# Sleuthing with Differential Equations Deliverable

###### by Alexandra Ramirez, Josie Meltzer, and Miranda Lyons

-----------------------------------------------
# Problem 1: Contested Speeding Violation and the Wheels of Justice

## In a $35$ miles per hour speed zone a car driver applies and holds the brakes to the max and thus decelerates the car at $12 \frac{ft}{s^2}$ (according to the manufacturer specifications and settings) until the car stops in $225 ft$.

**a) How fast was the car traveling at the instant the driver applied the brakes? Make a recommendation to the law enforcement agency. Defend your recommendation with analysis.**

We are asked to find the velocity, $v(t)$, that the car was traveling at the time the brakes were applied. Before doing any calculations, we are given the following information,

* $35 mph$ speed zone
* Acceleration, $a(t)=-12\frac{ft}{s^2}$
* Distance, $s(t) = 225 ft$

By some simple calculus, we can find $v(t)$ by taking the integeral of $a(t)$, which is given. Consider the following,

$$
v(t) = \int a(t) dt \\
= \int -12 dt \\
= -12t + C_1
$$

Note that when $t=0$, $v(t)=0$. Therefore, we can write
$$v(0)=C_1$$

Similar to as before, we can find distance of time $t$, $s(t)$, by taking the integral of $v(t)$. Consider the following,

$$
s(t) = \int v(t) dt \\
= \int -12t + C_1 dt \\
= -6t^2 + C_1 t + C_2
$$

Note that when $t=0$, $s(0)=C_2$. Therefore,
$$s(t) = -6t^2 + C_1t$$

Now consider $v(0)$,

$$
v(0) = 0 \\
0=-12t +C_1 \\
t = \frac{C_1}{12}
$$

Now we will plug our found value of $t$ into the $s(t)$. Observe the following,

$$
s(t) = 225 \\
225 = -6 \bigg(\frac{C_1}{12} \bigg)^2 + C_1 \bigg(\frac{C_1}{12} \bigg) \\
225 = -6 \bigg(\frac{{C_1}^2}{144} \bigg)+ \frac{{C_1}^2}{12} \\
225 = -\frac{C_1}{24}+ \frac{2 {C_1}^2}{24} \\
225 = \frac{{C_1}^2}{24} \\
{C_1}^2 = 5400 \\
C_1 = 73.485 \frac{ft}{s}\
$$

We previously stated that $C_1 = v(0)$, thus $v(0)=73.485 \frac{ft}{s}$, which is what we were asked to find. The last thing we must do is convert $\frac{ft}{s}$ into $mph$. Consider,
| 73.485 ft | 1 mi | 3600 s |
| ----- | ---- | --- |
| s | 5280 ft | 1 hr |

$=50.103 mph$

Therefore, the car was traveling about $50.10 mph$ the instant the driver applied the brakes. We are previously given that the speed zone is $35 mph$, which means the driver was going about $15 mph$ over the speed limit. Our recommendation to the law enforcement would be to give the driver a ticket due to the fact he was above the speed limit by $15 mph$.

**b) The driver hires a lawyer who has the brakes system tested, showing the maximum deceleration possible is only $6 \frac{ft}{s^2}$. What does this say about how fast the car was traveling at the instant the driver applied the brakes? What should the wheels of justice do to the driver in this situation? Explain.**

The driver hires a lawyer who has the brakes system tested. We are given the following information,
* $35 mph$ speed zone
* Acceleration, $a(t)=-6\frac{ft}{s^2}$
* Distance, $s(t) = 225 ft$
By some simple calculus, we can find $v(t)$ by taking the integeral of $a(t)$, which is given. Consider the following,

$$
v(t) = \int a(t) dt \\
= \int -6 dt \\
= -6t + C_1
$$

Note that when $t=0$, $v(t)=0$. Therefore, we can write

$$v(0)=C_1$$

Similar to as before, we can find distance of time $t$, $s(t)$, by taking the integral of $v(t)$. Consider the following,

$$
s(t) = \int v(t) dt \\
= \int -6t + C_1 dt \\
= -3t^2 + C_1 t + C_2
$$

Note that when $t=0$, $s(0)=C_2$. Therefore,
$$s(t) = -3t^2 + C_1t$$

Now consider $v(0)$,

$$
v(0) = 0 \\
0=-6t +C_1 \\
t = \frac{C_1}{6}
$$

Now we will plug our found value of $t$ into the $s(t)$. Observe the following,

$$
s(t) = 225 \\
225 = -3 \bigg(\frac{C_1}{6} \bigg)^2 + C_1 \bigg(\frac{C_1}{6} \bigg) \\
225 = -3 \bigg(\frac{{C_1}^2}{36} \bigg)+ \frac{{C_1}^2}{6} \\
225 = -\frac{C_1}{12}+ \frac{2{C_1}^2}{12} \\
225 = \frac{{C_1}^2}{12} \\
{C_1}^2 = 2700 \\
C_1 = 51.962 \frac{ft}{s}
$$

