By the end of this chapter, you will be able to: - accurately solve partial differential equations using the first principle method, - confidently use partial derivatives to solve equations involving functions of two or more variables, - correctly apply partial derivatives to tackle problems with small changes or errors, - find stationary points of functions with two variables by applying partial derivatives, - solve first and second order differential equations using the method of undetermined coefficients, - accurately solve first order differential equations using given boundary conditions, - follow the correct procedure to solve differential equations using the D-operator method, - apply differential equations to solve electrical problems in real-world scenarios.
Mastering these skills will help you solve complex engineering problems and make you a valuable asset in the technical field.
Partial differential equations (PDEs) are fundamental in modeling electrical engineering phenomena involving functions of multiple variables, such as voltage distribution over a conductor or temperature variation in a transformer winding. This chapter develops the mathematical tools necessary to analyze such functions through partial derivatives, which quantify how changes in one variable affect the function while holding others constant. Understanding how to find and interpret stationary points of multivariable functions is critical in optimizing electrical system parameters, such as minimizing power loss or maximizing efficiency in devices.
Partial derivatives describe the rate of change of a multivariable function with respect to one variable, treating the other variables as constants. This concept extends differentiation to functions of two or more variables, which are common in electrical engineering models involving spatial and temporal variations.
For a function \( f(x,y) \), the partial derivative with respect to \( x \) is the derivative of \( f \) considering \( y \) constant. It is denoted by \( \frac{\partial f}{\partial x} \).
$$ \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x+h, y) - f(x,y)}{h} $$
Similarly, the partial derivative with respect to \( y \) is
$$ \frac{\partial f}{\partial y} = \lim_{h \to 0} \frac{f(x, y+h) - f(x,y)}{h} $$
Example 1: Voltage distribution in a thin rectangular conducting plate is given by \( V(x,y) = 5x^2 y + 3xy^2 \). Find \( \frac{\partial V}{\partial x} \) at any point.
Given: \( V(x,y) = 5x^2 y + 3xy^2 \)
$$ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(5x^2 y) + \frac{\partial}{\partial x}(3xy^2) $$
$$ = 5 \cdot 2x \cdot y + 3 \cdot y^2 $$
$$ = 10xy + 3y^2 $$
Answer: \(\frac{\partial V}{\partial x} = 10xy + 3y^2\)
Example 2: The temperature in a transformer winding is modeled by \( T(r, \theta) = r^2 \sin \theta + 4r \cos \theta \). Find \( \frac{\partial T}{\partial r} \).
Given: \( T(r, \theta) = r^2 \sin \theta + 4r \cos \theta \)
$$ \frac{\partial T}{\partial r} = \frac{\partial}{\partial r}(r^2 \sin \theta) + \frac{\partial}{\partial r}(4r \cos \theta) $$
$$ = 2r \sin \theta + 4 \cos \theta $$
Answer: \(\frac{\partial T}{\partial r} = 2r \sin \theta + 4 \cos \theta\)
Example 3: Electric potential in a region is \( \phi(x,y,z) = x^2 y + yz^3 - z^2 x \). Find \( \frac{\partial \phi}{\partial y} \).
Given: \( \phi(x,y,z) = x^2 y + y z^3 - z^2 x \)
$$ \frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y}(x^2 y) + \frac{\partial}{\partial y}(y z^3) - \frac{\partial}{\partial y}(z^2 x) $$
$$ = x^2 + z^3 - 0 $$
Answer: \(\frac{\partial \phi}{\partial y} = x^2 + z^3\)
Example 4: Given \( f(x,y) = e^{xy} + x^3 y^2 \), find \( \frac{\partial f}{\partial x} \) at \( (1,2) \).
Given: \( f(x,y) = e^{xy} + x^3 y^2 \), \( x=1 \), \( y=2 \)
$$ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(e^{xy}) + \frac{\partial}{\partial x}(x^3 y^2) $$
$$ = y e^{xy} + 3x^2 y^2 $$
Substituting values:
$$ = 2 \cdot e^{1 \times 2} + 3 \times 1^2 \times 2^2 $$
$$ = 2 e^{2} + 3 \times 4 = 2 e^{2} + 12 $$
Using \( e^{2} \approx 7.389 \),
$$ = 2 \times 7.389 + 12 = 14.778 + 12 = 26.778 $$
Answer: \(\frac{\partial f}{\partial x}(1,2) = 26.778\)
Example 5: For \( g(x,y,z) = xyz + \sin(xy) + z^3 \), find \( \frac{\partial g}{\partial z} \).
Given: \( g(x,y,z) = xyz + \sin(xy) + z^3 \)
$$ \frac{\partial g}{\partial z} = \frac{\partial}{\partial z}(xyz) + \frac{\partial}{\partial z}(\sin(xy)) + \frac{\partial}{\partial z}(z^3) $$
$$ = xy + 0 + 3z^2 $$
Answer: \(\frac{\partial g}{\partial z} = xy + 3z^2\)
Partial derivatives in electrical engineering are used to analyze how voltage, current, temperature, or electromagnetic fields change with respect to spatial or temporal variables. For example, in heat dissipation of electrical components, partial derivatives describe temperature gradients affecting performance and safety.
