Solution:

Let A and B be two similar matrices of order n \times n. By definition of similarity, there exists an invertible matrix P such that:

B = P^{-1}AP

Let m_A(x) and m_B(x) be the minimal polynomials of A and B respectively. First, let us evaluate any polynomial f(x) = c_k x^k + \dots + c_1 x + c_0 at matrix B:

f(B) = f(P^{-1}AP) = P^{-1} f(A) P

Now, substituting the minimal polynomial of A, m_A(x), into B:

m_A(B) = P^{-1} m_A(A) P = P^{-1} (0) P = 0

Since m_A(B) = 0, by the definition of a minimal polynomial, the minimal polynomial of B must divide m_A(x), i.e., m_B(x) \mid m_A(x).

Conversely, since B = P^{-1}AP \implies A = PBP^{-1}, an identical argument yields that substituting m_B(x) into A results in m_B(A) = 0, which implies m_A(x) \mid m_B(x).

Since both are monic polynomials and divide each other, they must be equal:

m_A(x) = m_B(x)

Solution:

The given curve is of degree n = 3. Let us group the terms by degree to find the oblique asymptotes y = mx + c.

Put x = 1 and y = m in the highest degree terms (degree 3):

\phi_3(m) = 1 + m + m^2 + m^3 = (1+m)(1+m^2)

To find the slopes m, set \phi_3(m) = 0 \implies (1+m)(1+m^2) = 0. The real root is m = -1.

Next, extract the degree 2 terms by putting x = 1 and y = m:

\phi_2(m) = 2 + 3m - 4m^2

We need the derivative of \phi_3(m):

\phi_3'(m) = 1 + 2m + 3m^2

The intercept c for a given slope m is calculated using the formula c = -\frac{\phi_{n-1}(m)}{\phi_n'(m)}:

c = -\frac{\phi_2(-1)}{\phi_3'(-1)} = -\frac{2 + 3(-1) - 4(-1)^2}{1 + 2(-1) + 3(-1)^2} = -\frac{2 - 3 - 4}{1 - 2 + 3} = -\frac{-5}{2} = \frac{5}{2}

Thus, substituting m = -1 and c = \frac{5}{2} into y = mx + c, we get the oblique asymptote:

y = -x + \frac{5}{2} \implies 2x + 2y - 5 = 0

Since the curve is of degree 3 and has no other real roots for \phi_3(m)=0, and there are no parallel asymptotes to the axes (as the coefficients of x^3 and y^3 are constants), 2x + 2y - 5 = 0 is the only real asymptote.

Solution:

Let y' = p. The given differential equation is in Clairaut’s form:

y = xp + p^2

Differentiating both sides with respect to x:

\frac{dy}{dx} = p + x\frac{dp}{dx} + 2p\frac{dp}{dx}

Since \frac{dy}{dx} = p, we substitute and simplify:

p = p + (x + 2p)\frac{dp}{dx} \implies (x + 2p)\frac{dp}{dx} = 0

This gives two possible cases:

Case 1: \frac{dp}{dx} = 0 \implies p = C, where C is an arbitrary constant. Substituting p = C back into the original equation yields the general solution:

y = Cx + C^2

Case 2: x + 2p = 0 \implies p = -\frac{x}{2}. Substituting this into the differential equation gives the singular solution y = x\left(-\frac{x}{2}\right) + \left(-\frac{x}{2}\right)^2 = -\frac{x^2}{4}.

The general solution required is y = Cx + C^2.

Solution:

First, simplify the expression for the vector field \vec{V}:

\vec{V} = (xy^2z^2)\mathbf{i} + (x^2yz^2)\mathbf{j} + (x^2y^2z)\mathbf{k}

A vector field is conservative if and only if its curl is equal to the zero vector (\nabla \times \vec{V} = \vec{0}). Let’s calculate the curl:

\nabla \times \vec{V} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ xy^2z^2 & x^2yz^2 & x^2y^2z \end{vmatrix}

Evaluating component-by-component:

\mathbf{i} \text{ component:} \quad \frac{\partial}{\partial y}(x^2y^2z) - \frac{\partial}{\partial z}(x^2yz^2) = 2x^2yz - 2x^2yz = 0 \mathbf{j} \text{ component:} \quad \frac{\partial}{\partial z}(xy^2z^2) - \frac{\partial}{\partial x}(x^2y^2z) = 2xy^2z - 2xy^2z = 0 \mathbf{k} \text{ component:} \quad \frac{\partial}{\partial x}(x^2yz^2) - \frac{\partial}{\partial y}(xy^2z^2) = 2xyz^2 - 2xyz^2 = 0

Since \nabla \times \vec{V} = 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} = \vec{0}, the vector field \vec{V} is conservative.

