By the end of this chapter, you will be able to:
Mastering these skills will help you tackle complex engineering problems with confidence and precision in your trade.
Partial differential equations (PDEs) form the mathematical foundation for modeling and analyzing many phenomena in electronics engineering such as heat dissipation in circuits, electromagnetic wave propagation, and semiconductor device behavior. Mastery of partial derivatives and their applications allows engineers to predict system responses and optimize designs effectively. This chapter focuses on understanding partial derivatives of functions with multiple variables and solving related problems, a critical skill for electronics engineers working on advanced circuit simulations and signal processing in Kenya’s dynamic technology sector.
Partial derivatives measure how a multivariable function changes as one variable changes, holding other variables constant. They are essential in analyzing systems where multiple independent parameters influence outcomes, such as voltage and temperature effects on semiconductor performance. The formal definition of a partial derivative for a function \(f(x,y)\) is given by
$$ \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x+h,y) - f(x,y)}{h} $$
which represents the rate of change of \(f\) with respect to \(x\) at constant \(y\).
Calculating partial derivatives involves differentiating a function with respect to one variable while treating other variables as constants. For a function \(f(x,y)\), the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) describe how \(f\) changes along the \(x\) and \(y\) directions, respectively.
$$ \frac{\partial f}{\partial x} = \frac{d}{dx} f(x,y) \quad \text{with } y \text{ constant} $$
$$ \frac{\partial f}{\partial y} = \frac{d}{dy} f(x,y) \quad \text{with } x \text{ constant} $$
Example 1: A temperature distribution in a semiconductor chip is modeled by \(T(x,y) = 5x^{2}y + 3xy^{2}\). Find \(\frac{\partial T}{\partial x}\).
Given: \(T(x,y) = 5x^{2}y + 3xy^{2}\)
$$ \frac{\partial T}{\partial x} = \frac{\partial}{\partial x} (5x^{2}y + 3xy^{2}) $$
$$ = 5 \cdot 2x \cdot y + 3 \cdot y^{2} $$
$$ = 10xy + 3y^{2} $$
Answer: \(\frac{\partial T}{\partial x} = 10xy + 3y^{2}\)
Example 2: For the voltage across a resistor given by \(V(R, I) = IR^{2} + 4R + 2I\), find \(\frac{\partial V}{\partial R}\).
Given: \(V(R,I) = IR^{2} + 4R + 2I\)
$$ \frac{\partial V}{\partial R} = \frac{\partial}{\partial R} (IR^{2} + 4R + 2I) $$
$$ = I \cdot 2R + 4 + 0 $$
$$ = 2IR + 4 $$
Answer: \(\frac{\partial V}{\partial R} = 2IR + 4\)
Example 3: A signal strength function is \(S(t,f) = e^{-0.1t} \sin(2\pi f t)\). Find \(\frac{\partial S}{\partial t}\).
Given: \(S(t,f) = e^{-0.1t} \sin(2\pi f t)\)
Using product rule:
$$ \frac{\partial S}{\partial t} = \frac{\partial}{\partial t} (e^{-0.1t}) \sin(2\pi f t) + e^{-0.1t} \frac{\partial}{\partial t} \sin(2\pi f t) $$
$$ = (-0.1 e^{-0.1t}) \sin(2\pi f t) + e^{-0.1t} (2\pi f \cos(2\pi f t)) $$
Answer: \(\frac{\partial S}{\partial t} = -0.1 e^{-0.1t} \sin(2\pi f t) + 2\pi f e^{-0.1t} \cos(2\pi f t)\)
Example 4: An engineer at Kenya Bureau of Standards (KEBS) is analyzing a quality control function \(Q(x,y) = 2x^2y + 7xy^2\). Find \(\frac{\partial Q}{\partial x}\).
Given: \(Q(x,y) = 2x^2y + 7xy^2\)
$$ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (2x^2y + 7xy^2) $$$$ = 2 \cdot 2x \cdot y + 7 \cdot y^2 $$$$ = 4xy + 7y^2 $$
Answer: \(\frac{\partial Q}{\partial x} = 4xy + 7y^2\)
Partial derivatives are used extensively in electronics engineering to analyze how output variables depend on multiple input parameters. They enable optimization of device performance by studying sensitivity and gradients of functions representing physical quantities.
