You used 25 instead of 20. Here is a redo:[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?
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.