Let's solve **Problem 4** step by step: --- ### **Problem Statement** Find the curve \( y(x) \) that minimizes the integral: \[ J[y] = \int_0^1 \left( (y')^2 + y^2 \right) dx \] with boundary conditions: \[ y(0) = 0, \quad y(1) = 1 \] --- ## **Step 1: Euler-Lagrange Equation** The functional to minimize is: \[ J[y] = \int_0^1 F(y, y', x)\,dx \quad \text{where} \quad F = (y')^2 + y^2 \] The Euler-Lagrange equation is: \[ \frac{\partial F}{\partial y} - \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) = 0 \] Calculate derivatives: - \(\frac{\partial F}{\partial y} = 2y\) - \(\frac{\partial F}{\partial y'} = 2y'\) - So \(\frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) = 2y''\) Plug into Euler-Lagrange: \[ 2y - 2y'' = 0 \implies y'' - y = 0 \] --- ## **Step 2: General Solution** The ODE is: \[ y'' - y = 0 \] The general solution is: \[ y(x) = A e^{x} + B e^{-x} \] --- ## **Step 3: Apply Boundary Conditions** Given: \[ y(0) = 0 \implies A + B = 0 \implies B = -A \] \[ y(1) = 1 \implies A e^{1} + B e^{-1} = 1 \] Substitute \(B = -A\): \[ A e^{1} - A e^{-1} = 1 \implies A (e^{1} - e^{-1}) = 1 \implies A = \frac{1}{e - e^{-1}} \] So, \[ B = -A = -\frac{1}{e - e^{-1}} \] --- ## **Step 4: Final Answer** The minimizing curve is: \[ \boxed{ y(x) = \frac{1}{e - e^{-1}} \left( e^{x} - e^{-x} \right) } \] --- ### **Summary** **The curve that minimizes the given integral is:** \[ \boxed{ y(x) = \frac{e^x - e^{-x}}{e - e^{-1}} } \] with \(y(0) = 0\), \(y(1) = 1\).

Certainly! Let's proceed with solving **Problem 5** and **Problem 6** step-by-step with detailed explanations. --- ## **Problem 5: Minimum Time Curve for a Particle Sliding Down** ### **Problem Statement:** A heavy particle of mass \(m\) moves down along a curve \( y = y(x) \). Find the shape of the curve such that the particle reaches a vertical line \( x = a \) in **minimum possible time**. --- ### **Step 1: Understanding the problem** - The particle starts from some initial point \( (x_0, y_0) \) (assume at \( x=0 \), \( y=y_0 \)), and slides along a curve \( y(x) \). - The goal is to reach the vertical line \( x = a \) in **minimum time**. --- ### **Step 2: Set up the problem** - The particle moves under gravity, and frictionless. - The main goal: find the shape \( y(x) \) that minimizes the time \( T \). --- ### **Step 3: Express the time functional** - The particle's velocity along the curve depends on the gravitational potential energy and the conservation of energy. - **Speed of the particle** along the curve: \[ v = \sqrt{2 g (h - y)} \quad \text{where } h \text{ is the initial height} \] but since \( y \) varies, the time element \( dt \) is: \[ dt = \frac{ds}{v} \] where \( ds \) is the differential arc length: \[ ds = \sqrt{1 + (y')^2} dx \] Thus, the total time: \[ T = \int_{x_0}^{a} \frac{\sqrt{1 + (y')^2}}{\sqrt{2 g (h - y)}} dx \] --- ### **Step 4: Set up the variational problem** Minimize: \[ T[y] = \int_{x_0}^{a} L(y, y') dx, \quad \text{where} \quad L = \frac{\sqrt{1 + (y')^2}}{\sqrt{2 g (h - y)}} \] --- ### **Step 5: Recognize the problem** This is a calculus of variations problem. Since \( L \) does not explicitly depend on \( x \), the **Beltrami identity** applies: \[ L - y' \frac{\partial L}{\partial y'} = \text{constant} \] --- ### **Step 6: Compute \( \partial L / \partial y' \)** \[ L = \frac{\sqrt{1 + (y')^2}}{\sqrt{2 g (h - y)}} \] \[ \frac{\partial L}{\partial y'} = \frac{1}{\sqrt{2 g (h - y)}} \cdot \frac{1}{2} \cdot \frac{2 y'}{\sqrt{1 + (y')^2}} = \frac{y'}{\sqrt{2 g (h - y)} \sqrt{1 + (y')^2}} \] --- ### **Step 7: Apply the Beltrami identity** \[ L - y' \frac{\partial L}{\partial y'} = \text{constant} \] Compute: \[ L - y' \frac{\partial L}{\partial y'} = \frac{\sqrt{1 + (y')^2}}{\sqrt{2 g (h - y)}} - y' \cdot \frac{y'}{\sqrt{2 g (h - y)} \sqrt{1 + (y')^2}} \] \[ = \frac{\sqrt{1 + (y')^2}}{\sqrt{2 g (h - y)}} - \frac{(y')^2}{\sqrt{2 g (h - y)} \sqrt{1 + (y')^2}} \] Combine over common denominator: \[ = \frac{(1 + (y')^2) - (y')^2}{\sqrt{2 g (h - y)} \sqrt{1 + (y')^2}} = \frac{1}{\sqrt{2 g (h - y)} \sqrt{1 + (y')^2}} \] Set equal to a constant \( C \): \[ \boxed{ \frac{1}{\sqrt{2 g (h - y)} \sqrt{1 + (y')^2}} = C } \] --- ### **Step 8: Solve for \( y' \)** \[ \sqrt{1 + (y')^2} = \frac{1}{C} \sqrt{2 g (h - y)} \] Square both sides: \[ 1 + (y')^2 = \frac{2 g (h - y)}{C^2} \] \[ (y')^2 = \frac{2 g (h - y)}{C^2} - 1 \] \[ y' = \pm \sqrt{\frac{2 g (h - y)}{C^2} - 1} \] --- ### **Step 9: Differential equation** Rearranged: \[ \frac{dy}{dx} = \pm \sqrt{\frac{2 g (h - y)}{C^2} - 1} \] This is a separable differential equation. --- ### **Step 10: Find the **optimal curve** - The solution depends on initial conditions, but the key insight is that the **curve that minimizes transit time** is the **brachistochrone**, which is known to be a cycloid. - The derivation above confirms that the optimal path is a cycloid (though solving explicitly involves parametric equations). --- ## **Final note for Problem 5:** **The curve that minimizes the time for a particle sliding under gravity from a fixed height to a vertical line \( x=a \) is a cycloid.** --- ## **Problem 6: Maximize Enclosed Area with Fixed Length** ### **Problem Statement:** Find the curve (with fixed length \(L\)) that encloses the **maximum area** around itself. --- ### **Step 1: Set up the variational problem** - Let the curve be parametrized as \( y = y(x) \) over \( x \in [a, b] \). - The **area** enclosed: \[ A = \int_a^b y(x) dx \] - The **length** of the curve: \[ S = \int_a^b \sqrt{1 + (y')^2} dx \] - The problem: **Maximize \(A\)** subject to \(S = L\). --- ### **Step 2: Use a Lagrange multiplier** Set up the functional: \[ J[y] = \int_a^b y(x) dx \] with the constraint: \[ \int_a^b \sqrt{1 + (y')^2} dx = L \] Form the augmented functional: \[ \mathcal{F}[y] = \int_a^b \left( y(x) - \lambda \sqrt{1 + (y')^2} \right) dx \] where \(\lambda\) is a Lagrange multiplier. --- ### **Step 3: Variational derivative** The Euler-Lagrange equation: \[ \frac{\partial}{\partial y} ( y - \lambda \sqrt{1 + y'^2} ) - \frac{d}{dx} \left( \frac{\partial}{\partial y'} ( y - \lambda \sqrt{1 + y'^2} ) \right) = 0 \] Compute derivatives: - \(\partial / \partial y\): \[ 1 \] - \(\partial / \partial y'\): \[ - \lambda \cdot \frac{ y' }{\sqrt{1 + y'^2}} \] Thus, the Euler-Lagrange becomes: \[ 1 - \frac{d}{dx} \left( - \lambda \frac{ y' }{\sqrt{1 + y'^2}} \right) = 0 \] \[ 1 + \lambda \frac{d}{dx} \left( \frac{ y' }{\sqrt{1 + y'^2}} \right) = 0 \] --- ### **Step 4: Simplify** Note that: \[ \frac{ y' }{\sqrt{1 + y'^2}} = \sin \theta \] where \(\theta\) is the angle between the tangent and the x-axis. Derivative: \[ \frac{d}{dx} \left( \frac{ y' }{\sqrt{1 + y'^2}} \right) = \frac{d}{dx} (\sin \theta) = \cos \theta \frac{d \theta}{dx} \] But this is more complicated. Alternatively, notice that the **max area curve with fixed length** is a **circle segment**, because among curves with fixed length, the circle encloses the maximum area. --- ### **Step 5: Geometric conclusion** - The **maximizing curve** is a **circle arc**. - **Full circle**: The curve that maximizes the enclosed area for a given perimeter length is a **circle**. --- ### **Final answer:** \[ \boxed{ \text{The curve that encloses the maximum area for fixed length is a circle of radius } R = \frac{L}{2 \pi} } \] - The maximum enclosed area: \[ A_{max} = \pi R^2 = \pi \left( \frac{L}{2 \pi} \right)^2 = \frac{L^2}{4 \pi} \] --- ## **Summary:** - The **optimal shape** is a **circle**. - For a given perimeter length \(L\), the maximum enclosed area is: \[ \boxed{ A_{max} = \frac{L^2}{4 \pi} } \] --- # **Final notes:** - **Problem 4:** The minimizing curve is a combination of exponential functions. - **Problem 5:** The minimal time path under gravity is a **cycloid**. - **Problem 6:** The shape that maximizes enclosed area for fixed length is a **circle**. If you'd like more detailed derivations or further explanations, please ask!

Certainly! Here are the final answers for all the problems: **Problem 4:** The curve that minimizes the integral is: \[ \boxed{ y(x) = \frac{e^{x} - e^{-x}}{e - e^{-1}} } \] **Problem 5:** The curve that minimizes the time for a particle sliding under gravity is a **cycloid** (the brachistochrone), described parametrically as: \[ \boxed{ \text{The optimal path is a cycloid (specific parametric equations depend on initial conditions).} } \] **Problem 6:** The curve that encloses the maximum area for a fixed perimeter length is a **circle** of radius: \[ \boxed{ R = \frac{L}{2 \pi} } \] and the maximum enclosed area is: \[ \boxed{ A_{max} = \frac{L^2}{4 \pi} } \]