Example 1: The power dissipated in a transistor is given by \(P(V,I) = VI - 0.01V^{2}\). Find the rate of change of power with respect to voltage when \(V = 10\) V and \(I = 2\) A.
Given: \(P(V,I) = VI - 0.01V^{2}\), \(V=10\) V, \(I=2\) A
$$ \frac{\partial P}{\partial V} = \frac{\partial}{\partial V}(VI - 0.01V^{2}) = I - 0.02V $$
Substitute values:
$$ = 2 - 0.02 \times 10 $$
$$ = 2 - 0.2 = 1.8 $$
Answer: \(\frac{\partial P}{\partial V} = 1.8\) W/V
Example 2: The capacitance of a device depends on temperature and frequency as \(C(T,f) = 10 + 0.5T - 0.01f^{2}\). Find \(\frac{\partial C}{\partial T}\) at \(T = 30^\circ C\), \(f=50\) Hz.
Given: \(C(T,f) = 10 + 0.5T - 0.01f^{2}\)
$$ \frac{\partial C}{\partial T} = 0.5 $$
Answer: \(\frac{\partial C}{\partial T} = 0.5\) F/°C
Example 3: The resistance of a wire is modeled by \(R(L,A) = \rho \frac{L}{A}\), where \(\rho = 1.68 \times 10^{-8}\) \(\Omega\)m. Find \(\frac{\partial R}{\partial A}\) for \(L = 2\) m and \(A = 1 \times 10^{-6}\) m\(^2\).
Given: \(R(L,A) = \rho \frac{L}{A}\), \(\rho = 1.68 \times 10^{-8}\), \(L=2\), \(A=1 \times 10^{-6}\)
$$ \frac{\partial R}{\partial A} = \frac{\partial}{\partial A} ( \rho \frac{L}{A} ) = - \rho \frac{L}{A^{2}} $$
Substitute values:
$$ = -1.68 \times 10^{-8} \times \frac{2}{(1 \times 10^{-6})^{2}} $$
$$ = -1.68 \times 10^{-8} \times \frac{2}{1 \times 10^{-12}} = -1.68 \times 10^{-8} \times 2 \times 10^{12} $$
$$ = -3.36 \times 10^{4} \, \Omega/\text{m}^2 $$
Answer: \(\frac{\partial R}{\partial A} = -3.36 \times 10^{4}\) \(\Omega\)/m\(^2\)
Example 4: At Jomo Kenyatta University of Agriculture and Technology (JKUAT), the efficiency of a solar cell is modeled as \(E(T,I) = 0.8T + 0.05I^2\), where \(T\) is temperature in °C and \(I\) is current in A. Find \(\frac{\partial E}{\partial I}\) at \(T = 25\)°C, \(I = 3\) A.
Given: \(E(T,I) = 0.8T + 0.05I^2\)
$$ \frac{\partial E}{\partial I} = 0.1I $$ Substitute \(I = 3\):$$ = 0.1 \times 3 = 0.3 $$
Answer: \(\frac{\partial E}{\partial I} = 0.3\)
Stationary points occur where the partial derivatives of a function with respect to all variables equal zero. These points represent local maxima, minima, or saddle points critical in optimizing electronic circuit parameters.
The conditions for stationary points for \(f(x,y)\) are:
$$ \frac{\partial f}{\partial x} = 0, \quad \frac{\partial f}{\partial y} = 0 $$
Classification uses second-order partial derivatives and the discriminant:
$$ D = \frac{\partial^{2} f}{\partial x^{2}} \frac{\partial^{2} f}{\partial y^{2}} - (\frac{\partial^{2} f}{\partial x \partial y})^{2} $$
Example 1: Find and classify the 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 first partial derivatives:
$$ \frac{\partial f}{\partial x} = 2x - 4 $$
$$ \frac{\partial f}{\partial y} = 2y - 6 $$
Set derivatives to zero:
$$ 2x - 4 = 0 \Rightarrow x = 2 $$
$$ 2y - 6 = 0 \Rightarrow y = 3 $$
Calculate second derivatives:
$$ \frac{\partial^{2} f}{\partial x^{2}} = 2, \quad \frac{\partial^{2} f}{\partial y^{2}} = 2, \quad \frac{\partial^{2} f}{\partial x \partial y} = 0 $$
Calculate 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: Local minimum at \((2,3)\)
Example 2: Determine the stationary points of \(f(x,y) = x^{3} - 3xy^{2}\).
