answer:To determine the mean curvature of the surface z = x^2 + y^2 at the point (1, 1, 2), we first need to find the principal curvatures k1 and k2. The mean curvature H is then given by the formula: H = (k1 + k2) / 2 Let's find the principal curvatures k1 and k2: 1. Compute the first and second fundamental forms: First, find the tangent plane to the surface at the given point (1, 1, 2): ∂f/∂x = 2x, ∂f/∂y = 2y At (1, 1, 2), we have: ∂f/∂x = 2, ∂f/∂y = 2 Now, we can find the normal vector N to the surface at the given point: N = (∂f/∂x, ∂f/∂y, -1) = (2, 2, -1) Next, compute the second derivatives of f: ∂²f/∂x² = 2, ∂²f/∂y² = 2 ∂²f/∂x∂y = 0 2. Compute the Gaussian curvature K: K = (e * g - f^2) / (EG - F^2) Here, e = 1, f = 0, g = 1 (from the second fundamental form). Also, E = 1 + (2)^2 = 5, F = 2x = 2, and G = 1 + (2)^2 = 5 (from the first fundamental form). Plugging these values into the equation for K: K = (1 * 1 - 0^2) / (5 * 5 - 2^2) = 0.2 3. Compute the mean curvature H: Now that we have the Gaussian curvature K, we can compute the mean curvature H using the formula: H = (K * (E * G - F^2) -, To find the mean curvature of the surface, we first need to find the partial derivatives of the function and then compute the necessary quantities related to the first and second fundamental forms. The given surface is defined by the equation z = x^2 + y^2. Let F(x,y,z) = x^2 + y^2 - z. Then, nabla F = frac{partial F}{partial x}mathbf{i} + frac{partial F}{partial y}mathbf{j} + frac{partial F}{partial z}mathbf{k} = 2xmathbf{i} + 2ymathbf{j} - mathbf{k}. At the point (1,1,2), the normal vector is given by nabla F(1,1,2) = 2mathbf{i} + 2mathbf{j} - mathbf{k}. Now, let's compute the first and second fundamental forms. The first fundamental form is given by E = nabla F cdot nabla F = (2x)^2 + (2y)^2 + (-1)^2 = 4(x^2 + y^2) + 1. At the point (1,1,2), we have E(1,1,2) = 4(1^2 + 1^2) + 1 = 9. The second fundamental form is given by e = frac{partial^2 F}{partial x^2}, quad f = frac{partial^2 F}{partial x partial y}, quad g = frac{partial^2 F}{partial y^2}. Computing these second partial derivatives, we get frac{partial^2 F}{partial x^2} = 2, quad frac{partial^2 F}{partial x partial y} = 0, quad frac{partial^2 F}{partial y^2} = 2. Thus, at the point (1,1,2), we have e = 2, quad f = 0, quad g = 2. Now we can compute the mean curvature using the formula H = frac{eg - f^2}{2(EG - F^2)^{3/2}} mathbf{n}, where mathbf{n} = frac{nabla F}{|nabla F|} is the unit normal vector. We have already computed |nabla F(1,1,2)| = sqrt{4+4+1} = sqrt{9} = 3. So, mathbf{n} = frac{1}{3}(2mathbf{i} + 2mathbf{j} - mathbf{k}). Now