Solution:

Let V be an inner product space with inner product \langle \cdot, \cdot \rangle. Define a norm induced by the inner product as \|x\| = \sqrt{\langle x, x \rangle}. We verify the norm properties:

1. \|x\| \ge 0 and \|x\| = 0 \iff \langle x, x \rangle = 0 \iff x = 0.
2. \|\alpha x\| = \sqrt{\langle \alpha x, \alpha x \rangle} = \sqrt{\alpha \bar{\alpha} \langle x, x \rangle} = |\alpha| \|x\|.
3. Triangle inequality: \|x+y\|^2 = \langle x+y, x+y \rangle = \|x\|^2 + 2\text{Re}\langle x, y \rangle + \|y\|^2. By Cauchy-Schwarz Inequality (|\langle x, y \rangle| \le \|x\|\|y\|), we get \|x+y\|^2 \le \|x\|^2 + 2\|x\|\|y\| + \|y\|^2 = (\|x\| + \|y\|)^2 \implies \|x+y\| \le \|x\| + \|y\|.

Thus, every inner product space is a normed vector space.

Converse is not true: A normed space is induced by an inner product if and only if it satisfies the Parallelogram Law:

\|x+y\|^2 + \|x-y\|^2 = 2(\|x\|^2 + \|y\|^2)

Consider the space \mathbb{R}^2 equipped with the supremum norm \|x\|_\infty = \max(|x_1|, |x_2|). Let x = (1, 0) and y = (0, 1). Then:

\|x\|_\infty = 1, \quad \|y\|_\infty = 1 \implies 2(\|x\|_\infty^2 + \|y\|_\infty^2) = 2(1 + 1) = 4

Now, x+y = (1, 1) and x-y = (1, -1):

\|x+y\|_\infty = 1, \quad \|x-y\|_\infty = 1 \implies \|x+y\|_\infty^2 + \|x-y\|_\infty^2 = 1 + 1 = 2

Since 2 \neq 4, the parallelogram law fails. Hence, not all normed vector spaces can form an inner product space.

Solution:

We want to optimize the squared distance from the origin f(x, y, z) = x^2 + y^2 + z^2 subject to two constraints:

g(x, y, z) = x^2 + y^2 - 1 = 0 h(x, y, z) = x + y + z - 1 = 0

Using Lagrange multipliers, set \nabla f = \lambda \nabla g + \mu \nabla h:

2x = 2\lambda x + \mu \implies (2 - 2\lambda)x = \mu 2y = 2\lambda y + \mu \implies (2 - 2\lambda)y = \mu 2z = \mu

Equating the equations for x and y, we get (2 - 2\lambda)x = (2 - 2\lambda)y. This gives two possibilities:

Case 1: x = y
From g(x,y,z) = 0 \implies x^2 + x^2 = 1 \implies 2x^2 = 1 \implies x = \pm \frac{1}{\sqrt{2}}.
– If x = y = \frac{1}{\sqrt{2}}, then from h(x,y,z) = 0: z = 1 - \frac{2}{\sqrt{2}} = 1 - \sqrt{2}. Point: \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 1-\sqrt{2}\right).
– If x = y = -\frac{1}{\sqrt{2}}, then z = 1 - \left(-\frac{2}{\sqrt{2}}\right) = 1 + \sqrt{2}. Point: \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 1+\sqrt{2}\right).

Case 2: 2 - 2\lambda = 0 \implies \lambda = 1
If \lambda = 1, then \mu = 0 \implies z = 0.
From the constraint equations: x + y = 1 and x^2 + y^2 = 1 \implies x^2 + (1-x)^2 = 1 \implies 2x^2 - 2x = 0 \implies x = 0 or x = 1.
This gives points (1, 0, 0) and (0, 1, 0).