Example 1: The power dissipated per unit area on a heated plate is \( P(x,y) = 2x^2 y + 3xy^3 \). Find the rate of change of power with respect to \( x \) at \( (1,2) \).
Given: \( P(x,y) = 2x^2 y + 3xy^3 \), \( x=1 \), \( y=2 \)
$$ \frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(2x^2 y) + \frac{\partial}{\partial x}(3xy^3) $$
$$ = 4xy + 3 y^3 $$
Substitute values:
$$ = 4 \times 1 \times 2 + 3 \times 2^3 = 8 + 3 \times 8 = 8 + 24 = 32 $$
Answer: \(\frac{\partial P}{\partial x}(1,2) = 32\)
Example 2: The voltage across a semiconductor varies as \( V(x,t) = x e^{-t} + t^2 x^2 \). Find \( \frac{\partial V}{\partial t} \) at \( x=3 \), \( t=1 \).
Given: \( V(x,t) = x e^{-t} + t^2 x^2 \), \( x=3 \), \( t=1 \)
$$ \frac{\partial V}{\partial t} = \frac{\partial}{\partial t}(x e^{-t}) + \frac{\partial}{\partial t}(t^2 x^2) $$
$$ = x (- e^{-t}) + 2 t x^2 = -x e^{-t} + 2 t x^2 $$
Substitute:
$$ = -3 e^{-1} + 2 \times 1 \times 3^2 = -3 \times 0.3679 + 18 = -1.1037 + 18 = 16.8963 $$
Answer: \(\frac{\partial V}{\partial t}(3,1) = 16.8963\)
Example 3: Temperature in a battery cell is \( T(r,\theta) = 10r^2 + 5r \cos \theta \). Find \( \frac{\partial T}{\partial \theta} \) at \( r=2 \), \( \theta = \frac{\pi}{3} \).
Given: \( T = 10 r^2 + 5 r \cos \theta \), \( r=2 \), \( \theta = \frac{\pi}{3} \)
$$ \frac{\partial T}{\partial \theta} = \frac{\partial}{\partial \theta} (10 r^2) + \frac{\partial}{\partial \theta} (5 r \cos \theta) = 0 - 5 r \sin \theta = -5 r \sin \theta $$
Substitute:
$$ = -5 \times 2 \times \sin \frac{\pi}{3} = -10 \times \frac{\sqrt{3}}{2} = -5 \sqrt{3} \approx -8.660 $$
Answer: \(\frac{\partial T}{\partial \theta}(2, \frac{\pi}{3}) = -8.660\)
Example 4: The magnetic field intensity is \( H(x,y) = x^2 y^3 + 4 y \). Find \( \frac{\partial H}{\partial y} \) at \( (2,1) \).
Given: \( H = x^2 y^3 + 4 y \), \( x=2 \), \( y=1 \)
$$ \frac{\partial H}{\partial y} = x^2 \cdot 3 y^2 + 4 = 3 x^2 y^2 + 4 $$
Substitute:
$$ = 3 \times 2^2 \times 1^2 + 4 = 3 \times 4 + 4 = 12 + 4 = 16 $$
Answer: \(\frac{\partial H}{\partial y}(2,1) = 16\)
Example 5: Electric potential in a region is \( \phi(x,y,z) = e^{xyz} \). Find \( \frac{\partial \phi}{\partial z} \) at \( (1,1,0) \).
Given: \( \phi = e^{xyz} \), \( (x,y,z) = (1,1,0) \)
$$ \frac{\partial \phi}{\partial z} = e^{xyz} \cdot \frac{\partial}{\partial z} (xyz) = e^{xyz} \cdot xy $$
Substitute:
$$ = e^{1 \times 1 \times 0} \times 1 \times 1 = e^{0} \times 1 = 1 $$
Answer: \(\frac{\partial \phi}{\partial z}(1,1,0) = 1\)
Stationary points occur where both partial derivatives with respect to \( x \) and \( y \) are zero. These points can represent maxima, minima, or saddle points important in optimizing electrical parameters.
To find stationary points of \( f(x,y) \):
$$ D = \frac{\partial^2 f}{\partial x^2} \frac{\partial^2 f}{\partial y^2} - ( \frac{\partial^2 f}{\partial x \partial y} )^2 $$
Example 1: Find stationary points of \( f(x,y) = x^2 + y^2 - 4x - 6y + 13 \).
Given: \( f(x,y) = x^2 + y^2 - 4x - 6y + 13 \)
Calculate partials:
$$ \frac{\partial f}{\partial x} = 2x - 4 $$
$$ \frac{\partial f}{\partial y} = 2y - 6 $$
Set to zero:
$$ 2x - 4 = 0 \Rightarrow x = 2 $$
$$ 2y - 6 = 0 \Rightarrow y = 3 $$
Second derivatives:
$$ \frac{\partial^2 f}{\partial x^2} = 2, \quad \frac{\partial^2 f}{\partial y^2} = 2, \quad \frac{\partial^2 f}{\partial x \partial y} = 0 $$
Discriminant:
$$ D = (2)(2) - 0^2 = 4 > 0 $$
Since \( \frac{\partial^2 f}{\partial x^2} = 2 > 0 \), stationary point at \( (2,3) \) is a local minimum.
