How to approach this type of question in matlab?

A ship travels on a straight line course described by y = (2005x)/6, where distances are measured in kilometers. The ship starts when x = -20 and ends when x = 40. Calculate the distance at closest approach to a lighthouse located at the coordinate origin (0,0). Do not solve this using a plot.

[Mohit]

*y=200-5x/6

[Sam Chak]

Mohit, do not express the equation using implied multiplication or implied division. Use standard computing notation for math operators when expressing an inline equation.

Because we don't know whether the path trajectory equation is

200 - 5*(x/6)

Or

(200 - 5*x)/6.

Answer 1

[Image Analyst]

This looks like a homework problem. If you have any questions ask your instructor or read the link below to get started:

How do I get help on homework questions on MATLAB Answers? - MATLAB Answers - MATLAB Central

Obviously, if it is homework, we can't give you the full solution because you're not allowed to turn in our code as your own.

Hints: linspace to define x, and be sure to use enough points to get good resolution, like ten thousand or so. Use sqrt to compute distance of (x,y) to (0,0) origin. Use min to find the minimum distance and index for that distance.

[Mohit]

>>x = linspace(-10,30,200);

>>y = 2*x-10;

>>D = sqrt(x.^2+y.^2);

>>min_dist=min(D)

[Image Analyst]

OK, but like I said, use more resolution in x and use the correct formula "y = (2005x)/6". Not sure where you got that formula and why it's different than your original question.

x = linspace(-10, 30, 20000);
y = 2005 * x / 6;
D = sqrt(x.^2 + y.^2);
min_dist = min(D)

min_dist = 0.1671

[Torsten]

@Mohit

min: x^2 + y^2 = x^2 + (200-5*x/6)^2

Do you know how to compute the vertex of this parabola ?

[Image Analyst]

@Mohit using your updated/corrected formula

x = linspace(-10, 30, 20000);
y = 200 - 5 * x / 6;
D = sqrt(x.^2 + y.^2);
min_dist = min(D)

min_dist = 177.5528

or maybe

x = linspace(-10, 30, 20000);
y = (200 - 5 * x) / 6;
D = sqrt(x.^2 + y.^2);
[min_dist, indexOfClosest] = min(D)

min_dist = 25.6074

indexOfClosest = 13197

plot(x, y, 'b-', 'LineWidth', 2)
grid on;
line([0, x(indexOfClosest)], [0, y(indexOfClosest)], 'Color', 'r', 'LineWidth', 2)
axis equal

Still not exactly sure what the formula is. Please use parentheses to make it unambiguous.

Answer 2

[Sam Chak]

Hi @Mohit

The distance of the ship at the closest approach to a lighthouse should be the perpendicular distance from the lighthouse (0, 0) to the straight line trajectory of the ship.

https://byjus.com/maths/distance-of-a-point-from-a-line/