Shortest distance

Please explain what formula you are using, or what reasoning leads to your calculations.

I don't get the same result by a different method, but something like what you did may be correct, and you may even have it right. I just don't yet have a reason to trust it.
 
[MATH]x = \text {distance of car from cross roads in km.}[/MATH]
[MATH]y = \text {distance of lorry from cross roads in km to west.}[/MATH]
[MATH]z = \text {distance of car from lorry in km to south.}[/MATH]
[MATH]t = \text {time in hours.}[/MATH]
[MATH]z = \sqrt{x^2 + y^2}.[/MATH]
[MATH]x = 5 - 25t.[/MATH]
[MATH]y = 4 - 35t.[/MATH]
[MATH]\therefore z = \sqrt{(5 - 25t)^2 + (4 - 35t)^2} = \sqrt{41 - 530t + 1850t^2} \implies[/MATH]
[MATH]\dfrac{dz}{dt} = \dfrac{3700t - 530}{2 \sqrt{41 - 530t + 1850t^2} } = \dfrac{1850t - 265}{ \sqrt{41 - 530t + 1850t^2} } \implies[/MATH]
[MATH]\dfrac{dz}{dt} = 0 \iff t = \dfrac{265}{1850} = \dfrac{53}{370}.[/MATH]
[MATH]t = \dfrac{53}{370} \implies x = 5 - 25 * \dfrac{53}{370} = 5 - 5 * \dfrac{53}{74} = \dfrac{370 - 265}{74} = \dfrac{35}{74}.[/MATH]
[MATH]t = \dfrac{53}{370} \implies y = 4 - 35 * \dfrac{53}{370} = 4 - 7 * \dfrac{53}{74} = \dfrac{296 - 371}{74} = -\ \dfrac{75}{74}.[/MATH]
[MATH]\therefore \text {min}(z) = \sqrt{\left ( \dfrac{35}{74} \right )^2 + \left (-\ \dfrac{75}{74} \right)^2} = \dfrac{\sqrt{1225 + 5625}}{74} \approx 1.12.[/MATH]
I too get a different answer. I shall check my arithmetic, but I cannot follow what you are doing.
 
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[MATH]x = \text {distance of car from cross roads in km.}[/MATH]
[MATH]y = \text {distance of lorry from cross roads in km to west.}[/MATH]
[MATH]z = \text {distance of car from lorry in km to south.}[/MATH]
[MATH]t = \text {time in hours.}[/MATH]
[MATH]z = \sqrt{x^2 + y^2}.[/MATH]
[MATH]x = 5 - 25t.[/MATH]
[MATH]y = 4 - 35t.[/MATH]
[MATH]\therefore z = \sqrt{(5 - 25t)^2 + (4 - 35t)^2} = \sqrt{41 - 530t + 1850t^2} \implies[/MATH]
[MATH]\dfrac{dz}{dt} = \dfrac{3700t - 530}{2 \sqrt{41 - 530t + 1850t^2} } = \dfrac{1850t - 265}{ \sqrt{41 - 530t + 1850t^2} } \implies[/MATH]
[MATH]\dfrac{dz}{dt} = 0 \iff t = \dfrac{265}{1850} = \dfrac{53}{370}.[/MATH]
[MATH]t = \dfrac{53}{370} \implies x = 5 - 25 * \dfrac{53}{370} = 5 - 5 * \dfrac{53}{74} = \dfrac{370 - 265}{74} = \dfrac{35}{74}.[/MATH]
[MATH]t = \dfrac{53}{370} \implies y = 4 - 35 * \dfrac{53}{370} = 4 - 7 * \dfrac{53}{74} = \dfrac{296 - 371}{74} = -\ \dfrac{75}{74}.[/MATH]
[MATH]\therefore \text {min}(z) = \sqrt{\left ( \dfrac{35}{74} \right )^2 + \left (-\ \dfrac{75}{74} \right)^2} = \dfrac{\sqrt{1225 + 5625}}{74} \approx 1.12.[/MATH]
I too get a different answer. I shall check my arithmetic, but I cannot follow what you are doing.
You used 25 instead of 20. Here is a redo:
[MATH]x = 5 - 20t.[/MATH]
[MATH]y = 4 - 35t.[/MATH]
[MATH]\therefore z = \sqrt{(5 - 20t)^2 + (4 - 35t)^2} = \sqrt{41 - 480t + 1625t^2} \implies[/MATH]
[MATH]\dfrac{dz}{dt} = \dfrac{3250t - 480}{2 \sqrt{41 - 480t + 1625t^2} } = \dfrac{1625t - 240}{ \sqrt{41 - 530t + 1850t^2} } \implies[/MATH]
[MATH]\dfrac{dz}{dt} = 0 \iff t = \dfrac{240}{1625} = \dfrac{48}{325}.[/MATH]
[MATH]t = \dfrac{48}{325} \implies x = 5 - 20 * \dfrac{48}{325} = 5 - 4 * \dfrac{48}{65} = \dfrac{325 - 192}{65} = \dfrac{133}{65}.[/MATH]
[MATH]t = \dfrac{48}{325} \implies y = 4 - 35 * \dfrac{48}{325} = 4 - 7 * \dfrac{48}{65} = \dfrac{260 - 336}{65} = -\ \dfrac{76}{65}.[/MATH]
[MATH]\therefore \text {min}(z) = \sqrt{\left ( \dfrac{133}{65} \right )^2 + \left (-\ \dfrac{76}{65} \right)^2} = \dfrac{\sqrt{133^2 + 76^2}}{65} \approx 2.357.[/MATH]
That's the number I got by a similar method.

