(a) Show that there is no tangent plane to the sphere x² + y² + z² - 4x + 2y - 4z + 4 = 0 that can be passed through the straight…

(a)

Complete the square in the sphere equation: x² + y² + z² - 4x + 2y - 4z + 4 = 0 => (x-2)² + (y+1)² + (z-2)² = 5. So the sphere has centre C = (2,-1,2) and radius R = √5.

The given line is (x+6)/2 = y+3 = z+1 = t. Hence x = 2t - 6, y = t - 3, z = t - 1. So it passes through A = (-6,-3,-1) and has direction vector v = (2,1,1).

Suppose a tangent plane to the sphere contains this line. Then its distance from C must equal R = √5.

Any plane containing the line has normal n perpendicular to v. Its distance from C is

|n·(C-A)|/||n||. For n perpendicular to v, this distance is at most the perpendicular distance from C to the line.

Now C - A = (8,2,3). The distance from C to the line is

||(C-A) × v||/||v||. Compute (8,2,3) × (2,1,1) = (-1,-2,4). Thus distance = √(1+4+16)/√(4+1+1) = √21/√6 = √(7/2).

But √(7/2) < √5 since 7/2 < 5. Therefore every plane containing the given line is at distance strictly less than the radius √5 from the centre. Hence no such plane can be tangent to the sphere.

Final answer: no tangent plane exists.

(b)

Take the convention f_xy = (f_x)_y.

For (x,y) ≠ (0,0), f_x = [(3x²y - y³)(x²+y²) - (x³y - xy³)(2x)]/(x²+y²)². Putting x = 0, y ≠ 0 gives f_x(0,y) = -y. Also, f_x(0,0) = lim h→0 [f(h,0)-f(0,0)]/h = 0. Therefore f_xy(0,0) = lim k→0 [f_x(0,k)-f_x(0,0)]/k = lim k→0 (-k)/k = -1.

Similarly, f_y = [(x³ - 3xy²)(x²+y²) - (x³y - xy³)(2y)]/(x²+y²)². Putting y = 0, x ≠ 0 gives f_y(x,0) = x. Also, f_y(0,0) = 0. Hence f_yx(0,0) = lim h→0 [f_y(h,0)-f_y(0,0)]/h = lim h→0 h/h = 1.

Final answer: f_xy(0,0) = -1, f_yx(0,0) = 1. The mixed partials are unequal because f_xy and f_yx are not continuous at the origin, so the Schwarz (Clairaut) theorem does not apply. With the opposite convention f_xy = ∂/∂x(∂f/∂y), the two values simply swap.

(c)(i)

For A = [[1,2,0],[2,1,-6],[2,-2,3]], the characteristic equation is det(A - λI) = 0:

|1-λ, 2, 0; 2, 1-λ, -6; 2, -2, 3-λ| = 0. Expanding: (1-λ)[(1-λ)(3-λ)-12] - 2[2(3-λ)+12] = 0 => (1-λ)(λ²-4λ-9) - 2(18-2λ) = 0 => -λ³ + 5λ² + 9λ - 45 = 0 => -(λ-3)(λ-5)(λ+3) = 0. Thus the eigenvalues are λ = 3, 5, -3.

For λ = 3: A - 3I = [[-2,2,0],[2,-2,-6],[2,-2,0]]. The equations give x = y and z = 0. So an eigenvector is (1,1,0)^T.

For λ = 5: A - 5I = [[-4,2,0],[2,-4,-6],[2,-2,-2]]. The equations give y = 2x and z = -x. So an eigenvector is (1,2,-1)^T.

For λ = -3: A + 3I = [[4,2,0],[2,4,-6],[2,-2,6]]. The equations give y = -2x and z = -x. So an eigenvector is (1,-2,-1)^T.

Final answer: eigenvalues 3, 5, -3 with eigenvectors (1,1,0)^T, (1,2,-1)^T, (1,-2,-1)^T respectively.

(c)(ii)

P_2 is a subspace of P_4. We have dim P_4 = 5, dim P_2 = 3.

A natural basis of P_4 is {1, x, x², x³, x⁴}. Since P_2 = span{1, x, x²}, the quotient P_4/P_2 has cosets represented by the remaining basis elements: {x³ + P_2, x⁴ + P_2}.

These two cosets are linearly independent: if a(x³+P_2) + b(x⁴+P_2) = P_2, then ax³ + bx⁴ ∈ P_2. But P_2 contains no x³ or x⁴ term, so a = b = 0.

They also span P_4/P_2, because any polynomial in P_4 can be written as a0 + a1x + a2x² + a3x³ + a4x⁴, whose coset is a3(x³+P_2) + a4(x⁴+P_2).

Thus dim(P_4/P_2) = 2. Also, dim P_4 - dim P_2 = 5 - 3 = 2. Therefore, dim(P_4/P_2) = dim P_4 - dim P_2.