Let’s calculate f = x^2 + y^2 + z^2 = 1 + z^2 (since x^2+y^2=1):

• For (1,0,0) and (0,1,0): f = 1 + 0 = 1 (Minimum distance)
• For \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 1-\sqrt{2}\right): f = 1 + (1-\sqrt{2})^2 = 1 + 1 + 2 - 2\sqrt{2} = 4 - 2\sqrt{2} \approx 1.17
• For \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 1+\sqrt{2}\right): f = 1 + (1+\sqrt{2})^2 = 4 + 2\sqrt{2} \approx 6.83 (Maximum distance)

Hence, the closest points are (1, 0, 0) and (0, 1, 0), and the farthest point is \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 1+\sqrt{2}\right).

Solution:

The volume is enclosed by two intersecting cylinders of equal radius a (known as Steinmetz solid). By symmetry, we can calculate the volume in the first octant and multiply it by 8.

In the first octant, x, y, z range from 0 to a. The boundaries give z = \sqrt{a^2 - x^2} and the base region in the xy-plane is limited by y = \sqrt{a^2 - x^2}.

V = 8 \int_{0}^{a} \int_{0}^{\sqrt{a^2-x^2}} z \, dy \, dx = 8 \int_{0}^{a} \int_{0}^{\sqrt{a^2-x^2}} \sqrt{a^2-x^2} \, dy \, dx

Integrating with respect to y:

V = 8 \int_{0}^{a} \left[ \sqrt{a^2-x^2} \cdot y \right]_{0}^{\sqrt{a^2-x^2}} dx = 8 \int_{0}^{a} (a^2 - x^2) \, dx

Integrating with respect to x:

V = 8 \left[ a^2x - \frac{x^3}{3} \right]_{0}^{a} = 8 \left( a^3 - \frac{a^3}{3} \right) = 8 \cdot \frac{2a^3}{3} = \frac{16a^3}{3}

Thus, the total volume enclosed between the surfaces is \frac{16a^3}{3}.

Solution:

The equation of any plane passing through the line of intersection of the two given planes can be written as:

(a_1x + b_1y + c_1z + d_1) + \lambda(a_2x + b_2y + c_2z + d_2) = 0

Rearranging the terms to standard form Ax + By + Cz + D = 0:

(a_1 + \lambda a_2)x + (b_1 + \lambda b_2)y + (c_1 + \lambda c_2)z + (d_1 + \lambda d_2) = 0

The normal vector to this plane is \vec{n_1} = (a_1 + \lambda a_2)\mathbf{i} + (b_1 + \lambda b_2)\mathbf{j} + (c_1 + \lambda c_2)\mathbf{k}.

The given plane is perpendicular to the xy-plane. The equation of the xy-plane is z = 0, and its normal vector is \vec{n_2} = 0\mathbf{i} + 0\mathbf{j} + 1\mathbf{k} = \mathbf{k}.

Since the two planes are perpendicular, their normal vectors must be orthogonal (\vec{n_1} \cdot \vec{n_2} = 0):

(a_1 + \lambda a_2)(0) + (b_1 + \lambda b_2)(0) + (c_1 + \lambda c_2)(1) = 0 \implies c_1 + \lambda c_2 = 0 \implies \lambda = -\frac{c_1}{c_2}

Substituting \lambda = -\frac{c_1}{c_2} back into our family-of-planes equation:

(a_1x + b_1y + c_1z + d_1) - \frac{c_1}{c_2}(a_2x + b_2y + c_2z + d_2) = 0

Multiplying through by c_2 clears the fraction:

(a_1c_2 - a_2c_1)x + (b_1c_2 - b_2c_1)y + (d_1c_2 - d_2c_1) = 0

Solution:

This is proved constructive using the Gram-Schmidt Orthogonalization Process. Let V be an n-dimensional inner product space and B = \{v_1, v_2, \dots, v_n\} be any basis of V. We construct an orthogonal basis U = \{u_1, u_2, \dots, u_n\} as follows:

Step 1: Set u_1 = v_1.

Step 2: Set u_2 = v_2 - \frac{\langle v_2, u_1 \rangle}{\|u_1\|^2} u_1.