let's compute the numerator of the mean curvature formula: (eg - f^2)(1,1,2) = (2)(2) - (0)^2 = 4. Next, compute the denominator: (EG - F^2)(1,1,2) = (9) - (0)^2 = 9. Now we can find the mean curvature: H = frac{4}{2(9)^{3/2}}(2mathbf{i} + 2mathbf{j} - mathbf{k}) = frac{4}{54sqrt{3}}(2mathbf{i} + 2mathbf{j} - mathbf{k}) = boxed{frac{2sqrt{3}}{27}(mathbf{i} + mathbf{j} - frac{1}{2}mathbf{k})}.,To find the mean curvature of the surface z = x^2 + y^2 at the point (1, 1, 2), we first need to compute the first and second fundamental forms of the surface. Let's parametrize the surface as follows: S(u, v) = (u, v, u^2 + v^2) Now, we compute the first partial derivatives: S_u = (1, 0, 2u) S_v = (0, 1, 2v) Next, we compute the second partial derivatives: S_uu = (0, 0, 2) S_uv = (0, 0, 0) S_vv = (0, 0, 2) Now, we compute the coefficients of the first fundamental form: E = <S_u, S_u> = 1 + 4u^2 F = <S_u, S_v> = 2uv G = <S_v, S_v> = 1 + 4v^2 At the point (1, 1, 2), we have: E = 5 F = 2 G = 5 Now, we compute the coefficients of the second fundamental form: L = <S_uu, N> = 2 / sqrt(1 + 4u^2 + 4v^2) M = <S_uv, N> = 0 N = <S_vv, N> = 2 / sqrt(1 + 4u^2 + 4v^2) At the point (1, 1, 2), we have: L = 2 / sqrt(9) = 2 / 3 M = 0 N = 2 / sqrt(9) = 2 / 3 Finally, we compute the mean curvature H: H = (1/2) * ( (E * N + G * L - 2 * F * M) / (E * G - F^2) ) H = (1/2) * ( (5 * (2/3) + 5 * (2/3) - 2 * 2 * 0) / (5 * 5 - 2^2) ) H = (1/2) * ( (10/3) / 21 ) H = 5 / 6 So, the mean curvature of the surface at the point (1, 1, 2) is 5/6.

answer:To find the principal curvatures of the surface, we first need to compute the first and second fundamental forms. Let's start by computing the first fundamental form. Given the surface equation: z = x^3 + y^3 - 3xy We can compute the partial derivatives with respect to x and y: ∂z/∂x = 3x^2 - 3y ∂z/∂y = 3y^2 - 3x Now, let's find the length of the normal vector N at the point (1, 1, -1): N = (-∂z/∂x, -∂z/∂y, 1) = (-3(1)^2 + 3(1), -3(1)^2 + 3(1), 1) = (-3, -3, 1) Length of N = √((-3)^2 + (-3)^2 + 1^2) = √(18 + 18 + 1) = √37 Now, we can compute the components of the first fundamental form: E = (∂x/∂x)^2 + (∂y/∂x)^2 = 1^2 + 1^2 = 2 F = (∂x/∂x)(∂y/∂y) - (∂x/∂y)(∂y/∂x) = 1*1 - 1*(-1) = 2 G = (∂y/∂y)^2 + (∂y/∂x)^2 = 1^2 + (-1)^2 = 2 Now, let's compute the second fundamental form. We need to find the second partial derivatives: ∂²z/∂x² = 6x ∂²z/∂x∂y = -3 ∂²z/∂y² = 6y Now, let's compute the components of the second fundamental form: L = N, To find the principal curvatures of a surface, we first need to find the first and second derivatives of the function defining the surface. Given the surface z = f(x, y) = x^3 + y^3 - 3xy, we have: f_x = 3x^2 - 3y f_y = 3y^2 - 3x f_{xx} = 6x f_{xy} = -3 f_{yx} = -3 (note that f_{xy} = f_{yx} by Clairaut's theorem) f_{yy} = 6y Now let's compute the partial derivatives at the point (1, 1, -1): f_x(1, 1) = 3 - 3 = 0 f_y(1, 1) = 3 - 3 = 0 f_{xx}(1, 1) = 6 f_{xy}(1, 1) = f_{yx}(1, 1) = -3 f_{yy}(1, 1) = 6 The Gaussian curvature K can be computed using the formula: K = f_{xx}f_{yy} - (f_{xy})^2 = 6 * 6 - (-3)^2 = 33 To find the principal curvatures, we need to solve the following equation for kappa: (f_{xx} - kappa)(f_{yy} - kappa) - (f_{xy})^2 = 0 Substituting our known values: (6 - kappa)(6 - kappa) - (-3)^2 = 0 kappa^2 - 12kappa + 27 = 0 This quadratic factors as: (kappa - 3)(kappa - 9) = 0 So the principal curvatures are kappa_1 = 3 and kappa_2 = 9.,To find the principal curvatures of the surface at the point (1, 1, -1), we first need to find the first and second fundamental forms of the surface. Let's parameterize the surface by letting x = u and y = v. Then, the position vector of the surface is given by: r(u, v) = (u, v, u^3 + v^3 - 3uv) Now, we need to find the first partial derivatives of r with respect to u and v: r_u = (∂r/∂u) = (1, 0, 3u^2 - 3v) r_v = (∂r/∂v) = (0, 1, 3v^2 - 3u) Next, we find the second partial derivatives: r_uu = (∂²r/∂u²) = (0, 0, 6u) r_uv = (∂²r/∂u∂v) = (0, 0, -3) r_vv = (∂²r/∂v²) = (0, 0, 6v) Now, we need to find the coefficients of the first fundamental form: E = r_u • r_u = (1, 0, 3u^2 - 3v) • (1, 0, 3u^2 - 3v) = 1 + (3u^2 - 3v)^2 F = r_u • r_v = (1, 0, 3u^2 - 3v) • (0, 1, 3v^2 - 3u) = (3u^2 - 3v)(3v^2 - 3u) G = r_v • r_v = (0, 1, 3v^2 - 3u) • (0, 1, 3v^2 - 3u) = 1 + (3v^2 - 3u)^2 And the coefficients of the second fundamental form: L = r_uu • N = (0, 0, 6u) • N M = r_uv • N = (0, 0, -3) • N N = r_vv • N = (0, 0, 6v) • N Here, N is the unit normal vector to the surface, which can be found by taking the cross product of r_u and r_v: N = r_u × r_v = (3u^2 - 3v, 3v^2 - 3u, 1) Now, we need to evaluate the coefficients of the first and second fundamental forms at the point (1, 1, -1), which corresponds to u = 1 and v = 1: E = 1 + (3 - 3)^2 = 1 F = (3 - 3)(3 - 1) = 0 G = 1 + (3 - 3)^2 = 1 L = (0, 0, 6) • (0, 2, 1) = 6 M = (0, 0, -3) • (0, 2, 1) = -3 N = (0, 0, 6) • (0, 2, 1) = 6 Now, we can find the principal curvatures by solving the following equation: k^2 - (L + N)k + (LN - M^2) = 0 k^2 - 12k + (6*6 - (-3)^2) = 0 k^2 - 12k + 27 = 0 Solving this quadratic equation, we get the principal curvatures: k1 = 6 - 3√3 k2 = 6 + 3√3 So, the principal curvatures of the surface at the point (1, 1, -1) are k1 = 6 - 3√3 and k2 = 6 + 3√3.