Given: \(f(x,y) = x^{3} - 3xy^{2}\)
First derivatives:
$$ \frac{\partial f}{\partial x} = 3x^{2} - 3y^{2} $$
$$ \frac{\partial f}{\partial y} = -6xy $$
Set to zero:
$$ 3x^{2} - 3y^{2} = 0 \Rightarrow x^{2} = y^{2} $$
$$ -6xy = 0 \Rightarrow x y = 0 $$
From \(xy=0\), either \(x=0\) or \(y=0\).
Stationary point at \((0,0)\).
Second derivatives:
$$ \frac{\partial^{2} f}{\partial x^{2}} = 6x, \quad \frac{\partial^{2} f}{\partial y^{2}} = -6x, \quad \frac{\partial^{2} f}{\partial x \partial y} = -6y $$
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:
$$ D = (0)(0) - 0^{2} = 0 $$
Test inconclusive.
Answer: Stationary point at \((0,0)\) is a saddle point (from graphing or higher order test).
Example 3: Find and classify stationary points of \(f(x,y) = x^{4} + y^{4} - 4xy + 1\).
Given: \(f(x,y) = x^{4} + y^{4} - 4xy + 1\)
First derivatives:
$$ \frac{\partial f}{\partial x} = 4x^{3} - 4y $$
$$ \frac{\partial f}{\partial y} = 4y^{3} - 4x $$
Set to zero:
$$ 4x^{3} - 4y = 0 \Rightarrow y = x^{3} $$
$$ 4y^{3} - 4x = 0 \Rightarrow y^{3} = x $$
Substitute \(y = x^{3}\) into second:
$$ (x^{3})^{3} = x \Rightarrow x^{9} = x $$
$$ x^{9} - x = 0 \Rightarrow x(x^{8} - 1) = 0 $$
Solutions:
$$ x = 0, \quad x^{8} = 1 \Rightarrow x = \pm 1 $$
Corresponding \(y\):
Second derivatives:
$$ \frac{\partial^{2} f}{\partial x^{2}} = 12x^{2}, \quad \frac{\partial^{2} f}{\partial y^{2}} = 12y^{2}, \quad \frac{\partial^{2} f}{\partial x \partial y} = -4 $$
Calculate discriminant \(D\) at each point:
$$ D = (12 \times 0)(12 \times 0) - (-4)^{2} = 0 - 16 = -16 < 0 $$
Saddle point.
$$ D = (12 \times 1)(12 \times 1) - (-4)^{2} = 144 - 16 = 128 > 0 $$
Since \(\frac{\partial^{2} f}{\partial x^{2}} = 12 > 0\), local minimum.
$$ D = (12 \times 1)(12 \times 1) - (-4)^{2} = 128 > 0 $$
Since \(\frac{\partial^{2} f}{\partial x^{2}} = 12 > 0\), local minimum.
Answer: Saddle point at \((0,0)\); local minima at \((1,1)\) and \((-1,-1)\).
Find the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) for \(f(x,y) = 3x^{2}y + 4xy^{3}\) at \(x=2\), \(y=1\). (5 marks)
For the function \(P(V,I) = 2VI - 0.05V^{2}\), calculate \(\frac{\partial P}{\partial V}\) and \(\frac{\partial P}{\partial I}\) at \(V = 12\) V and \(I = 1.5\) A. (6 marks)
Determine and classify the stationary points of \(f(x,y) = x^{3} + y^{3} - 3xy\). (9 marks)
A signal amplitude is modeled by \(A(t,f) = t^{2} \cos(\pi f t)\). Find \(\frac{\partial A}{\partial t}\) at \(t=1\) s and \(f=2\) Hz. (5 marks)
Calculate \(\frac{\partial R}{\partial L}\) and \(\frac{\partial R}{\partial A}\) for \(R(L,A) = \rho \frac{L}{A}\) given \(\rho = 1.7 \times 10^{-8} \Omega m\), \(L=3\) m, \(A=2 \times 10^{-6} m^{2}\). (7 marks)
This chapter introduced the concept of partial derivatives, which are derivatives of functions with two or more variables taken with respect to one variable while holding the others constant. It began by explaining how to derive partial derivatives for functions involving multiple variables, demonstrating the step-by-step process of differentiation in such contexts. The chapter then explored the practical applications of partial derivatives in engineering, highlighting how they are used to analyze rates of change in various directions. Further, it examined stationary points of functions of two variables, showing how to determine where these points occur by setting partial derivatives to zero. The methods for classifying stationary points as maxima, minima, or saddle points using second-order partial derivatives were also covered. Throughout, the chapter emphasized the importance of these techniques in solving partial differential equations relevant to engineering problems. The knowledge gained forms a foundation for more advanced analysis and modelling in engineering mathematics.