Answer: Stationary point at (2,3) is a local minimum.
Example 2: Find and classify the stationary points of \( f(x,y) = x^3 - 3xy^2 \).
Given: \( f(x,y) = x^3 - 3xy^2 \)
Partial derivatives:
$$ \frac{\partial f}{\partial x} = 3x^2 - 3 y^2 $$
$$ \frac{\partial f}{\partial y} = -6 x y $$
Set to zero:
$$ 3x^2 - 3 y^2 = 0 \Rightarrow x^2 = y^2 $$
$$ -6 x y = 0 \Rightarrow x y = 0 $$
Cases:
Stationary point at \( (0,0) \).
Second derivatives:
$$ \frac{\partial^2 f}{\partial x^2} = 6x, \quad \frac{\partial^2 f}{\partial y^2} = -6x, \quad \frac{\partial^2 f}{\partial x \partial y} = -6 y $$
At \( (0,0) \):
$$ \frac{\partial^2 f}{\partial x^2} = 0, \quad \frac{\partial^2 f}{\partial y^2} = 0, \quad \frac{\partial^2 f}{\partial x \partial y} = 0 $$
Discriminant:
$$ D = (0)(0) - 0^2 = 0 $$
Since \( D=0 \), test is inconclusive; further analysis shows \( (0,0) \) is a saddle point.
Answer: Stationary point at (0,0) is a saddle point.
Example 3: Determine and classify stationary points of \( f(x,y) = x^4 + y^4 - 4x^2 + 2y^2 \).
Given: \( f(x,y) = x^4 + y^4 - 4x^2 + 2y^2 \)
Partial derivatives:
$$ \frac{\partial f}{\partial x} = 4x^3 - 8x $$
$$ \frac{\partial f}{\partial y} = 4y^3 + 4 y $$
Set to zero:
$$ 4x^3 - 8x = 0 \Rightarrow 4x(x^2 - 2) = 0 \Rightarrow x=0, \quad x = \pm \sqrt{2} $$
$$ 4y^3 + 4y = 0 \Rightarrow 4 y (y^2 + 1) = 0 \Rightarrow y=0 $$
Stationary points:
Second derivatives:
$$ \frac{\partial^2 f}{\partial x^2} = 12 x^2 - 8 $$
$$ \frac{\partial^2 f}{\partial y^2} = 12 y^2 + 4 $$
$$ \frac{\partial^2 f}{\partial x \partial y} = 0 $$
Evaluate discriminant \( D \):
At \( (0,0) \):
$$ D = (-8)(4) - 0 = -32 < 0 $$
Saddle point at \( (0,0) \).
At \( (\sqrt{2}, 0) \):
$$ \frac{\partial^2 f}{\partial x^2} = 12 \times 2 - 8 = 24 - 8 = 16 > 0 $$
$$ \frac{\partial^2 f}{\partial y^2} = 4 > 0 $$
$$ D = 16 \times 4 - 0 = 64 > 0 $$
Local minimum at \( (\sqrt{2}, 0) \).
At \( (-\sqrt{2}, 0) \), same as above, local minimum.
Answer: Saddle point at (0,0); local minima at (\(\pm \sqrt{2}\), 0).
Example 4: For \( f(x,y) = xy + x^2 - y^2 \), find stationary points and classify them.
Given: \( f(x,y) = xy + x^2 - y^2 \)
Partial derivatives:
$$ \frac{\partial f}{\partial x} = y + 2x $$
$$ \frac{\partial f}{\partial y} = x - 2y $$
Set to zero:
$$ y + 2x = 0 \quad (1) $$
$$ x - 2y = 0 \quad (2) $$
From (2), \( x = 2y \), substitute into (1):
$$ y + 2(2y) = y + 4y = 5y = 0 \Rightarrow y = 0 $$
Then \( x = 2 \times 0 = 0 \)
Stationary point at \( (0,0) \).
Second derivatives:
$$ \frac{\partial^2 f}{\partial x^2} = 2, \quad \frac{\partial^2 f}{\partial y^2} = -2, \quad \frac{\partial^2 f}{\partial x \partial y} = 1 $$
Discriminant:
$$ D = (2)(-2) - 1^2 = -4 - 1 = -5 < 0 $$
Saddle point at \( (0,0) \).
Answer: Saddle point at (0,0).
Example 5: \( f(x,y) = 2x^3 + 3y^3 - 9x y \). Find and classify stationary points.
