Linear Algebra

$$ \text{Problem: Find } (I + B)^{-1}, \ \text{where } B = (I + A)^{-1}(I - A), \ A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ -2 & 3 & 0 & 0 \\ 0 & -4 & 5 & 0 \\ 0 & 0 & -6 & 7 \end{bmatrix} \\\text{Step 1: Compute } I + A \ \text{and } I - A \\I + A = \begin{bmatrix} 2 & 0 & 0 & 0 \\ -2 & 4 & 0 & 0 \\ 0 & -4 & 6 & 0 \\ 0 & 0 & -6 & 8 \end{bmatrix}, \quad I - A = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 2 & -2 & 0 & 0 \\ 0 & 4 & -4 & 0 \\ 0 & 0 & 6 & -6 \end{bmatrix} \\\text{Step 2: Analyze } I + B \\I + B = I + (I + A)^{-1}(I - A) = (I + A)^{-1}(I + A) + (I + A)^{-1}(I - A) = (I + A)^{-1} [(I + A) + (I - A)] = 2 (I + A)^{-1} \\\text{Step 3: Compute } (I + B)^{-1} \\(I + B)^{-1} = [2 (I + A)^{-1}]^{-1} = \frac{1}{2} (I + A) \\\text{Step 4: Calculate the result} \\\frac{1}{2} (I + A) = \frac{1}{2} \begin{bmatrix} 2 & 0 & 0 & 0 \\ -2 & 4 & 0 & 0 \\ 0 & -4 & 6 & 0 \\ 0 & 0 & -6 & 8 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ -1 & 2 & 0 & 0 \\ 0 & -2 & 3 & 0 \\ 0 & 0 & -3 & 4 \end{bmatrix} \\\text{Final Answer: } (I + B)^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ -1 & 2 & 0 & 0 \\ 0 & -2 & 3 & 0 \\ 0 & 0 & -3 & 4 \end{bmatrix} $$

Calculus

Some rather tricky integrals

$$ \int x^2 \ln x \, dx\\ \text{解: } \text{使用分部积分法：} \text{令 } u = \ln x, \ dv = x^2 \, dx, \ \text{则 } du = \frac{1}{x} \, dx, \ v = \frac{x^3}{3}. \\I = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx \, \text{理论求导没有 0，但可以用表格积分法。}\\ = \frac{x^3}{3} \ln x - \frac{1}{3} \int x^2 \, dx = \frac{x^3}{3} \ln x - \frac{1}{3} \cdot \frac{x^3}{3} = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C \\ = \frac{x^3}{3} \left( \ln x - \frac{1}{3} \right) + C $$

$$ \text{How to Compute } \int \sec^3 \theta \, d\theta \\\text{1. Integration by Parts: } \sec^3 \theta = \sec \theta \cdot \sec^2 \theta, \ u = \sec \theta, \ du = \sec \theta \tan \theta \, d\theta, \ dv = \sec^2 \theta \, d\theta, \ v = \tan \theta, \\\int \sec^3 \theta \, d\theta = \sec \theta \tan \theta - \int \sec \theta \tan^2 \theta \, d\theta \\\text{2. Simplify: } \tan^2 \theta = \sec^2 \theta - 1, \ \int \sec \theta \tan^2 \theta \, d\theta = \int \sec^3 \theta \, d\theta - \int \sec \theta \, d\theta, \\\int \sec^3 \theta \, d\theta = \sec \theta \tan \theta - \int \sec^3 \theta \, d\theta + \int \sec \theta \, d\theta \\\text{3. Solve: } 2 \int \sec^3 \theta \, d\theta = \sec \theta \tan \theta + \int \sec \theta \, d\theta, \ \int \sec \theta \, d\theta = \ln |\sec \theta + \tan \theta|, \\\int \sec^3 \theta \, d\theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln |\sec \theta + \tan \theta| + C \\\text{Exam Tip: Memorize } \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln |\sec \theta + \tan \theta| + C. \ $$

