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Introduction to Programming with MATLAB A Resource for Mathematicians, Engineers, and Scientists

Geoffrey W. Cox, Ph.D.

🧑🏻‍💻 Class Activities 🧑🏻‍💻 Class Activities

1. Hello Function.

(a)

Create a function named hello with no inputs and outputs that displays 'Hello World' using the fprintf command. Test your function by typing hello in the command window.
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Solution.
function hello()

	fprintf('Hello World!\n');

end
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(b)

Modify the function to accept one input (name) and display a greeting, such as 'Hello Alice, nice to meet you!' where name is 'Alice'.
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Solution.
function hello(name)

	fprintf('Hello %s, nice to meet you!\n', name);

end
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(c)

Try typing hello with no input and run it. What error message do you get, and what is MATLAB telling you to fix?
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Solution.
The error message is:
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Not enough input arguments.

Error in hello (line 3)
	fprintf('Hello %s, nice to meet you!\n', name);
This means that the function hello expects one input argument, but none was provided when it was called.
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(d)

Now, test your function with an input: hello('Robyn').
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Solution.
When you run hello('Robyn'), the output is:
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Hello Robyn, nice to meet you!
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2. Projectile Height Function.

Package the projectile height model script as a function.
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(a)

Create a function named projectile_height that takes four inputs: initial height h0, initial velocity v0, gravity g, and time t.
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The function should return one output, the height h at time t, given by:
\begin{equation*} h(t) = h_0 + v_0 t - \frac{1}{2} g t^2\text{.} \end{equation*}
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Do not include any fprintf commands inside the function. The function’s body should contain only a single line that computes the height.
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Solution.
function h = projectile_height(h0, v0, g, t)

	h = h0 + v0*t - 0.5*g*t^2;

end
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(b)

Test your function using h0 = 1.5, v0 = 12, g = 9.81, and t = 0.8.
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Solution.
h = projectile_height(1.5, 12, 9.81, 0.8);
fprintf('h = %.2f m\n', h);
The computed height is h = 7.96 meters.
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3. Circle Metrics Function.

Write a function that returns more than one output.
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(a)

Create a function named circle_metrics that takes one input r (radius) and returns two outputs: C (circumference) and A (area).
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Solution.
function [C, A] = circle_metrics(r)

	C = 2*pi*r;
	A = pi*r^2;

end
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(b)

Test your function with r = 4.5 and print the results with three decimal places.
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Solution.
[C, A] = circle_metrics(4.5);
fprintf('C = %.3f\t A = %.3f\n', C, A);
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4. Coin Total Function.

In the piggy bank script, you computed a total value from the number of pennies, nickels, dimes, and quarters. Now, package that computation into a function so you can reuse it with different coin counts.
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(a)

Create a function named coin_total with four inputs: nPennies, nNickels, nDimes, and nQuarters. The function should return one output, totalDollars.
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Solution.
function totalDollars = ...
	coin_total(nPennies, nNickels, nDimes, nQuarters)

	% Compute total value in cents
	cents = nPennies + 5*nNickels + 10*nDimes + 25*nQuarters;
	totalDollars = cents/100;
end
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(b)

Test your function by running the following command in the Command Window.
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total = coin_total(1, 1, 1, 1)
Solution.
Your result should be a single dollar amount stored in total.
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(c)

In the Command Window, compute dollars and leftover cents from total (do not change the function).
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Solution.
nDollars = floor(total);
nCents = total - 100 * nDollars;
fprintf('Total value: $%i.%02i\n', nDollars, nCents);
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5. Quadratic Solver Function.

Write a function that computes the two solutions of \(a x^2 + b x + c = 0\) using the quadratic formula. This function may return complex solutions when the discriminant is negative.
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(a)

Create a function named quadratic with inputs a, b, and c. Return two outputs, x1 and x2.
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Solution.
function [x1, x2] = quadratic(a, b, c)

	discriminant = b^2 - 4*a*c;
	x1 = (-b + sqrt(discriminant)) / (2*a);
	x2 = (-b - sqrt(discriminant)) / (2*a);

end
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(b)

Test your function on each equation below, and print the results. Use the real-and-imaginary printing approach so your output also works for complex solutions.
  1. \(\displaystyle 2x^2 - 7x + 5 = 0\)
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  2. \(\displaystyle 15x^2 - 135x + 300 = 0\)
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  3. \(\displaystyle x^2 + 36 = 0\)
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Solution.
[x1, x2] = quadratic(2, -7, 5);
fprintf('Eqn 1: x1 = %g%+gi, x2 = %g%+gi\n', real(x1), imag(x1), real(x2), imag(x2));

[x1, x2] = quadratic(15, -135, 300);
fprintf('Eqn 2: x1 = %g%+gi, x2 = %g%+gi\n', real(x1), imag(x1), real(x2), imag(x2));

[x1, x2] = quadratic(1, 0, 36);
fprintf('Eqn 3: x1 = %g%+gi, x2 = %g%+gi\n', real(x1), imag(x1), real(x2), imag(x2));
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(c)

Are there any inputs for which this function fails? Give an example and explain why.
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Solution.
The function fails when a = 0, because the quadratic formula divides by \(2a\text{.}\) Handling a = 0 properly requires making a decision inside the function, which needs flow control.
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