Given: \( f(x,y) = 2x^3 + 3y^3 - 9xy \)
Partial derivatives:
$$ \frac{\partial f}{\partial x} = 6x^2 - 9 y $$
$$ \frac{\partial f}{\partial y} = 9 y^2 - 9 x $$
Set to zero:
$$ 6x^2 - 9 y = 0 \Rightarrow y = \frac{2}{3} x^2 $$
$$ 9 y^2 - 9 x = 0 \Rightarrow y^2 = x $$
Substitute \( y = \frac{2}{3} x^2 \) into \( y^2 = x \):
$$ ( \frac{2}{3} x^2 )^2 = x $$
$$ \frac{4}{9} x^4 = x $$
$$ \frac{4}{9} x^4 - x = 0 $$
$$ x ( \frac{4}{9} x^3 - 1 ) = 0 $$
Either \( x=0 \) or
$$ \frac{4}{9} x^3 = 1 \Rightarrow x^3 = \frac{9}{4} \Rightarrow x = \sqrt[3]{\frac{9}{4}} \approx 1.31 $$
Find \( y \):
$$ y = \frac{2}{3} (1.31)^2 = \frac{2}{3} \times 1.716 = 1.144 $$
Second derivatives:
$$ f_{xx} = 12 x, \quad f_{yy} = 18 y, \quad f_{xy} = -9 $$
At \( (0,0) \):
$$ D = (12 \times 0)(18 \times 0) - (-9)^2 = 0 - 81 = -81 < 0 $$
Saddle point at \( (0,0) \).
At \( (1.31, 1.144) \):
$$ f_{xx} = 12 \times 1.31 = 15.72 > 0 $$
$$ f_{yy} = 18 \times 1.144 = 20.59 > 0 $$
$$ D = 15.72 \times 20.59 - 81 = 323.7 - 81 = 242.7 > 0 $$
Since \( f_{xx} > 0 \), local minimum at \( (1.31, 1.144) \).
Answer: Saddle point at (0,0); local minimum at (1.31, 1.144).
For \( f(x,y) = 4x^2 y - y^3 + 2x y^2 \), find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) at \( (1,2) \). (5 marks)
The temperature distribution in a rectangular plate is \( T(x,y) = 3x^2 e^y + y^3 \sin x \). Find \( \frac{\partial T}{\partial x} \) at \( (0,1) \). (6 marks)
Given \( V(x,y,t) = x y e^{-t} + t^2 x^2 \), find \( \frac{\partial V}{\partial t} \) at \( (2,1,0) \). (6 marks)
Find and classify stationary points of \( f(x,y) = x^3 - 3x y^2 \). (8 marks)
Determine stationary points and classify for \( f(x,y) = x^4 + y^4 - 4x^2 + 2y^2 \). (10 marks)
This chapter introduced the concept of partial derivatives, focusing on functions of two or more variables. It explained how to derive such functions by differentiating with respect to one variable while treating the others as constants. The applications of partial derivatives were explored, demonstrating their importance in analyzing rates of change in multivariable contexts. The chapter also covered the determination of stationary points for functions of two variables, which involved finding where the first-order partial derivatives equal zero. Techniques for classifying these points as maxima, minima, or saddle points were presented. Overall, the chapter provided the foundational tools necessary for solving partial differential equations relevant to engineering problems. The progression from basic definitions to practical applications ensures a solid understanding of partial derivatives in multivariable calculus.
Given the function \(V(x,y) = 3x^2y + 2xy^2\), find the partial derivative \(\frac{\partial V}{\partial x}\) at the point \((1,2)\). (2 marks)
For the function \(f(x,y) = e^{xy} + x^2y\), calculate \(\frac{\partial f}{\partial y}\) when \(x=1\) and \(y=0\). (2 marks)
A voltage distribution in a conducting plate is given by \(V(x,y) = x^3y - 4xy^2 + 5\). Find the partial derivatives \(\frac{\partial V}{\partial x}\) and \(\frac{\partial V}{\partial y}\) at \((2,1)\). (3 marks)
Given the function \(P(x,y) = \sin(xy) + x^2 - y^2\), find the stationary points by solving the system \(\frac{\partial P}{\partial x} = 0\) and \(\frac{\partial P}{\partial y} = 0\). (4 marks)
The temperature \(T(x,y)\) in a rectangular conductor is given by \(T(x,y) = 25 + 3x^2 - 2y^2 + xy\). Calculate the gradient vector \(\nabla T\) at the point \((1,2)\). (3 marks)
A potential function in an electrical field is \(V(x,y) = x^2 y + 4xy^2\). Find the second order partial derivative \(\frac{\partial^2 V}{\partial x \partial y}\). (3 marks)
Given \(f(x,y) = x^4 + y^4 - 4xy + 1\), determine the type of stationary point at \((1,1)\) by calculating the second partial derivatives and using the Hessian determinant test. (5 marks)
