Calculus isn’t useless: When Velociraptors Attack

1. Some critics may sneer calculus is useless. Oh yeah? Well, we can solve this problem:

Question: The velociraptor spots you 40 \rm\ meters away and attacks, accelerating at 4 \rm\ m/s^2 from a stand start, up to its top speed of 25 \rm\ m/s. When it spots you, you begin to flee, quickly reaching your top speed of 6 \rm\ m/s. How far can you get before you’re caught and devoured?

2. Set up. Lets pick coordinates so you are standing at the origin on the x axis. The velociraptor is at (40,0), the x axis is measured in meters.

So let u(t) describe your position, and let r(t) describe the raptor’s position at time t.

You run at 6 meters/second, and that can be described by
(1)\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}u(t)=-6
(it’s negative because you are running away from the raptor).

However, the velociraptor is accelerating at a constant rate. So
(2)\displaystyle \frac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}r(t)=-4
We can assume that initial velocity for the raptor is zero and again r(0)=40.

The problem becomes: when will u(t)=r(t)?

3. Execute. We can solve for your position immediately since (1) gives us (integrating):
(3)\displaystyle u(t)=u(0)+\int^{t}_{0}\frac{\mathrm{d}}{\mathrm{d}\tau}u(\tau)\,\mathrm{d}\tau=-6t
Now we did something a little tricky here! We used a dummy variable in the integral.

Why didn’t we use t? Because it’s the boundary of the interval. It’d be like saying 0\leq t\leq t instead of 0\leq\tau\leq t.

The raptor is trickier, but lets break it up into parts. We know its acceleration. We want to find its velocity. Let v(t) be the raptor’s velocity, and v(0)=0. Then
(4)\displaystyle v(t)=v(0)+\int^{t}_{0}\frac{\mathrm{d^{2}}}{\mathrm{d}\tau^{2}}r(\tau)\,\mathrm{d}\tau=-4t
This is partially true. Really, since it has a maximum speed of 25 meters/second, we should write
(5)\displaystyle v(t)=\begin{cases}-4t & \mbox{if }t<25/4\\  -25 & \mbox{otherwise}\end{cases}
Observe that the velocity reaches its maximum at 25/4 = 6.25 seconds.

How far will the raptor go? Again, we just use the fundamental theorem of calculus writing
(6)\displaystyle r(t)=r(0)+\int^{t}_{0}v(\tau)\,\mathrm{\tau}=\begin{cases}40-2t^{2} & \mbox{if }t<6.25\\  40-2(6.25^{2})-25t & \mbox{if }t\geq6.25\end{cases}
We see that
(7)\displaystyle   \begin{array}{rl}r(6.25)  &=40-2(6.25^{2})\\  &=40-2(39.0625)\\  &=40-78.125=-38.125\end{array}
During this time you would be at u(6.25)=-37.5.

We now know that you’ll be devoured in under 6.25 seconds!

4. Solving. So since t\leq 6.25, we will use
(8)\displaystyle r(t)=40-2t^{2}
Setting this equal to Equation (3) we have
(9)\displaystyle \underbrace{0-6t}_{u(t)}=\underbrace{40-2t^{2}}_{r(t)}
This is a quadratic equation, and rearranging terms to make it prettier gives us
(10)\displaystyle 2t^{2}-6t-40=0
Using the quadratic formula, we can tell exactly when the raptor will devour you! We have two solutions
(11a)\displaystyle t_{1}=\frac{1}{2}(3-\sqrt{89})\approx-3.21699057
which doesn’t make sense (since the raptor spots you at time t=0 and everything takes place for t\geq0). The other solution is
(11b)\displaystyle t_{0}=\frac{1}{2}(3+\sqrt{89})\approx6.21699057
which is really close to when the raptor will achieve maximum velocity!

How far did you get? Well, we see that
(12)\displaystyle u\left(\frac{3+\sqrt{89}}{2}\right)=-3(3+\sqrt{89})\approx-37.3019434
In other words, you got 37 meters away! (About 122 feet) Bravo!


About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Applications, Calculus, Differential, Integral, Optimization. Bookmark the permalink.

2 Responses to Calculus isn’t useless: When Velociraptors Attack

  1. Pingback: [Index] Calculus of a Single Variable | My Math Blog

  2. Pingback: [Index] Calculus in a Single Variable | My Math Blog

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