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 
to 
. There are three main ways you can do this:
- to directly, diagonally across the river
- to
and then to
- At slight angle from to
and then to
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 is. To find the optimal path, we must find the position of which minimizes the travel time. This sounds like calculus!
Let 
be the distance between points and . In math terms, 
.
We first need an expression for the total time traveled in terms of . The basic equation for time traveled at constant speed is
.
The first part of the trip is in the water where we travel from to . We can use Pythagorean's theorem to get this distance as 
. Assuming that we can travel at a speed of 
in the water, the time for this part is
.
The land part of the trip involves traveling what's left over of 
after having already traveled , so 
. Traveling at 
on land, the time for this part of the trip is
.
The total trip time, 
, is just the sum of these two.
(1)
To optimize this function with respect to , 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 . In other words, we want to find a spot where the derivative is zero.
Taking the derivative gives:
Setting this equal to zero, we can solve for :
(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 
to be 7 blocks (1 block = 1 meter). Plugging in, we find that 
. 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:
Plugging in , we see that 
which, being positive, means that is a minimum point for . Visually:
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 that is 2.9 meters from point 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 
First, rewrite the square root as a power, 
Differentiating, using the chain rule on the first term, 
Finding 
Starting with 
, we differentiate with respect to x. Note that the derivative of 
with respect to is 0, so we only have to deal with the first term. Using the product rule and chain rule:
Cleaning up with some algebra:
14x14x3 quarry:
and the number of blocks to dig down at each diameter, 
.
blocks deep, where 
if 
if 
by 
.
, is found by simply adding up all the blocks at a given level and then adding all the levels. Noting that 
is related to 
. Now, the summation bit is straightforward to see, but the upper limit 