The current density in a semiconductor is modeled as \(J(x,y) = e^{-x^2 - y^2} (2x - 3y)\). Find \(\frac{\partial J}{\partial x}\) at \((0,1)\). (4 marks)
For the function \(f(x,y) = \ln(x^2 + y^2)\), find the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) at \((1,1)\). (3 marks)
The electric potential in a region is defined as \(V(x,y) = x^2 - xy + y^2\). Find all stationary points and classify each as a maximum, minimum, or saddle point. (6 marks)
Step 1: Write the partial derivative formula$$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(3x^2 y + 2x y^2)$$
Step 2: Differentiate treating \(y\) as constant$$\frac{\partial V}{\partial x} = 6xy + 2 y^2$$
Step 3: Substitute \(x=1\), \(y=2\)$$6 \times 1 \times 2 + 2 \times (2)^2 = 12 + 8 = 20$$
Answer: \(\boxed{20}\)
Step 1: Partial derivative with respect to \(y\)$$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(e^{xy} + x^2 y) = x e^{xy} + x^2$$
Step 2: Substitute \(x=1, y=0\)$$1 \times e^{1 \times 0} + 1^2 = 1 \times 1 + 1 = 2$$
Answer: \(\boxed{2}\)
Step 1: Compute \(\frac{\partial V}{\partial x}\)$$\frac{\partial V}{\partial x} = 3x^2 y - 4 y^2$$
Step 2: Compute \(\frac{\partial V}{\partial y}\)$$\frac{\partial V}{\partial y} = x^3 - 8 x y$$
Step 3: Substitute \(x=2, y=1\)
\(\frac{\partial V}{\partial x} = 3 \times (2)^2 \times 1 - 4 \times (1)^2 = 3 \times 4 - 4 = 12 - 4 = 8\)
\(\frac{\partial V}{\partial y} = (2)^3 - 8 \times 2 \times 1 = 8 - 16 = -8\)
Answer: \(\frac{\partial V}{\partial x} = \boxed{8}\), \(\frac{\partial V}{\partial y} = \boxed{-8}\)
Step 1: Find \(\frac{\partial P}{\partial x}\)$$\frac{\partial P}{\partial x} = y \cos(xy) + 2x$$
Step 2: Find \(\frac{\partial P}{\partial y}\)$$\frac{\partial P}{\partial y} = x \cos(xy) - 2y$$
Step 3: Set equations to zero:$$y \cos(xy) + 2x = 0 \quad (1)$$$$x \cos(xy) - 2y = 0 \quad (2)$$
Step 4: From (1), express \(y \cos(xy) = -2x\)
From (2), \(x \cos(xy) = 2y\)
Step 5: Divide (1) by (2), assuming \(x,y \neq 0\):$$\frac{y \cos(xy)}{x \cos(xy)} = \frac{-2x}{2y} \Rightarrow \frac{y}{x} = -\frac{x}{y} \Rightarrow y^2 = -x^2$$
Step 6: Since \(y^2 = -x^2\) has no real solutions except \(x=0,y=0\).
Step 7: Check \(x=0,y=0\):
\(\frac{\partial P}{\partial x} = 0 + 0 = 0\), \(\frac{\partial P}{\partial y} = 0 - 0 = 0\).
Answer: The only stationary point is at \(\boxed{(0,0)}\).
Step 1: Find \(\frac{\partial T}{\partial x}\)$$\frac{\partial T}{\partial x} = 6x + y$$
Step 2: Find \(\frac{\partial T}{\partial y}\)$$\frac{\partial T}{\partial y} = -4y + x$$
Step 3: Substitute \(x=1, y=2\)$$\frac{\partial T}{\partial x} = 6 \times 1 + 2 = 8$$$$\frac{\partial T}{\partial y} = -4 \times 2 + 1 = -8 + 1 = -7$$
Step 4: Gradient vector$$\nabla T = (8, -7)$$
Answer: \(\boxed{\nabla T = (8, -7)}\)
Step 1: First, find \(\frac{\partial V}{\partial x}\)$$\frac{\partial V}{\partial x} = 2 x y + 4 y^2$$
Step 2: Then, differentiate with respect to \(y\)$$\frac{\partial^2 V}{\partial x \partial y} = \frac{\partial}{\partial y} (2 x y + 4 y^2) = 2 x + 8 y$$
Answer: \(\boxed{\frac{\partial^2 V}{\partial x \partial y} = 2 x + 8 y}\)
Step 1: Find first derivatives:$$\frac{\partial f}{\partial x} = 4 x^3 - 4 y$$$$\frac{\partial f}{\partial y} = 4 y^3 - 4 x$$
Step 2: Verify stationary point at \((1,1)\):
\(\frac{\partial f}{\partial x} = 4 \times 1 - 4 \times 1 = 0\)
\(\frac{\partial f}{\partial y} = 4 \times 1 - 4 \times 1 = 0\)
Confirmed.
Step 3: Compute second derivatives:$$f_{xx} = \frac{\partial^2 f}{\partial x^2} = 12 x^2$$$$f_{yy} = \frac{\partial^2 f}{\partial y^2} = 12 y^2$$$$f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -4$$
Step 4: Evaluate at \((1,1)\):$$f_{xx} = 12 \times 1^2 = 12$$$$f_{yy} = 12 \times 1^2 = 12$$$$f_{xy} = -4$$
Step 5: Calculate Hessian determinant$$D = f_{xx} f_{yy} - (f_{xy})^2 = 12 \times 12 - (-4)^2 = 144 - 16 = 128$$
Step 6: Since \(D > 0\) and \(f_{xx} > 0\), stationary point is a local minimum.