Step k: Generally, for k = 2, \dots, n, compute:

u_k = v_k - \sum_{j=1}^{k-1} \frac{\langle v_k, u_j \rangle}{\|u_j\|^2} u_j

By construction, each u_k is orthogonal to all previous vectors u_1, \dots, u_{k-1}. Since a set of mutually orthogonal non-zero vectors is linearly independent, U = \{u_1, \dots, u_n\} forms an orthogonal basis of V.

To convert this into an orthonormal basis E = \{e_1, e_2, \dots, e_n\}, we normalize each vector by dividing by its norm:

e_i = \frac{u_i}{\|u_i\|} \quad \text{for } i = 1, 2, \dots, n

The resulting set E satisfies \langle e_i, e_j \rangle = \delta_{ij} (where \delta_{ij}=1 if i=j and 0 otherwise), forming an orthonormal basis.

Solution:

First, find the characteristic equation by computing \det(A - \lambda I) = 0:

\begin{vmatrix} -\lambda & 0 & -2 \\ 1 & 2-\lambda & 1 \\ 1 & 0 & 3-\lambda \end{vmatrix} = (2-\lambda)[-\lambda(3-\lambda) + 2] = (2-\lambda)(\lambda^2 - 3\lambda + 2) = 0 (2-\lambda)(\lambda-1)(\lambda-2) = 0 \implies \lambda = 1, 2, 2

Now find the eigenvectors:

For \lambda = 1: (A - I)X = 0 \implies \begin{pmatrix} -1 & 0 & -2 \\ 1 & 1 & 1 \\ 1 & 0 & 2 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix} \implies x = -2z, y = z. Eigenvector X_1 = \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}.

For \lambda = 2: (A - 2I)X = 0 \implies \begin{pmatrix} -2 & 0 & -2 \\ 1 & 0 & 1 \\ 1 & 0 & 1 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix} \implies x + z = 0. We obtain two linearly independent eigenvectors: X_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} and X_3 = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}.

Construct the modal matrix P and its diagonal matrix D:

P = \begin{pmatrix} -2 & 0 & -1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}

Find P^{-1} via row operations or formula: P^{-1} = \begin{pmatrix} -1 & 0 & -1 \\ 1 & 1 & 1 \\ 1 & 0 & 2 \end{pmatrix}. Since A = PDP^{-1}, we have A^5 = P D^5 P^{-1}:

D^5 = \begin{pmatrix} 1^5 & 0 & 0 \\ 0 & 2^5 & 0 \\ 0 & 0 & 2^5 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 32 & 0 \\ 0 & 0 & 32 \end{pmatrix} A^5 = \begin{pmatrix} -2 & 0 & -1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 32 & 0 \\ 0 & 0 & 32 \end{pmatrix} \begin{pmatrix} -1 & 0 & -1 \\ 1 & 1 & 1 \\ 1 & 0 & 2 \end{pmatrix} = \begin{pmatrix} -62 & 0 & -62 \\ 31 & 32 & 31 \\ 31 & 0 & 63 \end{pmatrix}

Solution:

This is a Cauchy-Euler homogeneous linear differential equation. Let x = e^z \implies z = \log x. Let D = \frac{d}{dz}. Then:

x\frac{dy}{dx} = Dy, \quad x^2\frac{d^2y}{dx^2} = D(D-1)y

Substituting these substitutions into the given equation yields:

[D(D-1) - 2D + 2]y = e^z + z e^{2z} + e^{3z} \implies (D^2 - 3D + 2)y = e^z + z e^{2z} + e^{3z}

1. Complementary Function (CF):
The auxiliary equation is m^2 - 3m + 2 = 0 \implies (m-1)(m-2) = 0 \implies m = 1, 2.

y_{cf} = C_1 e^z + C_2 e^{2z} = C_1 x + C_2 x^2

2. Particular Integral (PI):

y_{pi} = \frac{1}{(D-1)(D-2)} [e^z + z e^{2z} + e^{3z}] = P_1 + P_2 + P_3 P_1 = \frac{1}{(D-1)(D-2)}e^z = \frac{z}{1!} \frac{1}{1-2}e^z = -z e^z = -x\log x P_2 = \frac{1}{(D-1)(D-2)}(z e^{2z}) = e^{2z} \frac{1}{(D+2-1)(D+2-2)}z = e^{2z}\frac{1}{D(D+1)}z = e^{2z}\frac{1}{D}(1-D+\dots)z = e^{2z}\left(\frac{z^2}{2}-z\right) = x^2\left(\frac{(\log x)^2}{2} - \log x\right) P_3 = \frac{1}{(3-1)(3-2)}e^{3z} = \frac{1}{2}e^{3z} = \frac{x^3}{2}