answer:To find the principal curvatures at the point (1, 2, √2) on the surface defined by x^2 + y^2 + z^2 = 9, we first need to compute the first and second fundamental forms. The given surface is a sphere with radius R = 3 (since the equation of the surface represents a sphere of radius 3 centered at the origin). Let's find the first fundamental form (E, F, G) and the second fundamental form (L, M, N) at the point (1, 2, √2) on the surface. 1. First fundamental form: For a sphere with radius R, the coefficients of the first fundamental form are given by: E = 1 + (x/R)^2 = 1 + (1/3)^2 = 1 + 1/9 = 10/9 F = -x/R * y/R = -1/3 * 2/3 = -2/9 G = 1 + (y/R)^2 = 1 + (2/3)^2 = 1 + 4/9 = 13/9 So, the first fundamental form at the point (1, 2, √2) is (E, F, G) = (10/9, -2/9, 13/9). 2. Second fundamental form: For a sphere with radius R, the coefficients of the second fundamental form are given by: L = N = -1/R M = 0 N = -1/R So, the second fundamental form at the point (1, 2, √2) is (L, M, N) = (-1/3, 0, -1/3). 3. Principal curvatures: The principal curvatures k1 and k2 can be computed using the formula: k1, k2 = (L + N ± sqrt((L - N)^2 + 4M^2)) / 2G Substituting the values of L, M, N, and G, we get: k1, k2 = (-1/3 - (-1/3) ± sqrt((((-1/3, To find the principal curvatures, we first need to find the Gaussian curvature K and mean curvature H of the surface at the given point. The implicit function theorem allows us to compute the partial derivatives of x, y, and z with respect to u and v, where x = u, y = v, and z is given implicitly by the equation x^2 + y^2 + z^2 = 9. We have: ∂x/∂u = 1, ∂x/∂v = 0, ∂y/∂u = 0, ∂y/∂v = 1, ∂z/∂u = -(x/z), ∂z/∂v = -(y/z). Now let's compute the second-order partial derivatives needed for the Gaussian and mean curvatures: ∂²x/∂u² = 0, ∂²x/∂u∂v = 0, ∂²x/∂v∂u = 0, ∂²x/∂v² = 0, ∂²y/∂u² = 0, ∂²y/∂u∂v = 0, ∂²y/∂v∂u = 0, ∂²y/∂v² = 0, ∂²z/∂u² = (-x²/z - xz'(x²+y²+z²)), ∂²z/∂u∂v = (-xy/z - yz'(x²+y²+z²)), ∂²z/∂v∂u = (-yx/z - xz'(x²+y²+z²)), ∂²z/∂v² = (-y²/z - yz'(x²+y²+z²)). At the point (1, 2, √2), we have x = 1, y = 2, and z = √2. Thus, ∂²z/∂u² ≈ -1.515, ∂²z/∂u∂v ≈ -2.020, ∂²z/∂v∂u ≈ -2.020, ∂²z/∂v² ≈ -4.545. The Gaussian curvature K is given by: K = (LN - M²) / (EG - F²), where E = ∂x/∂u² = 0, F = ∂x/∂u∂v = 0, G = ∂y/∂v² = 1, L = ∂²z/∂u² ≈ -1.515, M = ∂²z/∂u∂v ≈ -2.020, N = ∂²z/∂v² ≈ -4.545. Thus, K ≈ (-1.515 * 1 - (-2.020)²) / (0 - 0) ≈ -4.082. The mean curvature H is given by: H = (EN + GL - 2FM) / 2(EG - F²), So, H ≈ (-1.515 * 1 + (-4.545) * 0 - 2 * 0 * (-2.020)) / (2 * 0) ≈ ∞. However, the mean curvature is not actually infinite; there seems to be an issue with the implicit function theorem calculation since the determinant of the Jacobian is zero at this point. To find the principal curvatures directly, we can use the following formulas: κ₁, κ₂ = (LN - M² + H(E±G) ± sqrt((L+N)²EG - 4(LN-M²)H² + (L-N)²EG)) / 2(EG-F²) Since H is approximately infinity, we can set H=0 to simplify the formula. We get: κ₁, κ₂ ≈ (LN - M² ± sqrt(L²EG + N²EG - 2LNEG)) / 2(EG-F²) Plugging in the values, we get: κ₁, κ₂ ≈ (−1.515 * -4.545 - (-2.020)² ± sqrt((-1.515)² * 1 * 0 - 4(-1.515) * -4.545 * 0 + (-4.545)² * 1 * 0)) / (2 * (1 - 0)) Since both terms under the square root are zero, we have: κ₁ ≈ -1.515 and κ₂ ≈ -4.545 So the principal curvatures at the point (1, 2, √2) are approximately κ₁ ≈ -1.515 and κ₂ ≈ -4.545.,To find the principal curvatures, we first need to find the normal vector, the shape operator, and the eigenvalues of the shape operator. The given surface is a sphere with radius 3 centered at the origin. The equation is x^2 + y^2 + z^2 = 9. 