Blogs, Calculus, WSDA

Mole rat hunt-Part 2 WSDA

In Mole rat hunting, we began our hunt and we were able to get 3 mole rats.  Even though this would be a successful hunt, we noticed that there was still one rat left.  This rat was running away from us, so we wanted to know if we would be able to catch up to it.  We saw last time, that, yes we were currently gaining ground on it.  With the hope that we would successfully catch one more mole rat, we find out today if we catch it or not.

If you would prefer the video version of this lesson in survival, I have embedded the video below.  You can also view it here.

Mole rat

We begin by finding the velocity and position functions for the mole rat.  We recall from last time that the rat is running at 4 mph down a hill.  Below, we have a visual representation of this.

Since we are running horizontally, and the mole rat is running along the diagonal, we will break this velocity into its vector components going horizontally in the \(w\) direction and vertically in the \(h\) direction.  In order to do this, we see that the mole rat’s triangle is similar to the original triangle we had for our bridge and billboard.  This is because the blue and yellow lines are both parallel, so the corresponding angles are equal.

We now let \(w_{r}\)  [\(h_{r},d_{r}\)] be the position of the mole rat in the \(w\) [\(h,d\)] directions, respectively.  This gives us that \[\begin{align*} \frac{\frac{dw_{r}}{dt}}{\frac{dd_{r}}{dt}}&=\frac{54}{59} \\ \frac{dw_{r}}{dt}&=5.4,\end{align*}\] and  \[\begin{align*} \frac{\frac{dh_{r}}{dt}}{\frac{dd_{r}}{dt}}&=\frac{-24}{59} \\ \frac{dw_{r}}{dt}&=-2.4.\end{align*}\]

With the directional velocities of the mole rat, we can find the directional positions since position is the integral of velocity.  Using the second part of the Fundamental Theorem of Calculus, we get that \[\begin{align*} w_{r}&=\int_{0}^{t}(\frac{dw_{r}(s)}{ds}ds)+w_{r}(0) \\ &=\int_{0}^{t}5.4ds+54 \\ &=5.4t+54.\end{align*}\]  In a similar manner, we find that \[\begin{align*} h_{r}&=\int_{0}^{t}(\frac{dh_{r}(s)}{ds}ds)+h_{r}(0) \\ &=\int_{0}^{t}-2.4ds+0\\ &=-2.4t.\end{align*}\]

Now that we know where the mole rat will be at any point in time, we will need to find our position functions so that we can compare.

My position

When finding my position, the problem becomes more complicated than it was for the mole rat.  This is because my movement will look different depending on whether I am still on the bridge, falling through the air, or am on the hill.  As such, we will find our position using a piece-wise defined function by looking at each of these cases separately.

 On bridge

On the bridge, my movement is fairly straightforward.  I am moving in the \(w\) direction at a rate of 6 mph, or 8.8 ft/sec.   That is, if we let \(w_{p}\) be my position in the \(w\) direction and \(h_{p}\) my position in the \(h\) direction, we get that \(\frac{dw_{p}}{dt}=8.8\) and \(\frac{dh_{p}}{dt}=0\).  We can again use the fundamental theorem to see that \[\begin{align*} w_{p}&=\int_{0}^{t}8.8ds+0=8.8t, \\ h_{p}&=\int_{0}^{t}0dt+24=24.\end{align*}\]

The next thing we need to find is how long we will be travelling on this path.  We will be on the bridge as long as we haven’t reached the end of the bridge.  That is, we need \(w \leq 54\).  Therefore, we will be on the bridge for \(\frac{54}{8.8}=6.14\) seconds.  Hence, we know our position for the first 6.4 seconds of the pursuit.