Combining the terms, the general complete solution is y = y_{cf} + y_{pi}:

y = C_1 x + C_2 x^2 - x\log x + x^2\left(\frac{(\log x)^2}{2} - \log x\right) + \frac{x^3}{2}

Solution:

Since x=0 is an ordinary point of the differential equation, we assume a power series solution of the form:

y = \sum_{n=0}^{\infty} a_n x^n \implies \frac{dy}{dx} = \sum_{n=1}^{\infty} n a_n x^{n-1} \implies \frac{d^2y}{dx^2} = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}

Substitute these expressions into the differential equation:

\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - \sum_{n=2}^{\infty} n(n-1) a_n x^n - 4\sum_{n=1}^{\infty} n a_n x^n + 2\sum_{n=0}^{\infty} a_n x^n = 0

Shifting indices to align terms to x^n:

\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^{\infty} n(n-1) a_n x^n - \sum_{n=0}^{\infty} 4n a_n x^n + \sum_{n=0}^{\infty} 2 a_n x^n = 0

Equating the coefficient of x^n to zero yields the recurrence relation:

(n+2)(n+1) a_{n+2} - [n(n-1) + 4n - 2] a_n = 0 a_{n+2} = \frac{n^2 + 3n - 2}{(n+2)(n+1)} a_n

Let’s find the coefficients for consecutive values of n:
– For n=0: a_2 = \frac{-2}{2\cdot 1} a_0 = -a_0
– For n=1: a_3 = \frac{1+3-2}{3\cdot 2} a_1 = \frac{2}{6} a_1 = \frac{1}{3} a_1
– For n=2: a_4 = \frac{4+6-2}{4\cdot 3} a_2 = \frac{8}{12}(-a_0) = -\frac{2}{3} a_0
– For n=3: a_5 = \frac{9+9-2}{5\cdot 4} a_3 = \frac{16}{20}\left(\frac{1}{3} a_1\right) = \frac{4}{15} a_1

Thus, the general power series expansion solution is:

y = a_0 \left(1 - x^2 - \frac{2}{3}x^4 - \dots\right) + a_1 \left(x + \frac{1}{3}x^3 + \frac{4}{15}x^5 + \dots\right)

Solution:

By Green’s Theorem in a plane, the line integral for work done along a closed contour \oint_C (M\,dx + N\,dy) equals \iint_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) dA. Here:

M = x^2 - y^2 \implies \frac{\partial M}{\partial y} = -2y N = x + y \implies \frac{\partial N}{\partial x} = 1 \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 1 - (-2y) = 1 + 2y

The region R is a sector bounded by the circular arc r = 4 between the lines y = x (which corresponds to \theta = \pi/4) and y = -x (which corresponds to \theta = -\pi/4). Converting to polar coordinates (x = r\cos\theta, y = r\sin\theta, dA = r\,dr\,d\theta):

W = \int_{-\pi/4}^{\pi/4} \int_{0}^{4} (1 + 2r\sin\theta) \, r \, dr \, d\theta = \int_{-\pi/4}^{\pi/4} \int_{0}^{4} (r + 2r^2\sin\theta) \, dr \, d\theta

Evaluating the inner integral with respect to r:

\int_{0}^{4} (r + 2r^2\sin\theta) \, dr = \left[ \frac{r^2}{2} + \frac{2r^3}{3}\sin\theta \right]_{0}^{4} = 8 + \frac{128}{3}\sin\theta

Now evaluate the outer integral with respect to \theta:

W = \int_{-\pi/4}^{\pi/4} \left(8 + \frac{128}{3}\sin\theta\right) d\theta = \left[ 8\theta - \frac{128}{3}\cos\theta \right]_{-\pi/4}^{\pi/4}

Since \cos(-\pi/4) = \cos(\pi/4), the cosine terms cancel out entirely:

W = 8\left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right) = 8\left(\frac{\pi}{2}\right) = 4\pi

Solution:

The cylinder equation is given by f(x,y,z) = x^2 + y^2 = 36. The outward unit normal vector \mathbf{n} is:

\mathbf{n} = \frac{\nabla f}{|\nabla f|} = \frac{2x\mathbf{i} + 2y\mathbf{j}}{\sqrt{4x^2 + 4y^2}} = \frac{2x\mathbf{i} + 2y\mathbf{j}}{2(6)} = \frac{x\mathbf{i} + y\mathbf{j}}{6}

Now calculate the dot product \vec{F} \cdot \mathbf{n}:

\vec{F} \cdot \mathbf{n} = (z^2\mathbf{i} + xy\mathbf{j} - y^2\mathbf{k}) \cdot \left(\frac{x\mathbf{i} + y\mathbf{j}}{6}\right) = \frac{xz^2 + xy^2}{6}

Using cylindrical coordinates: x = 6\cos\theta, y = 6\sin\theta. The differential area element for a cylinder of radius R=6 is dA = 6\,dz\,d\theta. In the first quadrant, \theta ranges from 0 to \pi/2.

Substitute these parametric values into the integral expression:

\vec{F} \cdot \mathbf{n} = \frac{(6\cos\theta)z^2 + (6\cos\theta)(36\sin^2\theta)}{6} = z^2\cos\theta + 36\cos\theta\sin^2\theta \iint_{S} \vec{F} \cdot \mathbf{n} \,dA = \int_{0}^{\pi/2} \int_{0}^{4} (z^2\cos\theta + 36\cos\theta\sin^2\theta) \cdot 6 \, dz \, d\theta

First integrate with respect to z:

6 \int_{0}^{4} (z^2\cos\theta + 36\cos\theta\sin^2\theta) \, dz = 6 \left[ \frac{z^3}{3}\cos\theta + 36z\cos\theta\sin^2\theta \right]_{0}^{4} = 6\left(\frac{64}{3}\cos\theta + 144\cos\theta\sin^2\theta\right) = 128\cos\theta + 864\cos\theta\sin^2\theta

Now integrate with respect to \theta:

\int_{0}^{\pi/2} (128\cos\theta + 864\cos\theta\sin^2\theta) \, d\theta = \left[ 128\sin\theta + 864\frac{\sin^3\theta}{3} \right]_{0}^{\pi/2} = 128(1) + 288(1) = 416

Solution:

The central attraction per unit mass is F = \frac{\mu}{r^7} \implies F = \mu u^7 where u = \frac{1}{r}. The angular momentum constant h is given by h = v_0 \cdot p = v_0 \cdot r_0\sin\alpha:

h = \sqrt{\frac{\mu}{3a^6}} \cdot a \cdot \sin(30^\circ) = \sqrt{\frac{\mu}{3a^4}} \cdot \frac{1}{2} \implies h^2 = \frac{\mu}{12a^4}

The differential equation of a central orbit in pedal form is:

h^2 u^2 \left( \frac{d^2u}{d\theta^2} + u \right) = F \implies \frac{\mu}{12a^4} u^2 \left( \frac{d^2u}{d\theta^2} + u \right) = \mu u^7 \implies \frac{d^2u}{d\theta^2} + u = 12a^4 u^5

Multiplying by 2\frac{du}{d\theta} and integrating gives the energy velocity equation:

\left(\frac{du}{d\theta}\right)^2 + u^2 = 4a^4 u^6 + C

At the point of projection, r = a \implies u = \frac{1}{a}, and \frac{du}{d\theta} = -u\cot\alpha = -\frac{1}{a}\cot(30^\circ) = -\frac{\sqrt{3}}{a}:

\left(-\frac{\sqrt{3}}{a}\right)^2 + \left(\frac{1}{a}\right)^2 = 4a^4\left(\frac{1}{a^6}\right) + C \implies \frac{3}{a^2} + \frac{1}{a^2} = \frac{4}{a^2} + C \implies C = 0

Thus, our differential equation simplifies directly to:

\left(\frac{du}{d\theta}\right)^2 = 4a^4 u^6 - u^2 = u^2(4a^4 u^4 - 1) \implies \frac{1}{u\sqrt{4a^4u^4 - 1}} du = d\theta

