1. **Problem:** Determine the convergence or divergence of the integrals: (a) $$\int_1^{+\infty} \frac{2 + \sin(x^2 + x - 1)}{x - \sqrt{x}} \, dx$$ (b) $$\int_1^{+\infty} \frac{1}{\sqrt{x^{4/3} - e^{-x}}} \, dx$$ **Step 1:** Analyze the behavior of the integrands as $x \to +\infty$. (a) For large $x$, $x - \sqrt{x} \sim x$ since $\sqrt{x}$ grows slower than $x$. The numerator $2 + \sin(x^2 + x - 1)$ oscillates between $1$ and $3$. So the integrand behaves roughly like $\frac{\text{bounded}}{x}$. Since $\int_1^{+\infty} \frac{1}{x} dx$ diverges (harmonic integral), the integral (a) diverges by comparison. (b) For large $x$, $e^{-x} \to 0$, so $x^{4/3} - e^{-x} \sim x^{4/3}$. Thus the integrand behaves like $\frac{1}{\sqrt{x^{4/3}}} = \frac{1}{x^{2/3}}$. Since $\int_1^{+\infty} x^{-2/3} dx = \int_1^{+\infty} x^{-p} dx$ with $p=2/3 < 1$, this integral diverges. **Answer:** Both integrals diverge. 2. **Problem:** Parametrized curve $x=2\ln t$, $y=t + \frac{1}{t}$, $t>0$. (a) Find $y=f(x)$. **Step 1:** Express $t$ in terms of $x$: $$x = 2 \ln t \implies \ln t = \frac{x}{2} \implies t = e^{x/2}.$$ **Step 2:** Substitute into $y$: $$y = e^{x/2} + e^{-x/2} = 2 \cosh\left(\frac{x}{2}\right).$$ So, $$f(x) = 2 \cosh\left(\frac{x}{2}\right).$$ (b) Find tangent line at $t=2$. **Step 1:** Compute $x(2)$ and $y(2)$: $$x(2) = 2 \ln 2, \quad y(2) = 2 + \frac{1}{2} = \frac{5}{2}.$$ **Step 2:** Compute derivatives: $$\frac{dx}{dt} = \frac{2}{t}, \quad \frac{dy}{dt} = 1 - \frac{1}{t^2}.$$ At $t=2$: $$\frac{dx}{dt} = 1, \quad \frac{dy}{dt} = 1 - \frac{1}{4} = \frac{3}{4}.$$ **Step 3:** Slope of tangent line: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3/4}{1} = \frac{3}{4}.$$ **Step 4:** Equation of tangent line: $$y - \frac{5}{2} = \frac{3}{4} (x - 2 \ln 2).$$ (c) Find $t$ for horizontal tangent line. **Step 1:** Horizontal tangent means $\frac{dy}{dx} = 0$. Since $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1 - \frac{1}{t^2}}{\frac{2}{t}} = \frac{t^2 - 1}{2t^2}.$$ Set numerator zero: $$t^2 - 1 = 0 \implies t = 1 \quad (t>0).$$ (d) Compute length from $t=1$ to $t=2$. **Step 1:** Arc length formula: $$L = \int_1^2 \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt.$$ **Step 2:** Compute derivatives: $$\frac{dx}{dt} = \frac{2}{t}, \quad \frac{dy}{dt} = 1 - \frac{1}{t^2}.$$ **Step 3:** Compute integrand: $$\sqrt{\left(\frac{2}{t}\right)^2 + \left(1 - \frac{1}{t^2}\right)^2} = \sqrt{\frac{4}{t^2} + 1 - \frac{2}{t^2} + \frac{1}{t^4}} = \sqrt{1 + \frac{2}{t^2} + \frac{1}{t^4}} = \sqrt{\left(1 + \frac{1}{t^2}\right)^2} = 1 + \frac{1}{t^2}.$$ **Step 4:** Integrate: $$L = \int_1^2 \left(1 + \frac{1}{t^2}\right) dt = \left[t - \frac{1}{t}\right]_1^2 = \left(2 - \frac{1}{2}\right) - (1 - 1) = \frac{3}{2}.$$ (e) Energy of curve portion for $t \in (0,1]$. **Step 1:** Energy is often proportional to integral of speed squared: $$E = \int_0^1 \left(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right) dt = \int_0^1 \left(\frac{4}{t^2} + \left(1 - \frac{1}{t^2}\right)^2\right) dt.$$ **Step 2:** Simplify integrand: $$\frac{4}{t^2} + 1 - \frac{2}{t^2} + \frac{1}{t^4} = 1 + \frac{2}{t^2} + \frac{1}{t^4}.$$ **Step 3:** Near $t=0$, dominant term is $\frac{1}{t^4}$ which is not integrable: $$\int_0^1 \frac{1}{t^4} dt$$ diverges. **Answer:** Energy is infinite. 3. **Problem:** Find center and radius of sphere: $$x^2 + y^2 + z^2 - 2x + 4z = -3.$$ **Step 1:** Complete the square for $x$ and $z$: $$x^2 - 2x = (x - 1)^2 - 1,$$ $$z^2 + 4z = (z + 2)^2 - 4.$$ **Step 2:** Rewrite equation: $$(x - 1)^2 - 1 + y^2 + (z + 2)^2 - 4 = -3,$$ $$(x - 1)^2 + y^2 + (z + 2)^2 = 2.$$ **Step 3:** Center and radius: Center: $(1, 0, -2)$ Radius: $\sqrt{2}$. 4. **Problem:** Points $A=(1,-1,2)$, $B=(3,-2,3)$, $C=(2,-2,2)$. (a) Find vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$. $$\overrightarrow{AB} = B - A = (3-1, -2+1, 3-2) = (2, -1, 1),$$ $$\overrightarrow{AC} = C - A = (2-1, -2+1, 2-2) = (1, -1, 0).$$ (b) Compute angle $\widehat{BAC}$. **Step 1:** Use dot product: $$\overrightarrow{AB} \cdot \overrightarrow{AC} = 2 \cdot 1 + (-1)(-1) + 1 \cdot 0 = 2 + 1 + 0 = 3.$$ **Step 2:** Magnitudes: $$|\overrightarrow{AB}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6},$$ $$|\overrightarrow{AC}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2}.$$ **Step 3:** Angle: $$\cos \theta = \frac{3}{\sqrt{6} \sqrt{2}} = \frac{3}{\sqrt{12}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}.$$ So, $$\theta = \frac{\pi}{6} = 30^\circ.$$ (c) Are points aligned? Check if $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are linearly dependent. Since $\overrightarrow{AB} = (2,-1,1)$ and $\overrightarrow{AC} = (1,-1,0)$ are not scalar multiples, points are not aligned. (d) Find plane equation through $A,B,C$. **Step 1:** Normal vector is cross product: $$\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & -1 & 0 \end{vmatrix} = \mathbf{i}((-1)(0) - 1(-1)) - \mathbf{j}(2(0) - 1(1)) + \mathbf{k}(2(-1) - (-1)(1)) = \mathbf{i}(0 + 1) - \mathbf{j}(0 - 1) + \mathbf{k}(-2 + 1) = (1,1,-1).$$ **Step 2:** Plane equation: $$1(x - 1) + 1(y + 1) - 1(z - 2) = 0,$$ $$x - 1 + y + 1 - z + 2 = 0,$$ $$x + y - z + 2 = 0.$$