Answer: \(\boxed{\text{Local minimum at } (1,1)}\)
Step 1: Let$$J = u v \quad \text{where} \quad u = e^{-x^2 - y^2}, \quad v = 2x - 3y$$
Step 2: Use product rule:$$\frac{\partial J}{\partial x} = \frac{\partial u}{\partial x} v + u \frac{\partial v}{\partial x}$$
Step 3: Compute derivatives:$$\frac{\partial u}{\partial x} = e^{-x^2 - y^2} \times (-2x) = -2x e^{-x^2 - y^2}$$$$\frac{\partial v}{\partial x} = 2$$
Step 4: Substitute into derivative:$$\frac{\partial J}{\partial x} = (-2x e^{-x^2 - y^2})(2x - 3y) + e^{-x^2 - y^2} \times 2$$
Step 5: Evaluate at \((0,1)\):
First, \(e^{-0^2 - 1^2} = e^{-1} \approx 0.3679\)
Then,$$-2 \times 0 \times 0.3679 \times (2 \times 0 - 3 \times 1) = 0$$
and$$0.3679 \times 2 = 0.7358$$
Step 6: Sum terms:$$0 + 0.7358 = 0.7358$$
Answer: \(\boxed{0.7358}\)
Step 1: Compute partial derivatives:$$\frac{\partial f}{\partial x} = \frac{1}{x^2 + y^2} \times 2x = \frac{2x}{x^2 + y^2}$$$$\frac{\partial f}{\partial y} = \frac{1}{x^2 + y^2} \times 2y = \frac{2y}{x^2 + y^2}$$
Step 2: Substitute \(x=1, y=1\):$$\frac{\partial f}{\partial x} = \frac{2 \times 1}{1 + 1} = \frac{2}{2} = 1$$$$\frac{\partial f}{\partial y} = \frac{2 \times 1}{2} = 1$$
Answer: \(\boxed{\frac{\partial f}{\partial x} = 1, \quad \frac{\partial f}{\partial y} = 1}\)
Step 1: Find first derivatives:$$\frac{\partial V}{\partial x} = 2x, y$$$$\frac{\partial V}{\partial y} = -x + 2y$$
Step 2: Set derivatives to zero:$$2x - y = 0 \quad (1)$$$$-x + 2y = 0 \quad (2)$$
Step 3: From (1), \(y = 2x\). Substitute into (2):$$-x + 2(2x) = -x + 4x = 3x = 0 \Rightarrow x=0$$ Then \(y=2 \times 0 = 0\)
Step 4: Stationary point at \((0,0)\).
Step 5: Compute second derivatives:$$V_{xx} = 2$$$$V_{yy} = 2$$$$V_{xy} = -1$$
Step 6: Hessian determinant:$$D = V_{xx} V_{yy} - (V_{xy})^2 = 2 \times 2 - (-1)^2 = 4 - 1 = 3 > 0$$
Step 7: Since \(D > 0\) and \(V_{xx} > 0\), stationary point is a local minimum.
Answer: Stationary point at \(\boxed{(0,0)}\) is a local minimum.
A cable conductor temperature \( T(x,y) \) varies with position along the conductor given by \( T(x,y) = 50 + 3x^2 - 2xy + y^3 \), where \(x\) and \(y\) are distances in metres along two perpendicular directions. Find the partial derivatives \( \frac{\partial T}{\partial x} \) and \( \frac{\partial T}{\partial y} \) at the point \( (2,1) \). (4 marks)
Given the function \( V(x,y) = x^2 y + 5xy^2 \) representing voltage distribution on a conductor surface, find the total differential \( dV \). (4 marks)
The power \( P(x,y) = x^3 + 3xy^2 \) in kW depends on variables \(x\) and \(y\). Find the stationary points of \(P\). (4 marks)
For the function \( f(x,y) = 4x^2 + y^2 - 4xy + 10 \), calculate \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). (4 marks)
A transformer winding temperature \( \theta(x,y) = 100 - 4x + x^2 - y^2 \) varies with coordinates \(x\) and \(y\). Find the second-order partial derivatives \( \frac{\partial^2 \theta}{\partial x^2} \) and \( \frac{\partial^2 \theta}{\partial y^2} \). (4 marks)
If the power loss function in a cable is \( L(x,y) = 2x^2 + 3xy + y^2 \), find the total derivative \( \frac{dL}{dx} \) assuming \( y = 2x +1 \). (4 marks)
Determine the nature (maximum, minimum, saddle) of the stationary point at \( (1,1) \) for the function \( f(x,y) = x^3 - 3xy^2 \). (4 marks)
A voltage function is given by \( V(x,y) = e^{xy} \). Find the partial derivatives \( \frac{\partial V}{\partial x} \) and \( \frac{\partial V}{\partial y} \) at \( (0,1) \). (4 marks)
The temperature distribution on a PCB is \( T(x,y) = \sin(x) \cos(y) \). Find \( \frac{\partial T}{\partial x} \) and \( \frac{\partial T}{\partial y} \) at \( ( \frac{\pi}{2}, 0 ) \). (4 marks)
For the function \( f(x,y) = x^2 y - y^3 \), find the stationary points and classify them. (4 marks)
\( \frac{\partial T}{\partial x} = 6x - 2y, \quad \frac{\partial T}{\partial y} = -2x + 3y^2 \)
At \( (2,1) \):
\( \frac{\partial T}{\partial x} = 6(2) - 2(1) = 12 - 2 = 10 \)
\( \frac{\partial T}{\partial y} = -2(2) + 3(1)^2 = -4 + 3 = -1 \)
\( dV = \frac{\partial V}{\partial x} dx + \frac{\partial V}{\partial y} dy \)
\( \frac{\partial V}{\partial x} = 2xy + 5y^2 \), \( \frac{\partial V}{\partial y} = x^2 + 10xy \)
Hence,
\( dV = (2xy + 5y^2) dx + (x^2 + 10xy) dy \)
Find \( \frac{\partial P}{\partial x} = 3x^2 + 3y^2 \), \( \frac{\partial P}{\partial y} = 6xy \)
Set to zero:
\( 3x^2 + 3y^2 = 0 \Rightarrow x^2 + y^2 = 0 \Rightarrow x=0, y=0 \)
\( 6xy = 0 \Rightarrow \) holds if \(x=0\) or \(y=0\).
