# Understanding distances, derivatives, and integrals help?

• Monkey D. Luffy
In summary, two trains, A and B, are moving at constant speeds of 50 km/h and 60 km/h, respectively, on two intersecting railroad tracks at station O. Train A leaves the station at 10AM and one hour later, train B passes through the station. The problem is to find the minimum distance between the two trains. To solve this, we set up equations for the position of each train, with x being the position of A and y being the position of B. We use the Pythagorean theorem to represent the distance between the trains, denoted as s. To find the minimum distance, we need to find the critical point where the derivative of s is equal to 0. This is found
Monkey D. Luffy

## 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

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.

Monkey D. Luffy said:

## 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

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.

Whoever did the analysis was doing it the hard way. It is easier to evaluate ##s^2## before taking the derivative: ##s^2 = (50 t)^2 + (60 t - 60)^2 = 6100\, t^2 - 7200\, t + 3600##. Now ## ds^2/dt = 12200 \, t - 7200##. This equals 0 when ##t = 36/61 \doteq 0.590## (hours). At that time the squared distance between the trains is obtained by substituting ##t = 36/61## into the expression for ##s^2##, and that simplifies to ##s^2_{\min} = 90 000/61## (km^2). The minimum distance is ##s_{\min} = \sqrt{90 000/61} \doteq 38.41## km. This is a bit different from the answer you gave above. The differences are likely due to excessive roundoff that the author above performed. It is better to keep more figures during the calculation, then round off the answer after all the work is finished. (In this case it makes quite a difference--- more than 1 km in the final answer.)

Ray Vickson said:
Whoever did the analysis was doing it the hard way. It is easier to evaluate ##s^2## before taking the derivative: ##s^2 = (50 t)^2 + (60 t - 60)^2 = 6100\, t^2 - 7200\, t + 3600##. Now ## ds^2/dt = 12200 \, t - 7200##. This equals 0 when ##t = 36/61 \doteq 0.590## (hours). At that time the squared distance between the trains is obtained by substituting ##t = 36/61## into the expression for ##s^2##, and that simplifies to ##s^2_{\min} = 90 000/61## (km^2). The minimum distance is ##s_{\min} = \sqrt{90 000/61} \doteq 38.41## km. This is a bit different from the answer you gave above. The differences are likely due to excessive roundoff that the author above performed. It is better to keep more figures during the calculation, then round off the answer after all the work is finished. (In this case it makes quite a difference--- more than 1 km in the final answer.)
OHHH i see! thanks i see your way of doing it and it makes a lot more sense now! my only remaining question is why we take the derivative and set it to 0, like what does setting it to 0 have to do with anything? i have done optimization before but it was usually for a dog running on a beach then swimming to get a stick in the water or something along those lines. in that situation, optimization gave us the fastest time to fetch the stick. why does it now give us "closest distance"?

CWatters said:
I'm not sure if that's where you are stuck but it should be covered in your course work. For example see..

http://www.themathpage.com/acalc/max.htm
https://www.mathsisfun.com/calculus/maxima-minima.html
ah see i understand derivatives pretty decently and your links brought back some ideas i remember practicing in Grade 12 Calculus, but now its integrals and the notation that confuses me. for example let's say we have -pdV/T what does that notation mean? (Im just using an example we had in class, one that confused me then as well because this is the first time i am doing integrals)

## 1. What is the importance of understanding distances, derivatives, and integrals?

Understanding distances, derivatives, and integrals is crucial in many fields of science, as it allows us to model and analyze various physical phenomena such as motion, growth, and change. It also provides the foundation for more advanced mathematical concepts and tools used in scientific research and problem-solving.

## 2. How do distances, derivatives, and integrals relate to each other?

Distances, derivatives, and integrals are all interconnected concepts that build upon each other. Distances are the basic building blocks of derivatives, which describe the rate of change of a function. Integrals, on the other hand, are the reverse process of derivatives and are used to calculate the total distance or area under a curve.

## 3. How can understanding these concepts help in real-world scenarios?

Understanding distances, derivatives, and integrals can be applied in various real-world situations, such as predicting the trajectory of a moving object, optimizing processes in engineering and economics, and analyzing data in various fields of science. It also helps in developing critical thinking and problem-solving skills.

## 4. What are some common misconceptions about distances, derivatives, and integrals?

One common misconception is that these concepts are only relevant in mathematics and have no practical applications. Another misconception is that they are only applicable in specific fields and not relevant to other areas of science. In reality, understanding these concepts can be beneficial in many different fields and industries.

## 5. How can one improve their understanding of distances, derivatives, and integrals?

To improve understanding, it is important to first have a strong foundation in basic mathematical concepts such as algebra and geometry. It is also helpful to practice solving problems and applying these concepts in various scenarios. Seeking out additional resources, such as textbooks or online tutorials, can also aid in improving understanding.

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