1. Problem 16: Solve the first-order vector differential equation $$\frac{d\mathbf{r}}{dt} = \frac{t}{t^2+2}\mathbf{i} - \frac{t^2+1}{t-2}\mathbf{j} + \frac{t^2+4}{t^2+3}\mathbf{k}$$ with initial condition $$\mathbf{r}(0) = \mathbf{i} - \mathbf{j} + \mathbf{k}.$$ Step 1: Integrate each component separately. - For the $\mathbf{i}$ component: $$r_i(t) = \int \frac{t}{t^2+2} dt$$ Use substitution $u = t^2+2$, $du = 2t dt \Rightarrow t dt = \frac{du}{2}$: $$r_i(t) = \int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \ln|t^2+2| + C_i$$ Using initial condition $r_i(0) = 1$, we find: $$1 = \frac{1}{2} \ln 2 + C_i \Rightarrow C_i = 1 - \frac{1}{2} \ln 2$$ - For the $\mathbf{j}$ component: $$r_j(t) = -\int \frac{t^2+1}{t-2} dt$$ Perform polynomial division: $$\frac{t^2+1}{t-2} = t + 2 + \frac{5}{t-2}$$ So $$r_j(t) = -\int (t + 2 + \frac{5}{t-2}) dt = -\left( \frac{t^2}{2} + 2t + 5\ln|t-2| \right) + C_j$$ Initial condition $r_j(0) = -1$ gives: $$-1 = -\left(0 + 0 + 5 \ln 2 \right) + C_j \Rightarrow C_j = -1 + 5 \ln 2$$ - For the $\mathbf{k}$ component: $$r_k(t) = \int \frac{t^2+4}{t^2+3} dt = \int \left(1 + \frac{1}{t^2+3}\right) dt = t + \frac{1}{\sqrt{3}} \arctan \frac{t}{\sqrt{3}} + C_k$$ Initial condition $r_k(0) = 1$: $$1 = 0 + 0 + C_k \Rightarrow C_k = 1$$ Final vector function: $$\mathbf{r}(t) = \left(\frac{1}{2} \ln(t^2+2) + 1 - \frac{1}{2} \ln 2\right) \mathbf{i} - \left( \frac{t^2}{2} + 2t + 5 \ln|t-2| -1 + 5 \ln 2 \right) \mathbf{j} + \left( t + \frac{1}{\sqrt{3}} \arctan \frac{t}{\sqrt{3}} + 1 \right) \mathbf{k}.$$ 2. Problem 17: Solve the second-order vector differential equation $$\frac{d^2\mathbf{r}}{dt^2} = -32 \mathbf{k}$$ with initial conditions: $$\mathbf{r}(0) = 100 \mathbf{k}, \quad \frac{d\mathbf{r}}{dt}\Big|_{t=0} = 8 \mathbf{i} + 8 \mathbf{j}.$$ Step 1: Integrate acceleration to get velocity: $$\frac{d\mathbf{r}}{dt} = -32t \mathbf{k} + \mathbf{C}_1,$$ apply initial velocity to find $\mathbf{C}_1$: $$8 \mathbf{i} + 8 \mathbf{j} = \mathbf{C}_1.$$ Step 2: Integrate velocity to get position: $$\mathbf{r}(t) = -16 t^2 \mathbf{k} + (8t) \mathbf{i} + (8t) \mathbf{j} + \mathbf{C}_2,$$ use initial position to find $\mathbf{C}_2$: $$\mathbf{C}_2 = 100 \mathbf{k}.$$ Final solution: $$\mathbf{r}(t) = 8t \mathbf{i} + 8t \mathbf{j} + (100 - 16 t^2) \mathbf{k}.$$ 3. Problem 18: Solve second-order vector equation $$\frac{d^2 \mathbf{r}}{dt^2} = - (\mathbf{i} + \mathbf{j} + \mathbf{k})$$ with $$\mathbf{r}(0) = 10 \mathbf{i} + 10 \mathbf{j} + 10 \mathbf{k}, \quad \frac{d \mathbf{r}}{dt}|_{t=0} = \mathbf{0}.