We previously stated that $C_1 = v(0)$, thus $v(0)=51.962 \frac{ft}{s}$, which is what we were asked to find. The last thing we must do is convert $\frac{ft}{s}$ into $mph$. Consider,

| 51.962 ft | 1 mi | 3600 s |
| ----- | ---- | --- |
| s | 5280 ft | 1 hr |

$=35.429 mph$

According to the brake system test, the driver must have been traveling about $35.429 mph$, which would be the appropriate speed for the given speed zone. Therefore, the wheels of justice should not give the driver a ticket because he was in fact going the correct speed the instant the brakes were applied.

-------------------------------------------------

# Problem 2: I Throw an Arrow into the Air, it Lands I Know Where!

## We shoot a small steel ball vertically into the air at a speed of $22 \frac{m}{s}$ in an upward direction, releasing it from ground level. Build a mathematical model of the altitude of the ball, $s(t)$, in feet at time $t$ in seconds, knowing that it experiences only constant acceleration due to gravity of $9.8 \frac{m}{s^2}$. Keep the model simple and state your assumptions.

Assumptions:

* No air resistance
* No wind

**a) How long is it until the ball strikes the ground?**

We are being asked to find time, $t$, the ball hits the ground. Before doing any calculations, we are given the following information:

* Ball is being thrown vertically into the air
* Acceleration, $a(t) = -9.8\frac{m}{s^2}$
* Ball is thrown at a speed of $22\frac{m}{s}$

We can find the velocity $v(t)$ by taking the integral of $a(t)$. This gives us

$$
v(t) = \int a(t) dt \\
= \int -9.8 dt \\
= -9.8t + C_1
$$

Note that when $t=0, v(0)=22$. We can write $C_1$ as

$$
v(0)=C_1 \\
v(t)=-9.8t+22
$$

To find the position of the ball at time $t$, we can take the integral of $v(t)$ to find $s(t)$.

$$
s(t) = \int v(t) dt \\
= \int -9.8t+22 dt \\
= -4.9t^2+22t+C_2
$$

Note that when $t=0, s(0)=0$. We can write $C_2$ as

$$
s(0)=C_2\\
s(t)=-4.9t^2+22t
$$

Now that we have all the information needed, we can find the time $t$ that the ball strikes the ground. To find this we can set $s(t)=0$.

$$
0=-4.9t^2+22t\\
=t(-4.9t+22)
$$

This gives us two different equations:

$$
0=t\\
and\\
0=-4.9t+22
$$

After solving for time $t$, we find that the ball strikes the ground at $t=0$ secs and $t=4.49$ secs which gives us the start time and end time.

**b) How fast is it traveling when striking the ground?**

To find how fast the ball is traveling we have to find the speed or the velocity of the ball when $t = 4.49$. We have:

$$
v(t) = -9.8t + 22 \\
v(4.49) = -9.8(4.49) + 22 \\
= -44 + 22 \\
= -22
$$

Therefore the ball is traveling $22 \frac{m}{s}$. Note that the negative above tells us the direction the ball is going in.

**c) What is its maximum altitude?**

In part (a) we found that the formula for the distance of the ball is $s(t) = -4.9t^2 + 22t$. We don't know the time but we do know that velocity would be zero at that time, $v(t) = 0$, at the maximum altitude. Consider this,

$$
v(t) = -9.8t + 22 \\
0 = -9.8t + 22 \\
-22 = -9.8t \\
t = 2.245
$$

Now that we have $t = 2.245$ we can find the maximum altitude. We have:

$$
s(2.245) = -4.9(2.245)^2 + 22(2.245) \\
= -4.9(5.04) + 49.39 \\
= -24.696 + 49.39 \\
= 24.694
$$

Therefore our maximum altitude is $s(t) = 24.694 m$

**d) How long does it stay at a height equal to 75% of its maximum height? 90% of its maximum height?**

Previously we found that the maximum height is 24.694 meters. We can then find 75% of its maximum height by completing the following,
$$ (24.694)(0.75)=18.521 $$
Now we must find how long it stays at a height equal to 18.521 meters. To do this, we will set our formula for $s(t)$ equal to 18.521 and solve for time, $t$. Consider,

$$ 18.521 = -4.9t^2 + 22t \\
0 = -4.9t^2+22t-18.521 \\
t=\frac{-22 \pm \sqrt{22^2-4(-4.9)(-18.521)}}{2(-4.9)} \\
t=\frac{-22 \pm 11}{-9.8} \\
t= 1.122, 3.367 $$

Therefore, the ball stays at 75% of it's maximum height when $t=1.122 s$ and $t=3.367 s$.