Given the function \( f(x,y) = 3x^2y + 2xy^2 \), find the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) at the point \((x,y) = (1,2)\). (2 marks)
Calculate \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) for \( z = x^3 y^2 + 4x y \) at \((x,y) = (2,1)\). (2 marks)
An electronic component’s temperature \(T(x,y)\) in °C varies over a circuit board according to \( T = 5xy + 3x^2 \). Find the rate of change of temperature with respect to \(x\) and \(y\) at point \((x,y) = (3,4)\). (2 marks)
For the voltage distribution \( V(x,y) = x^2 - xy + y^2 \) over a semiconductor surface, find the stationary points by solving \(\frac{\partial V}{\partial x} = 0\) and \(\frac{\partial V}{\partial y} = 0\). (3 marks)
Determine the nature (maximum, minimum, or saddle point) of the stationary point found in question 4 using the second derivative test. (3 marks)
An amplifier gain \( G(x,y) = x e^{-y} + y^2 \) depends on two variables \(x\) and \(y\). Calculate the partial derivatives \(\frac{\partial G}{\partial x}\) and \(\frac{\partial G}{\partial y}\) at \((x,y) = (1,0)\). (3 marks)
The charge density on a capacitor plate varies as \( \sigma(x,y) = 2x^2 y + 3y^3 \). Find the partial derivatives \(\frac{\partial \sigma}{\partial x}\) and \(\frac{\partial \sigma}{\partial y}\) at \((x,y) = (1,2)\). (3 marks)
Find the stationary points of the function \( f(x,y) = x^3 - 3xy^2 \) and classify each point using the second derivative test. (4 marks)
The electric potential on a rectangular plate is given by \( \phi(x,y) = 4x^2 + y^2 - 4xy \). Find the stationary points and determine their nature. (4 marks)
In a temperature gradient problem, the temperature \( T(x,y) = x^2 y + y^3 - 3x y \). Find all stationary points and classify them using partial derivatives. (5 marks)
$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (3x^2 y + 2xy^2) = 6xy + 2y^2$$ Substitute \(x=1, y=2\):$$6 \times 1 \times 2 + 2 \times 2^2 = 12 + 8 = 20$$
$$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (3x^2 y + 2xy^2) = 3x^2 + 4xy$$ Substitute \(x=1, y=2\):$$3 \times 1^2 + 4 \times 1 \times 2 = 3 + 8 = 11$$
Answer:
\(\frac{\partial f}{\partial x} = \mathbf{20}\), \(\frac{\partial f}{\partial y} = \mathbf{11}\)
Calculate \(\frac{\partial z}{\partial x}\):
$$\frac{\partial z}{\partial x} = 3x^2 y^2 + 4y$$ Substitute \(x=2, y=1\):$$3 \times 2^2 \times 1^2 + 4 \times 1 = 3 \times 4 + 4 = 12 + 4 = 16$$
Calculate \(\frac{\partial z}{\partial y}\):
$$\frac{\partial z}{\partial y} = 2x^3 y + 4x$$ Substitute \(x=2, y=1\):$$2 \times 2^3 \times 1 + 4 \times 2 = 2 \times 8 + 8 = 16 + 8 = 24$$
Answer:
\(\frac{\partial z}{\partial x} = \mathbf{16}\), \(\frac{\partial z}{\partial y} = \mathbf{24}\)
Calculate \(\frac{\partial T}{\partial x}\):
$$\frac{\partial T}{\partial x} = 5y + 6x$$ Substitute \(x=3, y=4\):$$5 \times 4 + 6 \times 3 = 20 + 18 = 38$$ °C/unit \(x\)
Calculate \(\frac{\partial T}{\partial y}\):
$$\frac{\partial T}{\partial y} = 5x$$ Substitute \(x=3\):$$5 \times 3 = 15$$ °C/unit \(y\)
Answer:
Rate of change with respect to \(x = \mathbf{38}\) °C/unit, with respect to \(y = \mathbf{15}\) °C/unit
Find stationary points by solving:
$$\frac{\partial V}{\partial x} = 2x, y = 0$$$$\frac{\partial V}{\partial y} = -x + 2y = 0$$
From the first:$$y = 2x$$
Substitute into second:$$-x + 2(2x) = -x + 4x = 3x = 0 \Rightarrow x=0$$
Then:$$y = 2 \times 0 = 0$$
Stationary point: \((0,0)\)
Calculate discriminant \(D\):
$$D = V_{xx} V_{yy} - (V_{xy})^2 = 2 \times 2 - (-1)^2 = 4 - 1 = 3 > 0$$
Since \(V_{xx} > 0\) and \(D > 0\), the stationary point at \((0,0)\) is a local minimum.