$$ \text{求半圆的参数方程：} \quad x^2 + y^2 = a^2, \ y > 0, \ \text{参数 } t = \frac{dy}{dx} \text{（切线斜率）} \\ \text{解：从 } x^2 + y^2 = a^2 \text{ 求导得 } \frac{dy}{dx} = -\frac{x}{y}, \ \text{即 } t = -\frac{x}{y}, \ x = -t y \\ \text{代入得 } (-t y)^2 + y^2 = a^2, \ y^2 (t^2 + 1) = a^2, \ y = \frac{a}{\sqrt{t^2 + 1}} \ (y > 0), \ x = -\frac{a t}{\sqrt{t^2 + 1}} \\ \text{参数方程：} \ x = -\frac{a t}{\sqrt{t^2 + 1}}, \ y = \frac{a}{\sqrt{t^2 + 1}} $$

$$ \text{求参数曲线 } x = -\sqrt{t + 1}, \ y = \sqrt{3t} \text{ 在 } t = 3 \text{ 时的切线方程和二阶导数 } \frac{d^2 y}{dx^2} \\\text{解：} 1. \ \text{求点坐标：当 } t = 3, \ x = -\sqrt{3 + 1} = -\sqrt{4} = -2, \ y = \sqrt{3 \cdot 3} = \sqrt{9} = 3, \ \text{点为 } (-2, 3) \\2. \ \text{求切线斜率 } \frac{dy}{dx}：\frac{dx}{dt} = \frac{d}{dt} (-\sqrt{t + 1}) = -\frac{1}{2\sqrt{t + 1}}, \ \frac{dy}{dt} = \frac{d}{dt} (\sqrt{3t}) = \sqrt{3} \cdot \frac{1}{2\sqrt{t}} \\\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\sqrt{3} \cdot \frac{1}{2\sqrt{t}}}{-\frac{1}{2\sqrt{t + 1}}} = -\sqrt{3} \cdot \frac{\sqrt{t + 1}}{\sqrt{t}}, \ t = 3 \text{ 时 } \frac{dy}{dx} = -\sqrt{3} \cdot \frac{\sqrt{4}}{\sqrt{3}} = -\sqrt{3} \cdot \frac{2}{\sqrt{3}} = -2 \\3. \ \text{切线方程：点 } (-2, 3), \text{ 斜率 } m = -2, \ y - 3 = -2 (x + 2) \Rightarrow y - 3 = -2x - 4 \Rightarrow y = -2x - 1 \\4. \ \text{求二阶导数一定想清楚！ } \frac{d^2 y}{dx^2}： \frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( -\sqrt{3} \cdot \frac{\sqrt{t + 1}}{\sqrt{t}} \right) = -\sqrt{3} \cdot \frac{\frac{1}{2\sqrt{t + 1}} \cdot \sqrt{t} - \sqrt{t + 1} \cdot \frac{1}{2\sqrt{t}}}{t}, \ t = 3 $$

"Proof problems related to limits.”

$$ \text{Prove: } \lim_{t \to t_0} (\mathbf{r}1(t) \times \mathbf{r}2(t)) = A \times B, \ \text{given } \lim{t \to t_0} \mathbf{r}1(t) = A, \ \lim{t \to t_0} \mathbf{r}2(t) = B \\\text{Let } \mathbf{r}1(t) = (f_1(t), f_2(t), f_3(t)), \ \mathbf{r}2(t) = (g_1(t), g_2(t), g_3(t)), \ A = (a_1, a_2, a_3), \ B = (b_1, b_2, b_3) \\\mathbf{r}1(t) \times \mathbf{r}2(t) = (f_2(t)g_3(t) - f_3(t)g_2(t)) \mathbf{i} + (f_3(t)g_1(t) - f_1(t)g_3(t)) \mathbf{j} + (f_1(t)g_2(t) - f_2(t)g_1(t)) \mathbf{k} \\\lim{t \to t_0} (\mathbf{r}1(t) \times \mathbf{r}2(t)) = \lim{t \to t_0} (f_2(t)g_3(t) - f_3(t)g_2(t)) \mathbf{i} + \lim{t \to t_0} (f_3(t)g_1(t) - f_1(t)g_3(t)) \mathbf{j} + \lim{t \to t_0} (f_1(t)g_2(t) - f_2(t)g_1(t)) \mathbf{k} \\\text{By the Limit Product Rule: } \lim{t \to t_0} f_i(t) = a_i, \ \lim{t \to t_0} g_i(t) = b_i, \\\text{so } \lim{t \to t_0} (f_2(t)g_3(t) - f_3(t)g_2(t)) = a_2 b_3 - a_3 b_2, \ \lim{t \to t_0} (f_3(t)g_1(t) - f_1(t)g_3(t)) = a_3 b_1 - a_1 b_3, \ \lim_{t \to t_0} (f_1(t)g_2(t) - f_2(t)g_1(t)) = a_1 b_2 - a_2 b_1 \\\text{Thus, } \lim_{t \to t_0} (\mathbf{r}_1(t) \times \mathbf{r}_2(t)) = (a_2 b_3 - a_3 b_2) \mathbf{i} + (a_3 b_1 - a_1 b_3) \mathbf{j} + (a_1 b_2 - a_2 b_1) \mathbf{k} = A \times B $$