1. Normal vector: The gradient of the function F(x, y, z) = x^2 + y^2 + z^2 - 9 at the point (1, 2, √2) will give us the normal vector. ∇F(x, y, z) = (2x, 2y, 2z) ∇F(1, 2, √2) = (2, 4, 2√2) 2. Shape operator: The shape operator S is defined as the derivative of the normal vector with respect to the surface. We can find the shape operator by taking the gradient of the normal vector and evaluating it at the given point. S = -d(∇F)/dF The partial derivatives of the normal vector components are: ∂(2x)/∂x = 2, ∂(2x)/∂y = 0, ∂(2x)/∂z = 0 ∂(4y)/∂x = 0, ∂(4y)/∂y = 4, ∂(4y)/∂z = 0 ∂(2√2z)/∂x = 0, ∂(2√2z)/∂y = 0, ∂(2√2z)/∂z = 2√2 So, the shape operator S is given by the matrix: S = - 2, 0, 0, 0, 4, 0, 0, 0, 2√2 3. Eigenvalues: To find the principal curvatures, we need to find the eigenvalues of the shape operator S. The characteristic equation is given by: det(S - λI) = 0 Expanding the determinant, we get: (2 - λ)((4 - λ)(2√2 - λ)) = 0 The eigenvalues are λ1 = 2, λ2 = 4, and λ3 = 2√2. So, the principal curvatures at the point (1, 2, √2) are 2, 4, and 2√2.

answer:To find the principal curvatures, we need to compute the Gaussian curvature (K) and the mean curvature (H) at the given point (1, 1, 2). First, let's find the first and second fundamental forms of the surface. The surface is defined by z = x^2 + y^2. So, we can write the parametric form as: r(u, v) = (u, v, u^2 + v^2) Now, we compute the first partial derivatives with respect to u and v: ru = dr/du = (1, 0, 2u) rv = dr/dv = (0, 1, 2v) The first fundamental form coefficients are: E = ru • ru = 1 + 4u^2 F = ru • rv = 2uv G = rv • rv = 1 + 4v^2 Now, we compute the second partial derivatives: ruu = d^2r/du^2 = (0, 0, 2) ruv = d^2r/dudv = (0, 0, 0) rvv = d^2r/dv^2 = (0, 0, 2) The second fundamental form coefficients are: L = ruu • N = 2 / sqrt(1 + 4u^2 + 4v^2) M = ruv • N = 0 N = rvv • N = 2 / sqrt(1 + 4u^2 + 4v^2) Now, we can compute the Gaussian curvature (K) and the mean curvature (H): K = (LN - M^2) / (EG - F^2) H = (LE + GN - 2MF) / 2(EG - F^2) At the point (1, 1, 2), we have u = 1 and v = 1: E = 1 + 4(1)^2 = 5 F = 2(1)(1) = 2 G = 1 + 4(1)^2 = 5 L = 2 / sqrt(1 + 4(1)^,To find the principal curvatures, we first need to compute the first and second fundamental forms. Let's start with the first fundamental form: Given the surface equation z = x^2 + y^2, we can define the position vector r(x, y) = (x, y, x^2 + y^2). Now, we compute the partial derivatives with respect to x and y: r_x = (1, 0, 2x) r_y = (0, 1, 2y) Now, we compute the first fundamental form coefficients: E = r_x • r_x = 1 + 4x^2 F = r_x • r_y = 2xy G = r_y • r_y = 1 + 4y^2 At the point (1, 1, 2), we have: E = 5 F = 2 G = 5 Now, let's compute the second fundamental form. We need the unit normal vector N: N = (r_x × r_y) / |r_x × r_y| r_x × r_y = (-2x, -2y, 1) |r_x × r_y| = sqrt((-2x)^2 + (-2y)^2 + 1^2) = sqrt(4x^2 + 4y^2 + 1) At the point (1, 1, 2), we have: N = (-2, -2, 1) / sqrt(9) = (-2/3, -2/3, 1/3) Now, we compute the second fundamental form coefficients: L = N • r_xx = -4/3 M = N • r_xy = 0 N = N • r_yy = -4/3 Now, we can compute the principal curvatures using the formula: k1, k2 = (EG - F^2) / (E^2 + 2EF + G^2) ± sqrt(((EG - F^2) / (E^2 + 2EF + G^2))^2 - (LN - M^2) / (E^2 + 2EF + G^2)) Plugging in the values, we get: k1, k2 = (25 - 4) / (50) ± sqrt(((25 - 4) / (50))^2 - (-8/9) / (50)) k1, k2 = 21/50 ± sqrt((21/50)^2 + 8/450) k1 ≈ 0.420 k2 ≈ -0.420 So, the principal curvatures at the point (1, 1, 2) are approximately 0.420 and -0.420., To find the principal curvatures, we first need to compute the gradient and Hessian matrix of the surface. The given surface is defined by the equation z = f(x, y) = x^2 + y^2. The gradient of f is given by: nabla f = (frac{partial f}{partial x}, frac{partial f}{partial y}) = (2x, 2y) At the point (1, 1), the gradient is nabla f = (2, 2). The Hessian matrix H_f is given by: H_f = begin{bmatrix} frac{partial^2 f}{partial x^2} & frac{partial^2 f}{partial x partial y} frac{partial^2 f}{partial y partial x} & frac{partial^2 f}{partial y^2} end{bmatrix} = begin{bmatrix} 2 & 0 0 & 2 end{bmatrix} Now let's find the normal vector N at the point (1, 1, 2). We have: N = frac{nabla f}{|nabla f|} = frac{(2, 2)}{sqrt{2^2 + 2^2}} = (frac{1}{sqrt{2}}, frac{1}{sqrt{2}}, 0) To compute the principal curvatures, we'll use the Weingarten map, which is a linear map that describes how the normal vector changes as we move along the surface. It's represented by the matrix W = -dN, where dN is the differential of the normal vector. Since the normal vector does not depend on z, we have that frac{partial N}{partial z} = 0. Thus, the Weingarten map simplifies to: W = -dN = -begin{bmatrix} frac{partial N_x}{partial x} & frac{partial N_x}{partial y} & frac{partial N_x}{partial z} frac{partial N_y}{partial x} & frac{partial N_y}{partial y} & frac{partial N_y}{partial z} frac{partial N_z}{partial x} & frac{partial N_z}{partial y} & frac{partial N_z}{partial z} end{bmatrix} = begin{bmatrix} -frac{partial N_x}{partial x} & -frac{partial N_x}{partial y} -frac{partial N_y}{partial x} & -frac{partial N_y}{partial y} end{bmatrix} Computing the partial derivatives of the normal vector components, we get: frac{partial N_x}{partial x} = -frac{1}{sqrt{2}}, quad frac{partial N_x}{partial y} = 0, quad frac{partial N_y}{partial x} = 0, quad frac{partial N_y}{partial y} = -frac{1}{sqrt{2}} Thus, the Weingarten map is: W = begin{bmatrix} frac{1}{sqrt{2}} & 0 0 & frac{1}{sqrt{2}} end{bmatrix} The principal curvatures are the eigenvalues of the Weingarten map. The characteristic polynomial of W is: det(W - lambda I) = (w_{11} - lambda)(w_{22} - lambda) - w_{12}w_{21} = (frac{1}{sqrt{2}} - lambda)^2 - 0 = frac{1}{2} - sqrt{2}lambda + lambda^2 Setting this equal to zero and solving for lambda, we find that the principal curvatures are: lambda_{1,2} = frac{sqrt{2} pm 1}{2} Thus, the principal curvatures at the point (1, 1, 2) are frac{sqrt{2}+1}{2} and frac{sqrt{2}-1}{2}.