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Minecraft Calculus: River Crossing Escape

This is a classic introductory calculus problem, with a Minecraft theme.  This particular variation is inspired by example 4 in section 4.5 of Stewart's calculus.

I'll set the scene:  You're weaponless.  You have half a heart and you're being chased by a skeleton archer.  There's a river between you and your cabin.  You need to get to your cabin as quickly as possible!

In more "mathy" terms, you want to get from A to B.  There are three main ways you can do this:

  1. A to B directly, diagonally across the river
  2. A to C and then C to B
  3. At slight angle from A to D and then D to B

We know that swimming is slower than running, so option 1 isn't the best.  Option 2 involves the smallest time in the water, but also the longest distance traveled.

Option 3 is somewhere in between - you spend a bit more time in the water, but the total distance is decreased.  The travel time in this case depends on where exactly point D is.  To find the optimal path, we must find the position of D which minimizes the travel time.  This sounds like calculus!

Let x be the distance between points C and D.  In math terms, |CD|=x.

We first need an expression for the total time traveled in terms of x.  The basic equation for time traveled at constant speed is

t=\frac{distance}{speed}.

The first part of the trip is in the water where we travel from A to D.  We can use Pythagorean's theorem to get this distance as |AD|=\sqrt{x^2+w^2}.  Assuming that we can travel at a speed of v_w in the water, the time for this part is

t_w=\frac{\sqrt{x^2+w^2}}{v_w}.

The land part of the trip involves traveling what's left over of l after having already traveled x, so |DB|=l-x.  Traveling at v_l on land, the time for this part of the trip is

t_l=\frac{l-x}{v_l}.

The total trip time, T(x), is just the sum of these two.

T(x)=\frac{\sqrt{x^2+w^2}}{v_w}+\frac{l-x}{v_l}     (1)

To optimize this function with respect to x, we need to find where it is stationary and then verify that this point is a minimum.  This means we want to find a place where the function isn't changing with small changes in x.  In other words, we want to find a spot where the derivative is zero.

Taking the derivative gives:

T'(x)=\frac{x}{v_w\sqrt{x^2+w^2}}-\frac{1}{v_l}

Setting this equal to zero, we can solve for x:

x=\frac{v_ww}{\sqrt{v_l^2-v_w^2}}     (2)

Let's plug in some numbers!  In Minecraft, you can swim at about 2.2 m/s and sprint at 5.6 m/s.  Let's take the river width w to be 7 blocks (1 block = 1 meter).  Plugging in, we find that x=2.9.  At this point, though, we can't tell if this is a maximum or a minimum.  One way to find out is to examine the curvature of the function at this point by using the 2nd derivative:

T''(x)=\frac{1}{v_w\sqrt{x^2+w^2}}(1-\frac{x^2}{x^2+w^2})

Plugging in x=2.9, we see that T''(2.9)=0.04 which, being positive, means that x=2.9 is a minimum point for T(x).  Visually:

crossing

 

So if you want to get to your cabin as quickly as possible, the fastest route is to swim across the river to a point D that is 2.9 meters from point C and then run the rest of the way.

This is actually a general problem.  Try replacing the speeds I used with speeds for soul sand, crouching and walking, a boat, etc. and see what happens!  Particularly, what happens to (2) if you can travel more quickly in the water than on land?

 

Scratch Work

Finding T'(x):  First, rewrite the square root as a power,  T(x)=\frac{(x^2+w^2)^{\frac{1}{2}}}{v_w}+\frac{l-x}{v_l}

Differentiating, using the chain rule on the first term,  T'(x)=\frac{\frac{1}{2}(x^2+w^2)^{-\frac{1}{2}}(2x)}{v_w}-\frac{1}{v_l}=\frac{x}{v_w\sqrt{x^2+w^2}}-\frac{1}{v_l}

Finding T''(x):  Starting with T'(x)=\frac{(x^2+w^2)^{-\frac{1}{2}}x}{v_w}+\frac{1}{v_l}, we differentiate with respect to x.  Note that the derivative of \frac{1}{v_l} with respect to x is 0, so we only have to deal with the first term.  Using the product rule and chain rule:

T''(x)=\frac{-\frac{1}{2}(x^2+w^2)^{-\frac{3}{2}}(2x^2)}{v_w}+\frac{(x^2+w^2)^{-\frac{1}{2}}}{v_w}

Cleaning up with some algebra:

T''(x)=\frac{-x^2}{v_w(x^2+w^2)^{\frac{3}{2}}}+\frac{1}{v_w\sqrt{x^2+w^2}} =\frac{1}{v_w\sqrt{x^2+w^2}}(1-\frac{x^2}{x^2+w^2})

 

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