Dr. Brajesh Kumar Jha

Cholesky Decomposition Method – Solved Problems The Cholesky Decomposition Method is used to solve systems of linear equations when the coefficient matrix is symmetric and positive definite . The matrix is decomposed as A = LL T where L is a lower triangular matrix. The solution is obtained in two steps: Forward substitution: LY = B Backward substitution: L T X = Y Problem 1 Solve using Cholesky decomposition 4x + 2y = 6 2x + 3y = 7 Show Solution Matrix form A = [4 2 2 3] L = [2 0 1 √2] Forward substitution: 2y₁ = 6 y₁ = 3 3 + √2 y₂ = 7 y₂ = 2√2 Backward substitution: √2 y = 2√2 y = 2 2x + 2 = 3 x = 1/2 Solution: x = 1/2 y = 2 Problem 2 Solve 4x + 2y + 2z = 8 2x + 5y + z = 3 2x + y + 3z = 5 Show Solution Matrix A = [4 2 2 2 5 1 2 1 3] Cholesky factor L = [2 0 0 1 2 0 1 0 √2] Forward substitution gives Y = [4 -1/2 1/√2] Backwar...

Total Derivatives – Theory and Solved Problems Definition If \(z=f(x,y)\) where \(x\) and \(y\) depend on parameter \(t\), then the total derivative is \[ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \] This formula is known as the **Chain Rule for multivariable functions**. Visual Understanding of the Chain Rule In problems involving total derivatives , variables often depend on other variables. The Chain Rule shows how a change in one variable affects another through intermediate variables. The following diagrams illustrate this dependency structure. Diagram 1: Basic Chain Rule Structure z x y t This diagram represents: \[ z = f(x,y), \quad x = x(t), \quad y = y(t) \] \[ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \] Diagram 2: Two-Level Dependency z x y u...

Double Integration in Polar Coordinates Animated Step-by-Step Solutions Example 1: Area of Circle $x^2+y^2 \le 4$ Evaluate: $$\iint_R 1 \, dA$$ ▶ Show Detailed Solution Step 1: Convert to Polar Coordinates $x = r\cos\theta$, $y = r\sin\theta$ $x^2+y^2 = r^2$ $dA = r\,dr\,d\theta$ Step 2: Limits $0 \le r \le 2$, $0 \le \theta \le 2\pi$ Step 3: Integral $$ \int_0^{2\pi}\int_0^2 r\,dr\,d\theta $$ $$ \int_0^2 r\,dr = \frac{r^2}{2}\Big|_0^2 = 2 $$ $$ \int_0^{2\pi} 2\,d\theta = 4\pi $$ Final Answer: $4\pi$ Example 2: Quarter Circle Integral Evaluate: $$\iint_R (x^2+y^2)\, dA$$ ▶ Show Detailed Solution Step 1: $x^2+y^2 = r^2$ Limits: $0 \le r \le 3$ $0 \le \theta \le \frac{\pi}{2}$ $$ \int_0^{\pi/2}\int_0^3 r^3\,dr\,d\theta $$ $$ \int_0^3 r^3 dr = \frac{81}{4} $$ $$ \frac{81}{4} \cdot \frac{\pi}{2} = \frac{81\pi}{8} $$ Final Answer: $\frac{81\pi}{8}$ Example 3: Annular Region Evaluate: $$\iint_R 1\,dA$$ ▶ ...

Limits of Functions of Several Variables Solved Examples with Collapsible Solutions The following problems illustrate different techniques used to evaluate limits of multivariable functions, including direct substitution, algebraic simplification, polar coordinates, and the path method. Example 1 $$ \lim_{(x,y)\to(0,0)} \frac{3x^2 - y^2 + 5}{x^2 + y^2 + 2} $$ Click to View Solution Since numerator and denominator are continuous at $(0,0)$, substitute directly: $$ \text{Numerator} = 5 $$ $$ \text{Denominator} = 2 $$ $$ \boxed{\text{Limit} = \frac{5}{2}} $$ Example 2 $$ \lim_{(x,y)\to(0,0)} \frac{x^2 - xy}{\sqrt{x}-\sqrt{y}} $$ Click to View Solution Factor the numerator: $$ x^2 - xy = x(x-y) $$ Use the identity: $$ x-y = (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) $$ After cancellation: $$ = x(\sqrt{x}+\sqrt{y}) $$ As $(x,y)\to(0,0)$: $$ \boxed{\text{Limit} = 0} $$ Example 3 $$ \lim_{(x,y)\to(0,0)} \frac{4xy^2}{x^2+y^2} $$ Method 1: P...