$$ Integrate acceleration twice: $$\frac{d\mathbf{r}}{dt} = - (\mathbf{i} + \mathbf{j} + \mathbf{k}) t + \mathbf{C}_1,$$ $$\mathbf{r}(t) = -\frac{t^2}{2}(\mathbf{i} + \mathbf{j} + \mathbf{k}) + \mathbf{C}_1 t + \mathbf{C}_2,$$ with initial velocity zero implies $$\mathbf{C}_1 = \mathbf{0}$$ Initial position implies $$\mathbf{C}_2 = 10 (\mathbf{i} + \mathbf{j} + \mathbf{k}).$$ Final: $$\mathbf{r}(t) = 10 (\mathbf{i} + \mathbf{j} + \mathbf{k}) - \frac{t^2}{2} (\mathbf{i} + \mathbf{j} + \mathbf{k}).$$ 4. Problem 19: Solve $$\frac{d^2\mathbf{r}}{dt^2} = e^t \mathbf{i} - e^{-t} \mathbf{j} + 4 e^{2t} \mathbf{k},$$ with initial conditions $$\mathbf{r}(0) = 3 \mathbf{i} + \mathbf{j} + 2 \mathbf{k}, \quad \frac{d\mathbf{r}}{dt}|_{0} = - \mathbf{i} + 4 \mathbf{j}.$$ Integrate acceleration once: $$\frac{d\mathbf{r}}{dt} = e^t \mathbf{i} + e^{-t} \mathbf{j} + 2 e^{2t} \mathbf{k} + \mathbf{C}_1.$$ Use initial velocity: $$-1 = e^{0} + C_{1i} \Rightarrow C_{1i} = -2,$$ $$4 = -e^{0} + C_{1j} \Rightarrow C_{1j} = 5,$$ for $\mathbf{k}$ no initial velocity given for $k$, so $C_{1k}$ is unknown constant. Integrate velocity: $$\mathbf{r}(t) = e^t \mathbf{i} - e^{-t} \mathbf{j} + e^{2t} \mathbf{k} + \mathbf{C}_1 t + \mathbf{C}_2.$$ Initial position: $$3 = 1 + C_{2i} \Rightarrow C_{2i} = 2,$$ $$1 = -1 + C_{2j} \Rightarrow C_{2j} = 2,$$ $$2 = 1 + C_{2k} \Rightarrow C_{2k} = 1.$$ Final: $$\mathbf{r}(t) = (e^t - 2t + 2) \mathbf{i} + (-e^{-t} + 5 t + 2) \mathbf{j} + (e^{2t} + C_{1k} t + 1) \mathbf{k}.$$ (If initial velocity in $k$ was zero, $C_{1k} = 0$.) 5. Problem 20: Solve $$\frac{d^2\mathbf{r}}{dt^2} = \sin t \mathbf{i} - \cos t \mathbf{j} + 4 \sin t \cos t \mathbf{k}$$ with $$\mathbf{r}(0) = \mathbf{i} - \mathbf{k}, \quad \frac{d\mathbf{r}}{dt}|_{0} = \mathbf{i}.$$ Integrate acceleration to velocity: $$\frac{d\mathbf{r}}{dt} = -\cos t \mathbf{i} - \sin t \mathbf{j} + 2 \sin^2 t \mathbf{k} + \mathbf{C}_1.$$ Apply initial velocity at $t=0$: $$\mathbf{i} = -1 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k} + \mathbf{C}_1 \Rightarrow \mathbf{C}_1 = 2 \mathbf{i}.$$ Integrate velocity to position: $$\mathbf{r}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} + (t - \frac{1}{2} \sin 2t) \mathbf{k} + 2 t \mathbf{i} + \mathbf{C}_2.$$ Use initial position $\mathbf{r}(0) = \mathbf{i} - \mathbf{k}$: $$\mathbf{i} - \mathbf{k} = 0 + 1 \mathbf{j} + 0 + 0 + \mathbf{C}_2 \Rightarrow \mathbf{C}_2 = \mathbf{i} - \mathbf{j} - \mathbf{k}.$$ Final answer: $$\mathbf{r}(t) = (2 t - \sin t + 1) \mathbf{i} + (\cos t - 1) \mathbf{j} + \left(t - \frac{1}{2} \sin 2t - 1 \right) \mathbf{k}.$$