\text{Problem: Find } \mathbf{r}(t) \text{ of a particle starting at } (1, -1, 2) \text{ moving toward } (3, 0, 3), \ \text{with } \mathbf{a} = 2\mathbf{i} + \mathbf{j} + \mathbf{k} \text{ and initial speed 2 at } t = 0. \\ \text{Flawed Reasoning: Assumed } \mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2} \mathbf{a} t^2, \ \mathbf{r}_0 = (1, -1, 2), \ \mathbf{v}_0 = 2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}, \\ \text{so } \mathbf{r}(t) = (1 + 2t + t^2)\mathbf{i} + (-1 + 2t + \frac{1}{2}t^2)\mathbf{j} + (2 + 2t + \frac{1}{2}t^2)\mathbf{k}. \\ \text{Why Wrong: Speed } |\mathbf{v}_0| = \sqrt{12} \approx 3.46 \neq 2, \ \text{and } \mathbf{v}_0 = 2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \text{ not parallel to direction } (2, 1, 1). \\ \text{Correct Approach: } \mathbf{v}(t) = (2t + c_1)\mathbf{i} + (t + c_2)\mathbf{j} + (t + c_3)\mathbf{k}, \ |\mathbf{v}(0)| = 2, \ \text{direction } (2, 1, 1), \\ (c_1, c_2, c_3) = k(2, 1, 1), \ \sqrt{6k^2} = 2, \ k = \frac{\sqrt{6}}{3}, \ \mathbf{v}(0) = \frac{2\sqrt{6}}{3}\mathbf{i} + \frac{\sqrt{6}}{3}\mathbf{j} + \frac{\sqrt{6}}{3}\mathbf{k}, \\ \mathbf{r}(t) = \left(t^2 + \frac{2\sqrt{6}}{3}t + 1\right)\mathbf{i} + \left(\frac{1}{2}t^2 + \frac{\sqrt{6}}{3}t - 1\right)\mathbf{j} + \left(\frac{1}{2}t^2 + \frac{\sqrt{6}}{3}t + 2\right)\mathbf{k}, \\ \text{Takeaway: Speed is magnitude; velocity aligns with } (2, 1, 1); \ \text{integrate with conditions.}

Parametric equations and polar coordinate equations

$$ \text{Find the polar equation for } x = e^{2t} \cos t, \ y = e^{2t} \sin t, \ -\infty < t < \infty \\\text{Step 1: Polar Coordinates: } x = r \cos \theta, \ y = r \sin \theta, \ r = \sqrt{x^2 + y^2}, \ \theta = \tan^{-1} \left( \frac{y}{x} \right) \\\begin{aligned}& \text{Step 2: Compute } r: \\& r = \sqrt{(e^{2t} \cos t)^2 + (e^{2t} \sin t)^2} = \sqrt{e^{4t} (\cos^2 t + \sin^2 t)} = e^{2t} \quad (r \geq 0) \\& \text{Step 3: Compute } \theta: \\& \tan \theta = \frac{e^{2t} \sin t}{e^{2t} \cos t} = \tan t, \ \theta = t \\& \text{Step 4: Eliminate } t: \\& r = e^{2\theta} \\& \text{Step 5: Verify: } x = e^{2\theta} \cos \theta, \ y = e^{2\theta} \sin \theta, \ \text{set } \theta = t, \ \text{matches } x = e^{2t} \cos t, \ y = e^{2t} \sin t \\& \text{Step 6: Range: } t \in (-\infty, \infty), \ \theta \in (-\infty, \infty), \ r = e^{2\theta} \text{ is a logarithmic spiral.} \\& \text{Final Answer: } r = e^{2\theta}\end{aligned} $$