How to approach this type of question in matlab?
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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.
2 件のコメント
Sam Chak
2024 年 8 月 3 日 19:40
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.
回答 (2 件)
Image Analyst
2024 年 8 月 3 日 18:59
編集済み: Image Analyst
2024 年 8 月 3 日 19:03
This looks like a homework problem. If you have any questions ask your instructor or read the link below to get started:
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.
4 件のコメント
Image Analyst
2024 年 8 月 3 日 21:11
@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)
or maybe
x = linspace(-10, 30, 20000);
y = (200 - 5 * x) / 6;
D = sqrt(x.^2 + y.^2);
[min_dist, indexOfClosest] = min(D)
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.
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