Advanced Numerical Problems on Partial Differentiation Problem 1: Laplace’s Equation Problem: Verify that \( u = x^3 - 3xy^2 \) satisfies Laplace’s equation. Show / Hide Solution \[ u_x = 3x^2 - 3y^2, \quad u_y = -6xy \] \[ u_{xx} = 6x, \quad u_{yy} = -6x \] \[ u_{xx} + u_{yy} = 6x - 6x = 0 \] Hence verified. Problem 2: Total Differential Problem: If \( z = e^{xy}\sin(x+y) \), find total differential \( dz \). Show / Hide Solution \[ dz = z_x dx + z_y dy \] \[ z_x = e^{xy}[y\sin(x+y)+\cos(x+y)] \] \[ z_y = e^{xy}[x\sin(x+y)+\cos(x+y)] \] \[ dz = e^{xy}[y\sin(x+y)+\cos(x+y)]dx \] \[ + e^{xy}[x\sin(x+y)+\cos(x+y)]dy \] Problem 3: Directional Derivative Problem: Find directional derivative of \( \phi=x^2y+yz^3 \) at (1,-1,1) in direction \(2\mathbf{i}-\mathbf{j}+2\mathbf{k}\). Show / Hide Solution \[ \nabla \phi = (2xy)\mathbf{i} + (x^2+z^3)\mathbf{j} + (3yz^2)\mathbf{k} \] At (1,-1,1): \[ \nabla \phi = (-2)\mathbf{i} + 2\mathbf{j} - 3\...

Numerical Examples on Partial Differentiation Example 1: First Order Partial Derivatives Problem: If \( z = 3x^2y + 2xy^3 - 5y \), find \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \). Show / Hide Solution Differentiate w.r.t \(x\): \[ \frac{\partial z}{\partial x} = 6xy + 2y^3 \] Differentiate w.r.t \(y\): \[ \frac{\partial z}{\partial y} = 3x^2 + 6xy^2 - 5 \] Example 2: Second Order Partial Derivatives Problem: If \( z = x^3y^2 + 4xy \), find \( \frac{\partial^2 z}{\partial x^2} \). Show / Hide Solution First derivative: \[ \frac{\partial z}{\partial x} = 3x^2y^2 + 4y \] Second derivative: \[ \frac{\partial^2 z}{\partial x^2} = 6xy^2 \] Example 3: Mixed Partial Derivatives Problem: If \( z = x^2y^3 \), verify mixed partial derivatives. Show / Hide Solution \[ \frac{\partial z}{\partial x} = 2xy^3 \] \[ \frac{\partial^2 z}{\partial y \partial x} = 6xy^2 \] \[ \frac{\partial z}{\partial y} = 3x^2y^2 \] ...

Applications of Double Integrals Double integration is a fundamental tool in multivariable calculus used to compute volumes, areas, mass, centroids, and physical quantities over two-dimensional regions. 1. Volume Under a Surface If $z = f(x,y)$ over region $R$, the volume is: $$ V = \iint_R f(x,y)\, dA $$ Illustration Surface 2. Area of a Plane Region The area of region $R$: $$ A = \iint_R 1\, dA $$ Example Region (Circle) R 3. Mass of a Lamina If density is $\rho(x,y)$: $$ M = \iint_R \rho(x,y)\, dA $$ If density is constant $\rho = k$: $$ M = k \cdot \text{Area}(R) $$ 4. Center of Mass (Centroid) $$ \bar{x} = \frac{1}{M} \iint_R x \rho(x,y)\, dA $$ $$ \bar{y} = \frac{1}{M} \iint_R y \rho(x,y)\, dA $$ Centroid Illustration (x̄, ȳ) 5. Moments of Inertia $$ I_x = \iint_R y^2 \rho(x,y)\, dA $$ $$ I_y = \iint_R x^2 \rho(x,y)\, dA $$ $$ I_0 = \iint_R (x^2 + y^2)\rho(x,y)\...