QQuestionAnatomy and Physiology
QuestionAnatomy and Physiology
1
Makerere
University
Department of Electrical and Computer Engineering
B.Sc. in Electrical Engineering, B.Sc. in Computer & Communications
Engineering and B.Sc. Biomedical Engineering
EMT^1101 – ENGINEERING MATHEMATICS I
Coursework Set 1
Instructions:
a) Work in groups of 2–3 members.
b) Each group must include members from different programmes (BELE, BCCE, BBI).
c) Clearly list each member’s name, registration number, and programme on the cover page.
d) Solutions may be handwritten or typed.
e) Submission deadline: 30th October 2025 at 8:00 AM (strictly).
1. For power system engineers, it is essential to ensure maximum power transfer from the
source to the load. Figure 1 shows a circuit in which a non-ideal voltage source is
connected to a variable load resistor with resistance 𝑅𝐿. The source voltage is 𝑉 and its
internal resistance is 𝑅𝑆. Calculate the value of 𝑅𝐿 which results in the maximum power
being transferred from the voltage source to the load resistor.
6 Marks
Figure 1
2. Find the dimensions of the right-circular cylinder of largest volume that can be inscribed
in a sphere of radius R.
4 Marks
3. Sketch graphs of the functions
2
(i) 𝑦 =
𝑥2−𝑥−6
𝑥+ 1
5 Marks
(ii)𝑦 =
𝑥−1
𝑥2−4
5 Marks
4. Given the system of linear equations
4𝑥 − 5𝑦 + 7𝑧 = −14
9𝑥 + 2𝑦 − 3𝑧 = 47
𝑥 − 𝑦 − 5𝑧 = 11
Solve the equation using
(i) Crammer’s rule
5 Marks
(ii)Gauss elimination method
5 Marks
5. (a) Sketch graphs of the following functions
(i) 𝑓(𝑥) =
𝑥
|𝑥|
2 Marks
(ii) 𝑓(𝑥) = √4 − 𝑥2
2 Marks
(iii)𝑓(𝑥) = {𝑥2,
𝑥 > 1
2,
𝑥 ≤ 1
2 Marks
(b) An open box is to be made from an 8𝑐𝑚 × 15𝑐𝑚 piece of sheet metal by cutting out
squares with sides of length 𝑥 from each of the four corners and bending up the sides.
Express the volume 𝑉 of the box as a function 𝑥, and state the domain and range of the
function.
4 Marks
6. Evaluate the following integrals
(i) ∫
sin^2 3𝑥 cos 3𝑥 𝑑𝑥
𝜋 2
⁄
0
4 Marks
(ii) ∫
cos 2𝑥
√7−3 sin 2𝑥 𝑑𝑥
𝜋 4
⁄
0
4 Marks
(iii)∫
𝑥2
√4−3𝑥 𝑑𝑥
1
0
4 Marks
(iv) ∫ √tan 𝑥 sec^2 𝑥 𝑑𝑥
4 Marks
(v) ∫ 𝑒𝑎𝑥 cos 𝑏𝑥 𝑑𝑥
4 Marks
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Step 1:: To calculate the value of $R\_L$ which results in the maximum power being transferred from the voltage source to the load resistor, we need to find the value of $R\_L$ that makes the voltage across the load resistor equal to half of the source voltage.
\frac{R\_L}{R\_S + R\_L} V = \frac{1}{2} V
This is because maximum power transfer occurs when the load resistance is equal to the source resistance, and the voltage is divided equally across the two resistors. The voltage across the load resistor is given by:
Step 2:: Solving for $R\_L$ gives:
R\_L = \frac{1}{2} R\_S
Step 3:: Therefore, the value of $R\_L$ which results in the maximum power being transferred from the voltage source to the load resistor is:
R\_L = \boxed{\frac{1}{2} R\_S}
Step 4:: To find the dimensions of the right-circular cylinder of largest volume that can be inscribed in a sphere of radius $R$, we need to maximize the volume of the cylinder subject to the constraint that the cylinder fits entirely within the sphere.
r + h \leq 2R
The volume of the cylinder is given by: The constraint that the cylinder fits entirely within the sphere is given by:
Step 5:: To maximize the volume of the cylinder, we need to find the values of $r$ and $h$ that satisfy the constraint and maximize the volume.