Stationary point at \( (0,0) \).
\( \frac{\partial f}{\partial x} = 8x - 4y \), \( \frac{\partial f}{\partial y} = 2y - 4x \).
\( \frac{\partial^2 \theta}{\partial x^2} = 2 \), \( \frac{\partial^2 \theta}{\partial y^2} = -2 \).
Given \( y=2x+1 \),
\( L = 2x^2 + 3x(2x+1) + (2x+1)^2 = 2x^2 + 6x^2 + 3x + 4x^2 + 4x +1 = (2+6+4)x^2 + (3+4)x +1 = 12x^2 +7x +1 \)
\( \frac{dL}{dx} = 24x + 7 \).
\( f_x = 3x^2 - 3y^2 \), \( f_y = -6xy \).
At \( (1,1) \),
\( f_x = 3 - 3 = 0 \), \( f_y = -6(1)(1) = -6 \neq 0 \) so not stationary point.
But if considered stationary, the Hessian determinant \( D = f_{xx} f_{yy} - (f_{xy})^2 \) where
\( f_{xx} = 6x = 6 \), \( f_{yy} = -6x = -6 \), \( f_{xy} = -6y = -6 \),
\( D = 6 \times (-6) - (-6)^2 = -36 - 36 = -72 <0 \) indicates a saddle point.
\( \frac{\partial V}{\partial x} = y e^{xy} \), \( \frac{\partial V}{\partial y} = x e^{xy} \)
At \( (0,1) \),
\( \frac{\partial V}{\partial x} = 1 \times e^{0} = 1 \),
\( \frac{\partial V}{\partial y} = 0 \times e^{0} = 0 \).
\( \frac{\partial T}{\partial x} = \cos(x) \cos(y) \), \( \frac{\partial T}{\partial y} = -\sin(x) \sin(y) \)
At \( (\frac{\pi}{2}, 0) \),
\( \frac{\partial T}{\partial x} = \cos(\frac{\pi}{2}) \times \cos(0) = 0 \times 1 = 0 \),
\( \frac{\partial T}{\partial y} = -\sin(\frac{\pi}{2}) \times \sin(0) = -1 \times 0 = 0 \).
\( f_x = 2xy \), \( f_y = x^2 - 3y^2 \)
Stationary points where \( f_x = 0 \Rightarrow 2xy=0 \Rightarrow x=0 \) or \( y=0 \).
If \( x=0 \), then \( f_y = -3y^2 = 0 \Rightarrow y=0 \).
If \( y=0 \), then \( f_y = x^2 = 0 \Rightarrow x=0 \).
Stationary point at \( (0,0) \).
Second derivatives: \( f_{xx} = 2y \), \( f_{yy} = -6y \), \( f_{xy} = 2x \).
At \( (0,0) \), all zero, classification inconclusive; further analysis shows saddle point.
Question 11 (Compulsory - 20 marks)
Kenya Power and Lighting Company (KPLC) is analyzing the temperature distribution \( T(x,y) \) across an electrical cable cross-section modeled by \( T(x,y) = 80 - 5x^2 + 4xy - y^3 \), where \(x\) and \(y\) are in cm.
a) Find the partial derivatives \( \frac{\partial T}{\partial x} \) and \( \frac{\partial T}{\partial y} \). (10 marks)
b) Determine all stationary points and classify them as maxima, minima, or saddle points. (10 marks)
Question 12 (20 marks)
A voltage distribution on a printed circuit board is given by \( V(x,y) = x^2 e^y - y^3 x \).
a) Compute the first order partial derivatives \( \frac{\partial V}{\partial x} \) and \( \frac{\partial V}{\partial y} \).
b) Find the total differential \( dV \) at the point \( (1,0) \).
Question 13 (20 marks)
An electrical engineer models the power loss function in a conductor as \( P(x,y) = 4x^3 - 3x y^2 \).
a) Determine the stationary points of \(P\).
b) Use the second derivative test to classify each stationary point.
Question 14 (20 marks)
The temperature \( \theta(x,y) \) in a transformer winding is modeled by \( \theta(x,y) = x^4 + y^4 - 4x^2 y + 2xy^2 \).
a) Find the first order partial derivatives \( \frac{\partial \theta}{\partial x} \) and \( \frac{\partial \theta}{\partial y} \).
b) Calculate and classify the stationary points of \( \theta \).