Calculate \(\frac{\partial G}{\partial x}\):
$$\frac{\partial G}{\partial x} = e^{-y}$$ Substitute \(y=0\):$$e^{0} = 1$$
Calculate \(\frac{\partial G}{\partial y}\):
$$\frac{\partial G}{\partial y} = -x e^{-y} + 2y$$ Substitute \(x=1, y=0\):$$-1 \times e^{0} + 2 \times 0 = -1 + 0 = -1$$
Answer:
\(\frac{\partial G}{\partial x} = \mathbf{1}\), \(\frac{\partial G}{\partial y} = \mathbf{-1}\)
Calculate \(\frac{\partial \sigma}{\partial x}\):
$$\frac{\partial \sigma}{\partial x} = 4xy$$ Substitute \(x=1, y=2\):$$4 \times 1 \times 2 = 8$$
Calculate \(\frac{\partial \sigma}{\partial y}\):
$$\frac{\partial \sigma}{\partial y} = 2x^2 + 9y^2$$ Substitute \(x=1, y=2\):$$2 \times 1^2 + 9 \times 2^2 = 2 + 36 = 38$$
Answer:
\(\frac{\partial \sigma}{\partial x} = \mathbf{8}\), \(\frac{\partial \sigma}{\partial y} = \mathbf{38}\)
Find stationary points:
$$\frac{\partial f}{\partial x} = 3x^2 - 3y^2 = 0$$$$\frac{\partial f}{\partial y} = -6xy = 0$$
From second:
Either \(x=0\) or \(y=0\)
From first:$$3 \times 0 - 3 y^2 = -3 y^2 = 0 \Rightarrow y=0$$
Stationary point at \((0,0)\)
From first:$$3x^2 - 0 = 0 \Rightarrow x=0$$
Same point \((0,0)\)
So only stationary point at \((0,0)\)
Second derivatives:
$$f_{xx} = 6x$$$$f_{yy} = -6x$$$$f_{xy} = -6y$$
At \((0,0)\):
$$f_{xx} = 0, \quad f_{yy} = 0, \quad f_{xy} = 0$$
Discriminant:
$$D = f_{xx} f_{yy} - (f_{xy})^2 = 0 \times 0 - 0 = 0$$
Since \(D=0\), test is inconclusive.
Further analysis shows \((0,0)\) is a saddle point (known from function shape).
Find stationary points:
$$\frac{\partial \phi}{\partial x} = 8x - 4y = 0 \Rightarrow 8x = 4y \Rightarrow y = 2x$$$$\frac{\partial \phi}{\partial y} = 2y - 4x = 0 \Rightarrow 2y = 4x \Rightarrow y = 2x$$
Both equations give the same condition, so infinite points satisfy \(y=2x\).
No unique stationary point, but points on line \(y=2x\) are stationary.
Second derivatives:
$$\phi_{xx} = 8, \quad \phi_{yy} = 2, \quad \phi_{xy} = -4$$
Discriminant:
$$D = 8 \times 2 - (-4)^2 = 16 - 16 = 0$$
Discriminant zero means the test is inconclusive; these stationary points form a ridge along the line \(y=2x\).
