1. Problem 44: A new tower S is located so that the distance from S to tower A is 2 miles greater than the distance from S to tower B. Towers A and B are 10 miles apart. Step 1: Let the coordinates of towers A and B be $(0,0)$ and $(10,0)$ respectively for simplicity. Step 2: Let $S=(x,y)$. Then the distances are $SA=\sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2+y^2}$ and $SB=\sqrt{(x-10)^2 + y^2}$. Step 3: The condition $SA = SB + 2$ implies: $$\sqrt{x^2 + y^2} = \sqrt{(x-10)^2 + y^2} + 2$$ Step 4: Square both sides: $$x^2 + y^2 = (x-10)^2 + y^2 + 4 \sqrt{(x-10)^2 + y^2} + 4$$ Step 5: Cancel $y^2$ and move terms: $$x^2 = (x-10)^2 + 4 \sqrt{(x-10)^2 + y^2} + 4$$ Step 6: Expand $(x-10)^2 = x^2 - 20x + 100$: $$x^2 = x^2 - 20x + 100 + 4 \sqrt{(x-10)^2 + y^2} + 4$$ Step 7: Cancel $x^2$: $$0 = -20x + 104 + 4 \sqrt{(x-10)^2 + y^2}$$ Step 8: Rearranged: $$4 \sqrt{(x-10)^2 + y^2} = 20x - 104$$ Step 9: Divide by 4: $$\sqrt{(x-10)^2 + y^2} = 5x - 26$$ Step 10: The expression $5x - 26$ must be nonnegative, so domain is restricted. Step 11: Square again: $$(x-10)^2 + y^2 = (5x - 26)^2$$ Step 12: Expand: $$x^2 - 20x + 100 + y^2 = 25x^2 - 260x + 676$$ Step 13: Bring all to one side: $$0 = 25x^2 - 260x + 676 - x^2 + 20x - 100 - y^2$$ Step 14: Simplify: $$0 = 24x^2 - 240x + 576 - y^2$$ Step 15: Rearrange: $$24x^2 -240x + 576 = y^2$$ Step 16: Divide all by 576: $$\frac{24x^2}{576} - \frac{240x}{576} + 1 = \frac{y^2}{576}$$ Step 17: Simplify coefficients: $$\frac{x^2}{24} - \frac{5x}{12} + 1 = \frac{y^2}{576}$$ Step 18: Completing the square for $x$-terms: $$\frac{x^2}{24} - \frac{5x}{12} = \frac{1}{24}(x^2 - 10x) = \frac{1}{24}[(x-5)^2 - 25]$$ Step 19: Substitute back: $$\frac{1}{24} (x-5)^2 - \frac{25}{24} + 1 = \frac{y^2}{576}$$ Step 20: Calculate: $$\frac{1}{24} (x-5)^2 - \frac{1}{24} = \frac{y^2}{576}$$ Step 21: Multiply both sides by 576: $$24 (x-5)^2 - 24 = y^2$$ Step 22: Rearranged: $$24 (x-5)^2 - y^2 = 24$$ Step 23: Divide both sides by 24: $$(x-5)^2 - \frac{y^2}{24} = 1$$ Step 24: This is the equation of a hyperbola with center $(5,0)$. Conclusion: The locus is a branch of a hyperbola. Answer: (A). 2. Problem 45: Given $x=f(u,v)$ and $y=g(u,v)$, find expression for $\frac{\partial u}{\partial x}$. Step 1: From implicit functions, the partial derivatives satisfy Jacobian relations. Step 2: The Jacobian determinant: $$J = \frac{\partial(f,g)}{\partial(u,v)} = \frac{\partial f}{\partial u} \frac{\partial g}{\partial v} - \frac{\partial f}{\partial v} \frac{\partial g}{\partial u}$$ Step 3: The chain rule implies: $$\frac{\partial (x,y)}{\partial (u,v)} \cdot \frac{\partial (u,v)}{\partial (x,y)} = 1$$ Step 4: The inverse function theorem shows: $$\frac{\partial u}{\partial x} = \frac{1}{J} \left( \frac{\partial g}{\partial v} \right)$$ Step 5: More precisely: $$\frac{\partial u}{\partial x} = \frac{\partial g/\partial v}{J} = \frac{\frac{\partial g}{\partial v}}{\frac{\partial f}{\partial u} \frac{\partial g}{\partial v} - \frac{\partial f}{\partial v} \frac{\partial g}{\partial u}}$$ Conclusion: The correct expression corresponds to option (E). 3. Problem 46: Solve the differential equation $$y'' + 2y' + 3y = t$$ Step 1: Solve the homogeneous equation: $$y'' + 2y' + 3y = 0$$ Step 2: Characteristic equation: $$r^2 + 2r + 3 = 0$$ Step 3: Roots: $$r = \frac{-2 \pm \sqrt{4 - 12}}{2} = -1 \pm i \sqrt{2}$$ Step 4: General solution of homogeneous part: $$y_h = C_1 e^{-t} \cos(\sqrt{2} t) + C_2 e^{-t} \sin(\sqrt{2} t)$$ Step 5: Find particular solution $y_p$. Since RHS is $t$ (a polynomial of degree 1), try polynomial: $$y_p = At + B$$ Step 6: Derivatives: $$y_p' = A, \quad y_p'' = 0$$ Step 7: Substitute into equation: $$0 + 2A + 3(At + B) = t$$ Step 8: Equate coefficients: For $t$: $3A = 1 \implies A = \frac{1}{3}$ Constant: $$2A + 3B = 0 \implies 2 \times \frac{1}{3} + 3B = 0 \implies \frac{2}{3} + 3B = 0 \implies B = -\frac{2}{9}$$ Step 9: Particular solution: $$y_p = \frac{1}{3} t - \frac{2}{9}$$ Step 10: General solution: $$y = C_1 e^{-t} \cos(\sqrt{2} t) + C_2 e^{-t} \sin(\sqrt{2} t) + \frac{1}{3} t - \frac{2}{9}$$ Answer: (E). 4. Problem 47: Evaluate line integral $$\int_C (5x + y^3) \, dx + (3xy^2 + 8y) \, dy$$ Step 1: Path $C$ from $(2,0)$ to $(0,3)$ along a straight line. Step 2: Parametrize line segment: $$x = 2 - 2t, \quad y = 3t, \quad t \in [0,1]$$ Step 3: Compute derivatives: $$\frac{dx}{dt} = -2, \quad \frac{dy}{dt} = 3$$ Step 4: Substitute into integral: $$I = \int_0^1 [(5x + y^3) (-2) + (3xy^2 + 8y)(3)] dt$$ Step 5: Substitute $x$ and $y$: $$5x = 5(2 - 2t) = 10 - 10t$$ $$y^3 = (3t)^3 = 27 t^3$$ $$3xy^2 = 3 (2 - 2t)(9 t^2) = 3 (2 - 2t)(9 t^2) = 27 (2 - 2t) t^2 = 54 t^2 - 54 t^3$$ $$8y = 8 (3t) = 24 t$$ Step 6: Substitute: $$I = \int_0^1 [ (10 - 10t + 27 t^3)(-2) + (54 t^2 - 54 t^3 + 24 t)(3) ] dt$$ Step 7: Expand: $$= \int_0^1 [ -20 + 20 t - 54 t^3 + 162 t^2 - 162 t^3 + 72 t ] dt$$ Step 8: Combine like terms: $$= \int_0^1 [ -20 + 20 t + 72 t + 162 t^2 - 54 t^3 - 162 t^3 ] dt$$ $$= \int_0^1 [ -20 + 92 t + 162 t^2 - 216 t^3 ] dt$$ Step 9: Integrate term-by-term: $$\int_0^1 -20 dt = -20$$ $$\int_0^1 92 t dt = 46$$ $$\int_0^1 162 t^2 dt = 54$$ $$\int_0^1 -216 t^3 dt = -54$$ Step 10: Sum integrals: $$-20 + 46 + 54 - 54 = 26$$ Answer: (C).