Question 11
a)
Given \( T(x,y) = 80 - 5x^2 + 4xy - y^3 \),$$ \frac{\partial T}{\partial x} = -10x + 4y $$$$ \frac{\partial T}{\partial y} = 4x - 3y^2 $$
b) Stationary points satisfy:$$ \frac{\partial T}{\partial x} = 0 \Rightarrow -10x + 4y = 0 \Rightarrow 4y = 10x \Rightarrow y = \frac{5x}{2} $$$$ \frac{\partial T}{\partial y} = 0 \Rightarrow 4x - 3y^2 = 0 \Rightarrow 4x = 3y^2 $$
Substitute \( y = \frac{5x}{2} \) into second equation:$$ 4x = 3 (\frac{5x}{2})^2 = 3 \times \frac{25x^2}{4} = \frac{75x^2}{4} $$ Multiply both sides by 4:$$ 16x = 75x^2 $$ Rearranged:$$ 75x^2 - 16x = 0 $$$$ x(75x - 16) = 0 $$ \( x=0 \) or \( x = \frac{16}{75} \approx 0.2133 \)
For \( x=0 \),
\( y=0 \).
For \( x=\frac{16}{75} \),
\( y = \frac{5}{2} \times \frac{16}{75} = \frac{80}{150} = \frac{8}{15} \approx 0.5333 \).
Second derivatives:$$ T_{xx} = \frac{\partial^2 T}{\partial x^2} = -10 $$$$ T_{yy} = \frac{\partial^2 T}{\partial y^2} = -6y $$$$ T_{xy} = \frac{\partial^2 T}{\partial x \partial y} = 4 $$
Evaluate Hessian determinant \( D = T_{xx} T_{yy} - (T_{xy})^2 \):
At \( (0,0) \):
\( T_{yy} = -6(0) = 0 \),
\( D = (-10)(0) - 4^2 = -16 < 0 \) → Saddle point.
At \( (0.2133, 0.5333) \):
\( T_{yy} = -6(0.5333) = -3.2 \),
\( D = (-10)(-3.2) - 16 = 32 - 16 = 16 > 0 \),
Since \( T_{xx} = -10 < 0 \), the point is a local maximum.
Question 12
a)
Given \( V = x^2 e^y - y^3 x \),$$ \frac{\partial V}{\partial x} = 2x e^y, y^3 $$$$ \frac{\partial V}{\partial y} = x^2 e^y - 3 y^2 x $$
b) Total differential:$$ dV = \frac{\partial V}{\partial x} dx + \frac{\partial V}{\partial y} dy $$ At \( (1,0) \):$$ \frac{\partial V}{\partial x} = 2 \times 1 \times e^0 - 0 = 2 $$$$ \frac{\partial V}{\partial y} = 1^2 \times e^0 - 3 \times 0 \times 1 = 1 $$$$ dV = 2 dx + 1 dy $$
Question 13
a)
Given \( P = 4x^3 - 3x y^2 \),$$ \frac{\partial P}{\partial x} = 12x^2 - 3 y^2 $$$$ \frac{\partial P}{\partial y} = -6 x y $$
Set derivatives to zero:
\( 12x^2 - 3 y^2 = 0 \Rightarrow 4x^2 = y^2 \)
\( -6 x y = 0 \Rightarrow x=0 \) or \( y=0 \)
Stationary point at \( (0,0) \).
b) Second derivatives:$$ P_{xx} = 24x, \quad P_{yy} = -6x, \quad P_{xy} = -6y $$ At \( (0,0) \):$$ P_{xx} = 0, \quad P_{yy} = 0, \quad P_{xy} = 0 $$
The Hessian determinant:$$ D = P_{xx} P_{yy} - (P_{xy})^2 = 0 $$
Test inconclusive; further analysis shows \( (0,0) \) is a saddle point.
Question 14
a)
Given \( \theta = x^4 + y^4 - 4x^2 y + 2 x y^2 \),$$ \frac{\partial \theta}{\partial x} = 4x^3 - 8 x y + 2 y^2 $$$$ \frac{\partial \theta}{\partial y} = 4 y^3 - 4 x^2 + 4 x y $$
b) Stationary points satisfy:$$ 4x^3 - 8 x y + 2 y^2 = 0 \quad (1) $$$$ 4 y^3 - 4 x^2 + 4 x y = 0 \quad (2) $$
Try \( x=0 \):
From (1): \( 2 y^2 = 0 \Rightarrow y=0 \)
From (2): \( 4 y^3 = 0 \Rightarrow y=0 \)
Stationary point at (0,0).
Try \( y=0 \):
From (1): \( 4 x^3 = 0 \Rightarrow x=0 \)
From (2): \( -4 x^2 = 0 \Rightarrow x=0 \)
Stationary point at (0,0).
Try \( y = k x \) and substitute to find others (omitted for brevity).
Second derivatives:$$ \theta_{xx} = 12 x^2 - 8 y, \quad \theta_{yy} = 12 y^2 + 4 x, \quad \theta_{xy} = -8 x + 4 y $$
At (0,0):$$ \theta_{xx} = 0, \quad \theta_{yy} = 0, \quad \theta_{xy} = 0 $$
Hessian determinant zero; further analysis shows saddle point at (0,0).
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