Find stationary points:
Calculate derivatives:
$$\frac{\partial T}{\partial x} = 2xy - 3y = y(2x - 3) = 0$$$$\frac{\partial T}{\partial y} = x^2 + 3y^2 - 3x = 0$$
From first:
Either \(y=0\) or \(2x - 3=0 \Rightarrow x = \frac{3}{2}\)
Substitute in second:
$$x^2 + 0 - 3x = x^2 - 3x = 0 \Rightarrow x(x-3) = 0$$
So \(x=0\) or \(x=3\)
Stationary points: \((0,0)\), \((3,0)\)
Substitute in second:
$$(\frac{3}{2})^2 + 3 y^2 - 3 \times \frac{3}{2} = 0$$$$\frac{9}{4} + 3 y^2 - \frac{9}{2} = 0$$$$3 y^2 = \frac{9}{2} - \frac{9}{4} = \frac{18}{4} - \frac{9}{4} = \frac{9}{4}$$$$y^2 = \frac{9}{12} = \frac{3}{4}$$$$y = \pm \frac{\sqrt{3}}{2}$$
Stationary points: \((\frac{3}{2}, \frac{\sqrt{3}}{2})\), \((\frac{3}{2}, -\frac{\sqrt{3}}{2})\)
Second derivatives:
$$T_{xx} = 2y$$$$T_{yy} = 6y$$$$T_{xy} = 2x - 3$$
Evaluate at each point:
$$T_{xx} = 0, \quad T_{yy} = 0, \quad T_{xy} = 2 \times 0 - 3 = -3$$
Discriminant:
$$D = 0 \times 0 - (-3)^2 = -9 < 0$$
Saddle point.
$$T_{xx} = 0, \quad T_{yy} = 0, \quad T_{xy} = 2 \times 3 - 3 = 3$$
Discriminant:
$$D = 0 \times 0 - 3^2 = -9 < 0$$
Saddle point.
$$T_{xx} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} > 0$$$$T_{yy} = 6 \times \frac{\sqrt{3}}{2} = 3 \sqrt{3} > 0$$$$T_{xy} = 2 \times \frac{3}{2} - 3 = 3 - 3 = 0$$
Discriminant:
$$D = (\sqrt{3})(3 \sqrt{3}) - 0^2 = 3 \times 3 = 9 > 0$$
Since \(T_{xx} > 0\) and \(D > 0\), local minimum.
$$T_{xx} = 2 \times (-\frac{\sqrt{3}}{2}) = -\sqrt{3} < 0$$$$T_{yy} = 6 \times (-\frac{\sqrt{3}}{2}) = -3 \sqrt{3} < 0$$$$T_{xy} = 0$$
Discriminant:
$$D = (-\sqrt{3})(-3 \sqrt{3}) - 0 = 3 \times 3 = 9 > 0$$
Since \(T_{xx} < 0\) and \(D > 0\), local maximum.
Summary:
\((0,0)\) saddle point
\((3,0)\) saddle point
\((\frac{3}{2}, \frac{\sqrt{3}}{2})\) local minimum
\((\frac{3}{2}, -\frac{\sqrt{3}}{2})\) local maximum
Question 11 (Compulsory - 20 marks)
A design engineer at Safaricom is modeling the signal attenuation over a 2D antenna array, where the signal strength \(S(x,y)\) satisfies the PDE \(\frac{\partial^2 S}{\partial x^2} + \frac{\partial^2 S}{\partial y^2} = 0\).
a) Use the method of separation of variables to find a general solution for \(S(x,y)\) assuming boundary conditions \(S(0,y) = S(L,y) = 0\) for \(0 \le y \le W\). (10 marks)
b) Given \(L=5\) m, \(W=3\) m, and the boundary condition \(S(x,0) = 0\), \(S(x,W) = V_0 \sin(\pi x / L)\), find the particular solution \(S(x,y)\). (10 marks)
Question 12 (20 marks)
An electronics engineer models the temperature \(T(x,y)\) on a chip surface by the function \(T(x,y) = x^3 - 3xy^2\).
a) Find all stationary points of \(T\). (10 marks)
b) Classify each stationary point as a maximum, minimum, or saddle point using the second derivative test. (10 marks)
Question 13 (20 marks)
A partial differential equation governing voltage distribution \(V(x,y)\) in a rectangular semiconductor wafer is given by \(\frac{\partial V}{\partial x} + \frac{\partial V}{\partial y} = e^{x+y}\).
a) Verify whether the function \(V(x,y) = e^{x+y} + C\) is a solution to the PDE. (10 marks)
b) Find the general solution of the PDE using the method of characteristics. (10 marks)
Question 14 (20 marks)
The heat dissipation on an integrated circuit is modeled by the function \(H(x,y) = x^4 - 4x^2 y + y^2\).
