## 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!

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.