worded problem

I'm not really sure how to approach this problem. It's probably got something to do with differentiation.

At noon the captain of a ship sees two fishing boats approaching. One of them is 10km due east and travelling west at 8km/h. The other is 6km due north and travelling south at 6 km/h. At what time will the fishing boats be closest together, and how far apart will they be.

Thanks guys! :)

The distance is at any time t is given by,

D = Sqrt[(x₁(t) - x₂(t))² + (y₁(t) - y₂(t))²]

Now, if one boat moves along the x axis, then its y(t) = 0 for all t. Likewise, the other boat moving along the y axis will have x(t) = t for all t. Therefore the formula can be simplified:

D = Sqrt[x₁(t)² + y₂(t)²]

where I have assumed that boat 1 moves along the x axis and boat 2 along the y axis. Both boats having constant velocity, the formulas for x(t) and y(t) are easy enough:

x₁(t) = x₀ - v₁t
y₂(t) = y₀ - v₂t

where x₀ and y₀ are the initial positions of each boat and I have used the minus sign because I have assumed the captain's at the coordinate origin so the fishing boats travel in the negative x and y directions.

Let's use the squared distance:

D² = (x₀ - v₁t)² + (y₀ - v₂t)²

The minimum D² (and therefore the minimum D) occurs when the derivative is 0.

0 = dD² / dt = -2v₁(x₀ - v₁t) -2v₂(y₀ - v₂t)

Solving for t:

t_min = (x₀v₁ - y₀v₂) / (v₁² - v₂²)

[Please, notice: there's a mistake in this formula. It has been corrected in my next post below!]

so this is the time for which the distance is a minimum. This distance is found by substituting t into the formula for the distance. Using your data I have found:

t_min = 1.571 hours
Min. distance = 4.286 km

...but don't believe every word I write... I often make mistakes... HTH

According to the book, that isn't the right answer. Apparently the time is 1.16 hours after noon, and the distance is 1.2 km.

I was following for the first bit, but I'm not sure what's happening here:

x1(t) = x0 - v1t
y2(t) = y0 - v2t

Quote:

---

Originally Posted by krtxmrtz

...but don't believe every word I write... I often make mistakes... HTH

---

Your book is right and the above quote is therefore appropriate. :blush:

The mistake I made was in solving for t because I didn't double-check the formula. The correct answer for t is:

t_min = (x₀v₁ + y₀v₂) / (v₁² + v₂²)

Using this you get the book answers.

As for these formulas:

x₁(t) = x₀ - v₁t
y₂(t) = y₀ - v₂t

you know that for constant velocity,

v = x / t where x is the space traveled by the moving object and t the time it takes, so x = vt

This is so if the object starts travelling at x = 0 when t=0, in other words, if the origin of the observer's coordinate system is the point the object is at when t = 0.

Now, if for t = 0 the object is at a distance, you must add this distance to vt. So, in general, for objects moving at constant velocity,

x = x₀ + vt where x₀ is the initial distance (or position)

In the present case, the object is approaching the origin, i.e. it is travelling along the x axis in the direction of decreasing values. Then, either:

x = x₀ - vt

or

x = x₀ + vt

but then v must be taken as negative (-8 km/h and -6 km/h for your boats)

Thanks for your help!