Let u^2 = \frac{1}{2a^2}\sec\phi. Substituting this transformation variables integrates beautifully to yield the explicit polar curve:

r^2 = 2a^2\cos(2\theta)

Solution:

Let the forces be represented as vectors \vec{P} and \vec{Q}. The initial resultant vector \vec{R} is defined by:

\vec{R} = \vec{P} + \vec{Q}

When force P is reversed, it becomes -\vec{P}. The new resultant vector \vec{R'} is:

\vec{R'} = -\vec{P} + \vec{Q} = \vec{Q} - \vec{P}

We are given that the two resultants \vec{R} and \vec{R'} are perpendicular to each other. Therefore, their vector dot product must equal zero:

\vec{R} \cdot \vec{R'} = 0

Substitute the expressions for \vec{R} and \vec{R'}:

(\vec{Q} + \vec{P}) \cdot (\vec{Q} - \vec{P}) = 0

Expanding the dot product using the distributive property:

\vec{Q} \cdot \vec{Q} - \vec{Q} \cdot \vec{P} + \vec{P} \cdot \vec{Q} - \vec{P} \cdot \vec{P} = 0

Since the dot product is commutative (\vec{P} \cdot \vec{Q} = \vec{Q} \cdot \vec{P}), the middle terms cancel out:

\|\vec{Q}\|^2 - \|\vec{P}\|^2 = 0 \implies Q^2 - P^2 = 0 \implies P^2 = Q^2

Taking the positive square root since magnitudes are non-negative scalars:

P = Q

Solution:

Let V be a real vector space with \dim(V) = n. Let B = \{v_1, v_2, \dots, v_n\} be an ordered basis for V. Any vector v \in V can be uniquely expressed as a linear combination:

v = c_1v_1 + c_2v_2 + \dots + c_nv_n \quad \text{where } c_i \in \mathbb{R}

Define a coordinate mapping function T: V \to \mathbb{R}^n by:

T(v) = (c_1, c_2, \dots, c_n)

We prove that T is an isomorphism by verifying linearity, injectivity, and surjectivity:

1. Linearity: Let u = \sum a_iv_i and v = \sum b_iv_i. For scalars \alpha, \beta \in \mathbb{R}:

T(\alpha u + \beta v) = T\left(\sum (\alpha a_i + \beta b_i)v_i\right) = (\alpha a_1 + \beta b_1, \dots, \alpha a_n + \beta b_n) = \alpha T(u) + \beta T(v)

2. Injectivity (One-to-One): Suppose T(u) = T(v). Then (a_1, \dots, a_n) = (b_1, \dots, b_n) \implies a_i = b_i for all i, which implies u = v. Alternatively, \text{Ker}(T) = \{0\}.

3. Surjectivity (Onto): For any arbitrary coordinate vector (k_1, \dots, k_n) \in \mathbb{R}^n, there clearly exists a vector v = \sum k_iv_i \in V such that T(v) = (k_1, \dots, k_n).

Since T is a bijective linear transformation, V \cong \mathbb{R}^n.

Solution:

Let the initial amplitude be A, and the new amplitude be A' = 2A. We evaluate the impacts on the system variables based on standard physics relationships:

1. Time Period (T):
The time period of a simple harmonic oscillator depends solely on the mass and stiffness constant, given by the formula:

T = 2\pi\sqrt{\frac{m}{k}}

Since this formula is completely independent of displacement scale, the time period remains unchanged.

2. Total Energy (E):
The total mechanical energy stored within a harmonic oscillator is directly proportional to the square of its maximum deflection amplitude:

E = \frac{1}{2}kA^2

Substituting A' = 2A into the energy expression:

E' = \frac{1}{2}k(2A)^2 = 4 \left(\frac{1}{2}kA^2\right) = 4E

Hence, the total energy quadruples (increases by a factor of 4).

3. Maximum Velocity (v_{\max}):
The velocity peaks as the object crosses equilibrium, calculated using angular dynamic frequency \omega:

v_{\max} = \omega A

Substituting the doubled value A' = 2A:

v_{\max}' = \omega (2A) = 2v_{\max}

Thus, the maximum velocity is doubled.