Applied Mathematics I - Assignment

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1. Vector Analysis - Airplane Motion a. Ground Velocity Airplane velocity relative to air: $\vec{V}_A = 400\mathbf{i} + 100\mathbf{j}$ Wind velocity: $\vec{V}_W = 60\mathbf{i} + 80\mathbf{j}$ Ground velocity: $\vec{V}_G = \vec{V}_A + \vec{V}_W$ Calculation: $\vec{V}_G = (400\mathbf{i} + 100\mathbf{j}) + (60\mathbf{i} + 80\mathbf{j}) = (400+60)\mathbf{i} + (100+80)\mathbf{j} = 460\mathbf{i} + 180\mathbf{j}$ km/hr b. Angle Between Directions Airplane's direction (relative to air): $\vec{V}_A = 400\mathbf{i} + 100\mathbf{j}$ Actual ground path direction: $\vec{V}_G = 460\mathbf{i} + 180\mathbf{j}$ Magnitude of $\vec{V}_A$: $|\vec{V}_A| = \sqrt{400^2 + 100^2} = \sqrt{160000 + 10000} = \sqrt{170000} \approx 412.31$ Magnitude of $\vec{V}_G$: $|\vec{V}_G| = \sqrt{460^2 + 180^2} = \sqrt{211600 + 32400} = \sqrt{244000} \approx 493.96$ Dot product: $\vec{V}_A \cdot \vec{V}_G = (400)(460) + (100)(180) = 184000 + 18000 = 202000$ Angle $\theta$: $\cos \theta = \frac{\vec{V}_A \cdot \vec{V}_G}{|\vec{V}_A| |\vec{V}_G|} = \frac{202000}{412.31 \times 493.96} \approx \frac{202000}{203606.8} \approx 0.9921$ $\theta = \arccos(0.9921) \approx 7.96^\circ$ c. Rate of Change of Distance (Component of Ground Velocity) Radar station position: $\vec{r}_S = 10\mathbf{i} + 30\mathbf{j}$ km Airplane position: $\vec{r}_P = 20\mathbf{i} + 10\mathbf{j}$ km Vector from radar to airplane: $\vec{r}_{SP} = \vec{r}_P - \vec{r}_S = (20-10)\mathbf{i} + (10-30)\mathbf{j} = 10\mathbf{i} - 20\mathbf{j}$ km Unit vector in direction of $\vec{r}_{SP}$: $\hat{u}_{SP} = \frac{\vec{r}_{SP}}{|\vec{r}_{SP}|} = \frac{10\mathbf{i} - 20\mathbf{j}}{\sqrt{10^2 + (-20)^2}} = \frac{10\mathbf{i} - 20\mathbf{j}}{\sqrt{100+400}} = \frac{10\mathbf{i} - 20\mathbf{j}}{\sqrt{500}} = \frac{1}{\sqrt{5}}( \mathbf{i} - 2\mathbf{j})$ Ground velocity: $\vec{V}_G = 460\mathbf{i} + 180\mathbf{j}$ km/hr Rate of change of distance (component of $\vec{V}_G$ along $\vec{r}_{SP}$): $\vec{V}_G \cdot \hat{u}_{SP}$ Calculation: $(460\mathbf{i} + 180\mathbf{j}) \cdot \frac{1}{\sqrt{5}}(\mathbf{i} - 2\mathbf{j}) = \frac{1}{\sqrt{5}}((460)(1) + (180)(-2)) = \frac{1}{\sqrt{5}}(460 - 360) = \frac{100}{\sqrt{5}} = 20\sqrt{5} \approx 44.72$ km/hr 2. Matrix Algebra - Production Planning Consumption matrix $A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 2 \end{pmatrix}$, Availability matrix $B = \begin{pmatrix} 100 \\ 180 \\ 80 \end{pmatrix}$ Let $x_1$ be units of product $P_1$ and $x_2$ be units of product $P_2$. The system can be written as $A \mathbf{x} = B$, where $\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$. This gives the equations: $2x_1 + x_2 = 100$ (Material $M_1$) $3x_1 + 2x_2 = 180$ (Material $M_2$) $x_1 + 2x_2 = 80$ (Material $M_3$) a. Maximum Units (Full Utilization) We need to solve the system of linear equations for $x_1$ and $x_2$. We have 3 equations and 2 unknowns. We can use any two equations to find a solution and then check if it satisfies the third. From $M_1$: $x_2 = 100 - 2x_1$ Substitute into $M_3$: $x_1 + 2(100 - 2x_1) = 80$ $x_1 + 200 - 4x_1 = 80$ $-3x_1 = 80 - 200$ $-3x_1 = -120 \implies x_1 = 40$ Then, $x_2 = 100 - 2(40) = 100 - 80 = 20$ Check with $M_2$: $3(40) + 2(20) = 120 + 40 = 160$. This does NOT equal 180. This implies that it's not possible to fully use all materials for both products simultaneously (i.e., the system has no solution if we insist on full utilization of all three materials). However, the question asks for "maximum number of units" if all materials are fully used. This implies we need to find a solution that satisfies as many constraints as possible, or perhaps that the problem implies a feasible region where the maximum occurs. Given the direct formulation, the system $A\mathbf{x}=B$ has no exact solution where all materials are fully used. If we interpret "fully used" as meaning we need to find a solution to $A\mathbf{x}=B$, then based on the check above, there is no such solution. If the question implies finding the maximum production within the bounds, it would be a linear programming problem, which is beyond simple matrix inversion/solving for this setup. Let's re-evaluate: if a solution $(x_1, x_2)$ exists that satisfies all three equations, then $x_1=40, x_2=20$ is the only candidate from the first and third equations. Since it fails the second, there is no solution where all materials are *fully* used. Therefore, based on the strict interpretation of "fully used (no leftovers)", no such $x_1, x_2$ combination exists that uses exactly 100kg of $M_1$, 180kg of $M_2$, and 80kg of $M_3$. b. Production with $P_1 = 20$ units If $x_1 = 20$, we substitute this into the material equations: $2(20) + x_2 \le 100 \implies 40 + x_2 \le 100 \implies x_2 \le 60$ (from $M_1$) $3(20) + 2x_2 \le 180 \implies 60 + 2x_2 \le 180 \implies 2x_2 \le 120 \implies x_2 \le 60$ (from $M_2$) $20 + 2x_2 \le 80 \implies 2x_2 \le 60 \implies x_2 \le 30$ (from $M_3$) To not exceed material limits, $x_2$ must satisfy all these conditions. The most restrictive condition is $x_2 \le 30$. Therefore, if the company produces 20 units of $P_1$, it can produce a maximum of 30 units of $P_2$ without exceeding material limits. c. Uniqueness of Solution for System in (a) The system of equations from part (a) is: $2x_1 + x_2 = 100$ $3x_1 + 2x_2 = 180$ $x_1 + 2x_2 = 80$ This is an overdetermined system (3 equations, 2 unknowns). As shown in part (a), solving the first and third equations gives $x_1=40, x_2=20$. When these values are substituted into the second equation, $3(40) + 2(20) = 120 + 40 = 160 \ne 180$. Since the values obtained from two equations do not satisfy the third equation, the system is inconsistent. Therefore, the system in (a) has no solution (no combination of $x_1, x_2$ allows all materials to be fully used as specified).