Finding the Minimum Distance Between Two Points Using Graphing Calculator

[xzibition8612]

Homework Statement

Two points P and Q have position vectors in a reference frame that are given by r_OP = 50ti and r_OQ = 40i - 20tj meters. Find the minimum distance between P and Q and the time at which this occurs.

Homework Equations

The Attempt at a Solution

I suppose I take the magnitude of each vector, but then what? They both share the same origin, how do i manipulate it so that they give me the PQ length?

 

[Villyer]

What path do each of the points move on?

 

[xzibition8612]

the x and y-axis respectively

 

[Villyer]

What I would do here is just look at the two paths. Choose a few values of t, and see if they are getting closer together or farther apart.

Since their paths are very simple, I don't think it would be that hard to find the point where they switch from approaching to getting more distant.Edit, actually I just looked at them more closely and maybe it would be easier to plug the equations into the distance formula, since you know that the triangle between them will always have its right angle on the same point.

 

[xzibition8612]

right. but i would rather there be a methodical way instead of plug and chug. can you tell me a general formula i should follow, such as take the magnitude and then subtract them ...etc. thanks

 

[Villyer]

I haven't attempted to solve a problem such as this before, but as a general rule I would define the equations as x₁(t), y₁(t), x₂(t), and y₂(t).

The distance between two points is √{[x₁(t) - x₂(t)]² + [y₁(t) - y₂(t)]²}.

And then find the minimum value of the resulting function.

 

[RTW69]

I suggest using the distance formula however the coordinates will be function of time so d=f(t). You can graph d vs t and select smallest d or take the derivative d[d]/dt and set equal to zero

 

[xzibition8612]

alright i tried out the distance formula. sqrt [ (50t-40)^2+(20t)^2 ]

I get sqrt [1900t^2-2400]

then what? i don't think i can set it equal to anything...
thanks

 

[Villyer]

alright i tried out the distance formula. sqrt [ (50t-40)^2+(20t)^2 ]

This is right.

I get sqrt [1900t^2-2400]

I don't believe this is right.

sqrt [ (50t - 40)² + (20t)² ]
sqrt [ 2500t² - 4000t + 1600 + 400t² ]
sqrt [ 2900t² - 4000t + 1600 ]

This is the function for the distance between the two points, and the question asks for the minimum distance.

 

[gneill]

A small suggestion. If you are minimizing the distance d, then you are also minimizing d². This may be a useful consideration when it comes to minimizing the effort required to apply calculus...

 

[xzibition8612]

ok so far i got d^2=2900t^2-4000t+1600. Now the question is to find a t such that d would be at a minimum. How do I do that? Is taking the derivative with respect to t possible? But since d depends on t, wouldn't I get 2dd'?
Thanks.

 

[RTW69]

d=Sqrt(2900t²-4000t +1600)

Take the the derivative using chain rule, simplify, set numerator = 0, solve for t. The math will be easy

 

[xzibition8612]

so d will become zero since it's constant? and what does set numerator=0 mean? thanks

 

[gneill]

As I suggested, you can ignore the square root and minimize d², that will result in a minimum d, too.

So how do you go about finding a minimum (or a maximum for that matter) of a function f(t) with respect to t?

 

[RTW69]

If you are not familiar with differential Calculus, use your graphing calculator to graph the distance formula (Distance vs. time). Find the point on the graph where d is the smallest. What is the t value at that point? You can use a trace command or other features on your calculator to find the minimum d.