Now, my guess as to @ncchieh's method is that he is transforming to a relative frame of reference and finding the distance to the path, but forgetting that distances are not preserved. That method could perhaps be used to find where the closest point will be, and then find the distance in the original frame. But I would want to know what methods have been used in his class -- vectors, or derivatives?
 
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Dr. P

I fussed about at all the arithmetic and the LaTeX coding and never looked back to make sure that I had copied the original problem correctly. Thanks for the fix
 
You may notice I never said what number I got until I got the same value from your work. Calculate twice, commit once.

Do I ever make mistakes? Yup.
 
You used 25 instead of 20. Here is a redo:
[MATH]x = 5 - 20t.[/MATH]
[MATH]y = 4 - 35t.[/MATH]
[MATH]\therefore z = \sqrt{(5 - 20t)^2 + (4 - 35t)^2} = \sqrt{41 - 480t + 1625t^2} \implies[/MATH]
[MATH]\dfrac{dz}{dt} = \dfrac{3250t - 480}{2 \sqrt{41 - 480t + 1625t^2} } = \dfrac{1625t - 240}{ \sqrt{41 - 530t + 1850t^2} } \implies[/MATH]
[MATH]\dfrac{dz}{dt} = 0 \iff t = \dfrac{240}{1625} = \dfrac{48}{325}.[/MATH]
[MATH]t = \dfrac{48}{325} \implies x = 5 - 20 * \dfrac{48}{325} = 5 - 4 * \dfrac{48}{65} = \dfrac{325 - 192}{65} = \dfrac{133}{65}.[/MATH]
[MATH]t = \dfrac{48}{325} \implies y = 4 - 35 * \dfrac{48}{325} = 4 - 7 * \dfrac{48}{65} = \dfrac{260 - 336}{65} = -\ \dfrac{76}{65}.[/MATH]
[MATH]\therefore \text {min}(z) = \sqrt{\left ( \dfrac{133}{65} \right )^2 + \left (-\ \dfrac{76}{65} \right)^2} = \dfrac{\sqrt{133^2 + 76^2}}{65} \approx 2.357.[/MATH]
That's the number I got by a similar method.

Now, my guess as to @ncchieh's method is that he is transforming to a relative frame of reference and finding the distance to the path, but forgetting that distances are not preserved. That method could perhaps be used to find where the closest point will be, and then find the distance in the original frame. But I would want to know what methods have been used in his class -- vectors, or derivatives?
IMG_20190816_084505.jpg

My school use this method, not sure is vectors or directive
 
Thanks.

The fact that you put this question under calculus led me (us?) to assume you were supposed to use the method we've shown, and that your vector-based method was wrong since it gave a different result. What you have been taught is, in fact, just what I imagined; and now that I see it written out, I see I was wrong to think that the relative motion would distort distances. The method is valid. (Their wording is not: there is no such thing as "distance between a point and a relative velocity". What they mean is distance between the point and the path in the new frame of reference.)

But you have applied it incorrectly. When I apply it as it should be, I get exactly the answer we got above, which is in fact [MATH]\frac{19}{\sqrt{65}}[/MATH]. (I tried before, but not hard enough!)

Your error is in using the wrong hypotenuse in your final calculation. You followed the example too closely, not taking into account that the original relative positions are not along an axis as in the example.

Give it another try, sketching a diagram like theirs (but it will look quite different) so you can see the triangle you are using in that last step. Then we can help you fix that if needed.
 
Thanks.

The fact that you put this question under calculus led me (us?) to assume you were supposed to use the method we've shown, and that your vector-based method was wrong since it gave a different result. What you have been taught is, in fact, just what I imagined; and now that I see it written out, I see I was wrong to think that the relative motion would distort distances. The method is valid. (Their wording is not: there is no such thing as "distance between a point and a relative velocity". What they mean is distance between the point and the path in the new frame of reference.)

But you have applied it incorrectly. When I apply it as it should be, I get exactly the answer we got above, which is in fact [MATH]\frac{19}{\sqrt{65}}[/MATH]. (I tried before, but not hard enough!)

Your error is in using the wrong hypotenuse in your final calculation. You followed the example too closely, not taking into account that the original relative positions are not along an axis as in the example.

Give it another try, sketching a diagram like theirs (but it will look quite different) so you can see the triangle you are using in that last step. Then we can help you fix that if needed.
IMG_20190816_112058.jpg
I redo it and i get this, the 60.26 is my bigger angle, and i need the smaller angle with will be perpendicular with my L, so i get the angle by using 180 - 60.26 - 38.66 = 21.60 and i manage to get shortest distance
 
Good work!

This would have been easier to follow if you had labeled a little more, but your thinking is correct. You just had to find the right angle to apply the final step to. I used a somewhat different method in detail (finding another side in your figure, rather than an angle), but yours is good.
 
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