is the intersection of the two given surfaces, we need to solve the system of equations and eliminate one of the variables. Let's first solve the second equation for one of the variables, say y^2: y^2 = x^2 - z^2 Now, substitute this expression for y^2 into the first equation: x^2 + (x^2 - z^2) + z^2 = 81 This simplifies to: 2x^2 = 81 Now, solve for x^2: x^2 = 40.5 Since x^2 cannot be negative, we can take the positive root of 40.5: x = sqrt{40.5} approx 6.36 Now, substitute x^2 back into the equation y^2 = x^2 - z^2: y^2 = 40.5 - z^2 Since we are looking for a plane curve, we can separate the variables and obtain the parametric equations: x = 6.36 y = sqrt{40.5 - z^2} And z can take any values such that 40.5 - z^2 geq 0. Therefore, the equation of the plane curve that is the intersection of the two surfaces is given by the parametric equations: x = 6.36, y = sqrt{40.5 - z^2}, where -4.95 leq z leq 7.95, To find the equation of the plane curve that is the intersection of the two surfaces, we first need to solve one of the equations for one variable in terms of the other two variables. Let's solve the second equation for z: z^2 = y^2 + x^2 z = pmsqrt{y^2 + x^2} Since both surfaces are circles centered at the origin with radius 9, their intersection must be a circle as well. The sign of z does not matter because it represents the same point in 3D space. So we can choose the positive square root for simplicity: z = sqrt{y^2 + x^2} Now we have an expression for z in terms of x and y, which defines our plane curve. To write this as an equation, we can square both sides to eliminate the square root: z^2 = y^2 + x^2 This is the equation of the plane curve formed by the intersection of the two surfaces.

answer:To find the intersection of the surfaces defined by these equations, we need to solve the system of equations analytically or numerically. The equations are: 1) xy - z^2 = 1 2) x^2 + y^2 + z^2 = 9 First, we can rewrite the second equation as a circle equation: x^2 + y^2 + z^2 - 9 = 0 Now, clear the denominator in equation 1: xy - frac{1}{z^2} = 1 Multiply the second equation by z^2 to try to eliminate z^2: z^2(x^2 + y^2 + z^2 - 9) = 0 Now substitute z^2 from the second equation into the first equation: xy - frac{1}{z^2} = 1 xy - frac{1}{x^2 + y^2 + (9-x^2-y^2)} = 1 Simplifying, we get: xy = 1 Now we have (x, y, z) as the coordinates of the intersection points of the surfaces. However, finding the explicit solution is not trivial, so we can only say that the intersection of these surfaces is the set of points (x, y, z) which satisfy both equations. Unfortunately, there is no easy analytical solution for this problem.,To find the intersection of the surfaces, we need to solve the system of equations: (1) xy - z^2 = 1 (2) x^2 + y^2 + z^2 = 9 First, we can solve equation (1) for one of the variables, say z^2. We get: z^2 = xy - 1 Now, substitute this expression for z^2 into equation (2): x^2 + y^2 + (xy - 1) = 9 Rearrange the terms: x^2 + y^2 + xy - 1 = 9 Subtract 9 from both sides: x^2 + y^2 + xy - 10 = 0 Now we have a quadratic equation in two variables. To solve it, we can use substitution or elimination method. Let's use substitution. Solve equation (1) for y: y = frac{z^2 + 1}{x} Substitute this expression for y into the quadratic equation: x^2 + left(frac{z^2 + 1}{x}right)^2 + xleft(frac{z^2 + 1}{x}right) - 10 = 0 Multiply both sides by x^2 to eliminate the fractions: x^4 + (z^2 + 1)^2 + x^2(z^2 + 1) - 10x^2 = 0 Now we have a quartic equation in x and z. It's difficult to solve it analytically, so we can use numerical methods or graphical methods to find the intersection points. However, we can try to simplify it further by making a substitution. Let u = x^2 and v = z^2. Then the equation becomes: u^2 + (v + 1)^2 + u(v + 1) - 10u = 0 Now we have a quartic equation in u and v. We can try to solve it numerically or graphically to find the intersection points. Once we find the values of u and v, we can substitute back x^2 and z^2 to find the corresponding x, y, and z values., To find the intersection of the given surfaces, we first solve the first equation for z: xy - 1 = z^2 Rightarrow z = pm sqrt{xy - 1} Now, substitute this expression for z into the second equation: x^2 + y^2 + (pm sqrt{xy - 1})^2 = 9 Rightarrow x^2 + y^2 pm 2sqrt{x^2y^2 - y^2} + xy - 1 = 9 We can eliminate the square root term by squaring both sides of the equation: (x^2 + y^2 pm 2sqrt{x^2y^2 - y^2} + xy - 1)^2 = (9)^2 Rightarrow x^4 + 2x^2y^2 + y^4 pm 4x^2y^2sqrt{1 - frac{y^2}{x^2}} + 2x^3y - 2x^2 + 2xy^3 - 2y^2 + x^2y^2 - 2xy + 1 = 81 Next, we isolate the square root term: 4x^2y^2sqrt{1 - frac{y^2}{x^2}} = 81 - x^4 - 2x^2y^2 - y^4 - 2x^3y + 2x^2 - 2xy^3 + 2y^2 - x^2y^2 + 2xy - 1 Squaring both sides again: 16x^4y^4left(1 - frac{y^2}{x^2}right) = left(81 - x^4 - 2x^2y^2 - y^4 - 2x^3y + 2x^2 - 2xy^3 + 2y^2 - x^2y^2 + 2xy - 1right)^2 Rightarrow 16x^4y^4 - 16x^4y^6 = 6561 - 1458x^2 - 1218x^4 - 1458y^2 - 1218y^4 + dots At this point, it becomes difficult to isolate the intersection curve explicitly. Instead, we can analyze the equation obtained above to get some information about the intersection curve. Notice that the intersection curve lies on the cylinder x^2 + y^2 = 9, since when we substitute x = 3cos{theta} and y = 3sin{theta}, the equation becomes: 16(27cos^2{theta}sin^2{theta})(1 - sin^2{theta}) = 6561 - 1458(9cos^2{theta}) - dots This shows that the intersection curve lies on the cylinder x^2 + y^2 = 9. Furthermore, by using polar coordinates, we can rewrite the original equations as: r^2cos{theta}sin{theta} - z^2 = 1 r^2 + z^2 = 9 Solving for z in terms of r and θ, we get: z = pm sqrt{r^2 - r^2cos{theta}sin{theta}} The intersection curve is obtained by substituting this expression for z into the equation of the cylinder. The exact equation may not be easily obtained, but we now have a better understanding of the nature and location of the intersection curve.