First Order Differential Equations using Separation of Variables
Advanced Application Problems (Separation of Variables)
The following examples illustrate applications of first order differential equations in population growth, cooling law, chemical reactions, and physics.
Example 11 (Population Growth): If population grows according to
\( \frac{dP}{dt}=kP \), find \(P(t)\).
View Solution
Separate variables
\[
\frac{dP}{P}=k\,dt
\]
Integrate
\[
\int \frac{dP}{P}=\int k\,dt
\]
\[
\ln P = kt + C
\]
\[
P = Ce^{kt}
\]
Example 12 (Newton's Law of Cooling):
\[
\frac{dT}{dt}=-k(T-T_s)
\]
Find \(T(t)\).
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 $f(0,0)=0$:
$$
\boxed{\text{Continuous at } (0,0)}
$$