L(r, h, \lambda) = \pi r^2 h - \lambda (r + h - 2R)
To do this, we can use the method of Lagrange multipliers. The Lagrangian function is:
Step 6:: Taking the partial derivatives of the Lagrangian function with respect to $r$, $h$, and $\lambda$, and setting them equal to zero gives:
\frac{\partial L}{\partial \lambda} = r + h - 2R = 0
Step 7:: Solving these equations simultaneously gives:
h = \frac{4R}{3}
Step 8:: Therefore, the dimensions of the right-circular cylinder of largest volume that can be inscribed in a sphere of radius $R$ are:
h = \boxed{\frac{4R}{3}}
Step 9:: The volume of the cylinder is:
V = \pi r^2 h = \frac{16 \pi R^3}{9}
Step 10:: Therefore, the volume of the cylinder is:
V = \boxed{\frac{16 \pi R^3}{9}}
Step 11:: For part (iii), we need to sketch the graph of the function:
f(x) = \begin{cases} x^2, & x > 1 \ 2, & x \leq 1 \end{cases}
Step 12:: The graph of the function consists of two parts:
(ii) For $x \leq 1$, the function is $f(x) = 2$, which is a horizontal line.
Step 13:: The graph of the function is shown below:
[Insert graph here]
Step 14:: The volume of the box is given by:
V(x) = x(8-2x)(15-2x)
Step 15:: The domain of the function is:
0 < x < 4
Step 16:: The range of the function is:
0 \leq V(x) \leq 300
Step 17:: For part (vi), we need to evaluate the following integrals:
(i) $$\int\_0^{\pi/2} \sin^2 3x \cos 3x dx
Step 18:: Using the double angle identity for cosine, we can rewrite the integral as:
\int\_0^{\pi/2} \frac{1}{2} \sin 6x (1 - \sin^2 3x) dx
Step 19:: Let $u = \sin 3x$, then $du = 3 \cos 3x dx$, and the integral becomes:
\int\_0^1 \frac{1}{6} (1 - u^2) du = \boxed{\frac{1}{24}}
Step 20:: For part (vi), we need to evaluate the following integrals:
(ii) $$\int\_0^{\pi/4} \frac{\cos 2x}{\sqrt{7 - 3 \sin 2x}} dx
Step 21:: Let $u = \sin 2x$, then $du = 2 \cos 2x dx$, and the integral becomes:
\int\_0^{\sqrt{2}/2} \frac{1}{2 \sqrt{7 - 3u}} du = \boxed{\frac{1}{3} \left( \sqrt{7 - 3 \sqrt{2}} + \sqrt{7 + 3 \sqrt{2}} \right)}
Step 22:: For part (vi), we need to evaluate the following integrals:
(iii) $$\int\_0^1 \frac{x^2}{\sqrt{4 - 3x}} dx
Step 23:: Let $u = \sqrt{4 - 3x}$, then $du = -\frac{3}{2} dx$, and the integral becomes:
-\frac{2}{9} \int\_2^1 u^2 du = \boxed{\frac{2}{27} \left( 8 - 3 \sqrt{3} \right)}
Step 24:: For part (vi), we need to evaluate the following integrals:
(iv) $$\int \sqrt{\tan x} \sec^2 x dx
Step 25:: Let $u = \sqrt{\tan x}$, then $du = \frac{\sec^2 x}{2 \sqrt{\tan x}} dx$, and the integral becomes:
2 \int u^2 du = \boxed{\frac{2}{3} \tan^{3/2} x + C}
Step 26:: For part (vi), we need to evaluate the following integrals:
(v) $$\int e^{ax} \cos bx dx
Step 27:: Using integration by parts, we can rewrite the integral as:
\frac{1}{b} \int e^{ax} (b \cos bx + a \sin bx) dx - \frac{a}{b} \int e^{ax} \sin bx dx
Step 28:: Using integration by parts again, we can rewrite the second integral as:
\frac{1}{b^2} \int e^{ax} (b \cos bx + a \sin bx) dx
Step 29:: Therefore, the integral is:
\int e^{ax} \cos bx dx = \boxed{\frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) + C}
Step 30::
Final Answer
The solutions to the given problems are presented in steps above.
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