a) Determine the stationary points of \(H\). (10 marks)
b) Use the Hessian matrix to classify the stationary points. (10 marks)
Question 11
a) Assume \(S(x,y) = X(x)Y(y)\). Substituting into the PDE gives:$$X''(x)Y(y) + X(x)Y''(y) = 0 \Rightarrow \frac{X''(x)}{X(x)} = - \frac{Y''(y)}{Y(y)} = -\lambda$$
Solve:$$X'' + \lambda X = 0, \quad Y'' - \lambda Y = 0$$
Apply boundary conditions \(X(0) = X(L) = 0\), solution:$$X_n = \sin(\frac{n\pi x}{L}), \quad \lambda_n = (\frac{n\pi}{L})^2$$
General solution:$$S(x,y) = \sum_{n=1}^\infty A_n \sin(\frac{n\pi x}{L}) \sinh(\frac{n\pi y}{L})$$
b) Use \(S(x,0) = 0\) and \(S(x,W) = V_0 \sin(\pi x / L)\), so:$$S(x,W) = A_1 \sin(\frac{\pi x}{L}) \sinh(\frac{\pi W}{L}) = V_0 \sin(\frac{\pi x}{L})$$
Therefore:$$A_1 = \frac{V_0}{\sinh(\frac{\pi W}{L})}$$
Final solution:$$S(x,y) = \frac{V_0}{\sinh(\frac{\pi W}{L})} \sin(\frac{\pi x}{L}) \sinh(\frac{\pi y}{L})$$
Question 12
a) Find stationary points by solving:$$\frac{\partial T}{\partial x} = 3x^2 - 3y^2 = 0 \Rightarrow x^2 = y^2$$$$\frac{\partial T}{\partial y} = -6xy = 0 \Rightarrow x=0 \text{ or } y=0$$
From these, stationary points are at \((0,0)\).
b) Second derivatives:$$T_{xx} = 6x, \quad T_{yy} = -6x, \quad T_{xy} = T_{yx} = -6y$$
At (0,0):$$D = T_{xx}T_{yy} - (T_{xy})^2 = (0)(0) - 0 = 0$$
Test inconclusive, but by inspection \(T\) has saddle behavior at (0,0).
Question 13
a) Compute derivatives:$$\frac{\partial V}{\partial x} = e^{x+y}, \quad \frac{\partial V}{\partial y} = e^{x+y}$$
Sum:$$\frac{\partial V}{\partial x} + \frac{\partial V}{\partial y} = e^{x+y} + e^{x+y} = 2e^{x+y}
eq e^{x+y}$$
So \(V = e^{x+y} + C\) is not a solution.
b) Characteristics:$$\frac{dx}{1} = \frac{dy}{1} = \frac{dV}{e^{x+y}}$$
Integrate:$$y, x = c_1$$
Along characteristic:$$\frac{dV}{dx} = e^{x + y} = e^{x + (x + c_1)} = e^{2x + c_1}$$
Integrate:$$V = \frac{e^{2x + c_1}}{2} + c_2 = \frac{e^{y + x}}{2} + F(y, x)$$
General solution:$$V(x,y) = \frac{e^{x+y}}{2} + F(y, x)$$
Question 14
a) Stationary points satisfy:$$\frac{\partial H}{\partial x} = 4x^3 - 8xy = 0 \Rightarrow 4x(x^2 - 2y) = 0$$$$\frac{\partial H}{\partial y} = -4x^2 + 2y = 0 \Rightarrow y = 2x^2$$
Substitute \(y\) into first:$$4x(x^2 - 2(2x^2)) = 4x(x^2 - 4x^2) = 4x(-3x^2) = -12 x^3 = 0$$
So \(x=0\), then \(y=0\)
Stationary point at (0,0)
b) Hessian matrix:$$H = \begin{bmatrix} \frac{\partial^2 H}{\partial x^2} & \frac{\partial^2 H}{\partial x \partial y} \\[6pt] \frac{\partial^2 H}{\partial y \partial x} & \frac{\partial^2 H}{\partial y^2} \end{bmatrix} = \begin{bmatrix} 12x^2 - 8y & -8x \\ -8x & 2 \end{bmatrix}$$
At (0,0):$$H = \begin{bmatrix} 0 & 0 \\ 0 & 2 \end{bmatrix}$$
Determinant \(D = 0 \times 2 - 0 = 0\), test inconclusive. Further analysis shows saddle behavior.
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