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 = LLT
where L is a lower triangular matrix.
The solution is obtained in two steps:
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
This diagram represents:
\[
z = f(x,y), \quad x = x(t), \quad y = y(t)
\]
\[
\frac{dz}{dt}
=
\frac{\partial z}{\partial x}\frac{dx}{dt}
+
\frac{\partial z}{\partial y}\frac{dy}{dt}
\]
The following problems illustrate different techniques used to evaluate limits of multivariable functions,
including direct substitution, algebraic simplification, polar coordinates, and the path method.
Question 2: Discuss the Continuity of the following functions at given points.
Solved Problems with Collapsible Solutions
To check continuity at (0,0), we verify whether:
$$
\lim_{(x,y)\to(0,0)} f(x,y) = f(0,0)
$$
If the limit does not exist or is not equal to the function value,
then the function is not continuous at (0,0).
Problem I
$$
f(x,y)=
\begin{cases}
\dfrac{2x^2y}{x^4+y^2}, & (x,y)\ne(0,0) \\
0, & (x,y)=(0,0)
\end{cases}
$$
Click to View Solution
Take the path $y=mx^2$:
Numerator:
$$
2x^2(mx^2)=2mx^4
$$
Denominator:
$$
x^4+m^2x^4=x^4(1+m^2)
$$
Thus,
$$
f(x,y)=\frac{2m}{1+m^2}
$$
This depends on $m$.
Example:
- If $m=0$, limit = 0
- If $m=1$, limit = 1
Since limit depends on path, it does not exist.
$$
\boxed{\text{Not Continuous at } (0,0)}
$$
Problem II
$$
f(x,y)=
\begin{cases}
\dfrac{2xy}{x^2+y^2}, & (x,y)\ne(0,0) \\
0, & (x,y)=(0,0)
\end{cases}
$$
Click to View Solution
Take the path $y=mx$:
$$
f(x,y)=\frac{2mx^2}{x^2(1+m^2)}
=\frac{2m}{1+m^2}
$$
This depends on $m$.
Therefore, the limit does not exist.
$$
\boxed{\text{Not Continuous at } (0,0)}
$$
Problem III
$$
f(x,y)=
\begin{cases}
\dfrac{x^2-y^2}{x^2+y^2}, & (x,y)\ne(0,0) \\
0, & (x,y)=(0,0)
\end{cases}
$$
Click to View Solution
Take the path $y=mx$:
$$
\frac{x^2-m^2x^2}{x^2+m^2x^2}
=
\frac{1-m^2}{1+m^2}
$$
Different values for different $m$:
- If $m=0$, limit = 1
- If $m=1$, limit = 0
Hence limit does not exist.
$$
\boxed{\text{Not Continuous at } (0,0)}
$$
Problem IV
$$
f(x,y)=
\begin{cases}
\cos\!\left(\dfrac{x^3-y^3}{x^2+y^2}\right), & (x,y)\ne(0,0) \\
1, & (x,y)=(0,0)
\end{cases}
$$
Click to View Solution
Use polar coordinates:
$$
x=r\cos\theta, \quad y=r\sin\theta
$$
Then,
$$
\frac{x^3-y^3}{x^2+y^2}
=
r(\cos^3\theta-\sin^3\theta)
$$
As $r\to0$:
$$
r(\cos^3\theta-\sin^3\theta)\to0
$$
Thus,
$$
\cos(0)=1
$$
Since limit equals function value:
$$
\boxed{\text{Continuous at } (0,0)}
$$
Problem V
$$
f(x,y)=
\begin{cases}
\dfrac{x^3y}{x^2+y^2}, & (x,y)\ne(0,0) \\
0, & (x,y)=(0,0)
\end{cases}
$$
Click to View Solution
Use polar coordinates:
$$
x=r\cos\theta, \quad y=r\sin\theta
$$
Then,
$$
\frac{r^3\cos^3\theta \cdot r\sin\theta}{r^2}
=
r^2\cos^3\theta\sin\theta
$$
As $r\to0$:
$$
\to0
$$
Since limit equals $f(0,0)=0$:
$$
\boxed{\text{Continuous at } (0,0)}
$$
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
2. Area of a Plane Region
The area of region $R$:
$$
A = \iint_R 1\, dA
$$
Example Region (Circle)
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
5. Moments of Inertia
$$
I_x = \iint_R y^2 \rho(x,y)\, dA
$$
$$
I_y = \iint_R x^2 \rho(x,y)\, dA
$$
$$
I_0 = \iint_R (x^2 + y^2)\rho(x,y)\, dA
$$
6. Probability Applications
If $f(x,y)$ is a joint probability density function:
$$
P((x,y)\in R) = \iint_R f(x,y)\, dA
$$
$$
\iint_{\mathbb{R}^2} f(x,y)\, dA = 1
$$
7. Surface Area
If $z = f(x,y)$:
$$
A = \iint_R
\sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2
+ \left(\frac{\partial f}{\partial y}\right)^2} \, dA
$$