As we begin to fall, we note that, as long as we ignore air resistance there is no force slowing us down in the \(w\) direction.  However, in the \(h\) direction, gravity will pull us toward the ground causing us to accelerate at 32 \(\frac{ft}{sec^{2}}\).  Therefore, in the \(w\) direction, we have that \[\begin{align*} w_{p}&=\int_{6.14}^{t}8.8ds+w_{p}(6.14) \\ &=8.8t-54+54=8.8t.\end{align*}\]  In the \(h\) direction we find that \(\frac{d^{2}h_{p}}{dt^{2}}=-32\).  We will therefore have to integrate twice to find that \[\begin{align*} \frac{dh_{p}}{dt}&=\int_{6.14}^{t}-32ds+0 \\ &=-32t+196.48, \text{ so } \\ h_{p}&=\int_{6.14}^{t}-32s+196.48ds+h_{p}(0) \\ &=-16t^{2}+196.48t-579.2.\end{align*}\]

Now that we have our position functions while we are falling, we now need to determine when we stop falling.  We will clearly stop falling when we hit then ground, so we need to find the time when this happens.

Taking another look at our picture, we should note that the height of the ground changes as w changes.  Therefore, we cannot just find when the above \(h_{p}=0\).  We instead have to find the height of the hill as a function of \(w\) and determine when the \(h_{p}\) would be the same for the given \(w_{p}\).  In order to do this, we start by seeing that the hill’s height is linearly related \(w\).  In fact, we can see that the \(w\)-intercept is \(24\) and that at \(w=54\), we have \(h=0\).  Therefore, we get that \(h=24-\frac{24}{54}w\).  Since we know \(w_{p}=8.8t\), we then find \[\begin{align*}h_{p}&=h \\ -16t^{2}+196.48t-579.2&=24-\frac{24}{54}(8.8t) \\ t&=5.03,7.5.\end{align*}\]

We know that we will still be on the bridge at 5.03 seconds, so this solution doesn’t make sense within the context of the problem.  Therefore, we will be falling from the bridge until 7.5 seconds after we started running.

On the ground

When we finally land on the ground, we will assume that we will continue running at the 8.8 ft/sec that we were running on  the bridge.  However, this will be along the hill-side instead of the bridge.  If we want to find our horizontal and vertical positions, we then need to break this velocity into its vector components as we did for the mole rat.  Following the same outline using similar triangles, we find the \(\frac{dw_{p}}{dt}=8.05\) and \(\frac{dh_{p}}{dt}=-3.58\).  We can then find that \[\begin{align*}w_{p}&=\int_{7.5}^{t}8.05ds+8.8*7.5 \\ &=8.05t+5.63, \\ h_{p}&=\int_{7.5}^{t}-3.58ds+(24-\frac{24}{54}(8.8*7.5)) \\&=-3.58t+21.5.\end{align*}\]

Combining this information

In order to make it easier to look at, we will condense our finding.  We then get that \[\begin{align*}w_{r}&=5.4t+54. \\h_{r}&=-2.4t. \\w_{p}&=\begin{cases} 8.8t \text{ if } 0 \leq t \leq 7.5, \\8.05t+5.63 \text{ if } 7.5 < t. \end{cases}\\h_{p}&=\begin{cases}24 \text{ if } 0 \leq t \leq 6.14, \\-16t^{2}+196.48t-579.2 \text{ if } 6.14 < t \leq 7.5, \\-3.58t+21.5 \text{ if } 7.5 < t. \end{cases} \end{align*}\]

In order to determine when we catch up to the mole rat, we could let either \(w_{p}=w_{r}\) or \(h_{p}=h_{r}\).  If we do this, we note that we will catch up to the mole rat after 18.25 seconds.  In this time we would have run 152.5 feet horizontally and falling or run 43.8 feet vertically.


Now that we’ve seen this, would it be worth running after the mole rat?  That seems like a good distance to run in order to catch up with the rat, so maybe it would be worth staying back and taking the 3 we already have.  Otherwise, maybe you are really hungry, so it would be worth chasing after him.

Note that, if you view the video, you will see that I did chase after the mole rat, with some devastating results.  If you’d like to see what happened, head over to the end of the video.  This happened not because we did anything wrong in our calculations, but rather because we forgot to consider anything other than our food.

I hope you’ve learned something from the post about survival, integrals and some trig.  If you have and if you’ve enjoyed, let me know by liking the post below, or by sharing it on social media.  I hope you follow us here or subscribe to the YouTube page so that you will be updated when we release new survival videos!

1 thought on “Mole rat hunt-Part 2 WSDA”

We'd love to hear your thoughts!

This site uses Akismet to reduce spam. Learn how your comment data is processed.