- #1

Monkey D. Luffy

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## Homework Statement

Two railroad tracks intersect at right angles at station O. At 10AM the train A, moving west with constant speed of 50 km/h, leaves the station O. One hour later train B, moving south with the constant speed of 60 km/h, passes through the station O. Find minimum distance between these trains

## Homework Equations

dx/dt

dy/dt

## The Attempt at a Solution

I got up to the point where i formulated two equations for the trains:

Xa= 50t

Yb= 60t -60

From here i was lost as to how to figure out the minimum distance because today before i attempted the problem, in a tutorial class we learned that minimum distance is (Yb-Xa) in this case and i do not understand how to do it that way. So i searched for a way to do it online and got this as an answer:

Let x be the position of train A, let y be the position of train B, and let O be the origin. West will be the positive x direction and south will be the positive y direction. We are given that (dx/dt)=50 and (dy/dt)=60. Train A starts at time t=0, so its position is x=50t. Train B starts at time t=1, so its position is y=60(t-1) when t is greater than or equal to 1. (y should equal 0 when t is 1 hour.) The formula we need to minimize is the distance between the trains given by the pythagorean theorem. We denote the distance between the trains s, and write s^2=x^2+y^2. The mimimum will be at a critical point where the derivative is equal to 0, so we differentiate with regards to t. 2s(ds/dt)=2x(dx/dt)+2y(dy/dt). We are looking for the point at which (ds/dt)=0, so we need to find: 0=2x(dx/dt)+2y(dy/dt). We can divide by 2 and not change a thing (0/2=0). 0=x(dx/dt)+y(dy/dt). Now we substitute our equations for x and y, and our numbers for (dx/dt) and (dy/dt). 0=(50t)(50)+(60t-60)(60). Simplify: 0=2500t+3600t-3600. Add the t terms: 0=6100t-3600. Add 3600 to both sides: 3600=6100t. This gives us t=0.59. Now we know the time at which the shortest distance was at, so now we find the distance. x=50(0.59)=29.51km and y=60(0.59-1)=-24.6km. Now we use s^2=(29.51)^2+(-26.6)^2=39.73km. The shortest distancw between the trains was 39.73km.

I understand a lot of it but i do not get when this person said the derivative is equal to 0, this is the equation:

2x(dx/dt)+2y(dy/dt)

why would it not just be Xa-Yb still? why do we multiply the X and Y equations with their derivatives? i am very confused by the notation used here for derivatives as well as the method and why these steps (after the one i just listed) was taken... Could anyone help me please? Also, i know i am asking a lot but could you also explain the integrals notation? since its a derivative taken backwards, i thought it would be simple enough except i don't get what it means when you "take the integral with respect to something else."

Thank you so much guys and the class is principles of physics so i thought this would be the place to create this forum.