Similarly, we can find 90% of its maximum height by completing the same process as above. Observe,
$$(24.694)(0.90)=22.225 $$

Now we must find how long it stays at a height equal to 22.225 meters. To do this, we will set out formula for $s(t)$ equal to 22.225 and solve for time, $t$. Consider,

$$ 22.225 = -4.9t^2 + 22t \\
0 = -4.9t^2+22t - 22.225 \\
t = \frac{-22 \pm \sqrt{22^2-4(-4.9)(-22.225)}}{2(-4.9)} \\
t = \frac{-22 \pm 6.96}{-9.8} \\
t= 1.535, 2.955 $$

Therefore the ball stays at 90% of it's maximum height when $t= 1.535 s$ and $t=2.955 s$.

**e) When is the ball traveling at one half its maximum speed?**
We know that the ball's maximum speed is $22 m/s$ and $-22 m/s$. To find when the ball is traveling one half it's maximum speed, we must first find what half of the maximum speed is. Observe the following,
$$(22)(0.5)=11 m/s \\
(-22)(0.5)=-11 m/s$$

Let's first consider when $v(t)=11 m/s$, we can set the formula $v(t)$ equal to 11 and solve for time, $t$. Consider,
$$11 = -9.8t + 22 \\
-11 = -9.8t \\
t = 1.122 s $$

This tells us that the ball is traveling one half it's maximum speed at $t=1.122$. However, we must now look at when $v(t)=-11 m/s$. We can set the formula for $v(t)$ equal to -11 and solve for time, $t$. Consider the following,
$$ -11 = -9.8t + 22 \\
-33 = -9.8t \\
t = 3.367 s $$
This tells us that the ball is traveling one half it's maximum speed at $t=3.367 s$. Therefore, the times that the ball is traveling at one half it's maximum speed is when $t=1.122 s$ and $t=3.367 s$.

**f) Over what time interval(s) does the ball have an average velocity equal to 20% of its maximum velocity? Characterize these intervals completely.**

We know that the maximum velocity is $22 \frac{m}{s^2}$ therefore, 20% of the maximum velocity is $$ 22(.20) = 4.4$$

We're trying to find the interval $[t_1, t_2]$ in which the average velocity is equal to $4.4$. We'll use the average formula, $\frac{s(t_2) - s(t_1)}{t_2 - t_1}$ to get:

$$
= \frac{(-4.9(t_2)^2+22(t_2)) - (-4.9(t_1)^2+22(t_1))}{t_2 - t_1} \\
= \frac{-4.9(t_2)^2 + 4.9(t_1)^2 + 22(t_2) - 22(t_1)}{t_2 - t_1} \\
=\frac{-4.9(t_2^2 - t_1^2) + 22(t_2 - t_1)}{t_2 - t_1} \\
= \frac{-4.9(t_2 + t_1)(t_2 - t_1)}{t_2 - t_1} + \frac{22(t_2 - t_1)}{t_2 - t_1} \\
= 22 - 4.9(t_2 + t_1)
$$

Solving for $t_2$ to get:

$$
22 - 4.9(t_2 + t_1) = 4.4 \\
-4.9(t_2 + t_1) = -17.6 \\
t_2 + t_1 = 3.592 \\
t_2 = 3.592 - t_1
$$

A few other things we know:
* With the help of part (a), the start time is $t = 0$ and the end time is $t = 4.49$
* $t_ 2 > t_1$
* $t_2 \leq 4.49$

Now we can solve for $t_1$. Consider this,

$$
t_2 > t_1 \\
3.592 - t_1 > t_1 \\
3.592 > 2t_1 \\
1.796 > t_1
$$

Therefore the best way to characterize the interval $[t_1, t_2]$ is $0 < t_1 < 1.796$ and $t_2 = 3.592 - t_1$.

-------------------------------------------------

# Problem 3: Death of an Actor

### The great detective Sherlock Holmes and his assistant, Dr. Watson, are discussing the murder of actor Cornelius McHam. McHam was shot in the head, and his understudy, Barry Moore, was found standing over the body with the murder weapon in hand. Let’s listen in:

###### Watson: Open - and - shut case, Holmes. Moore is the murderer.

###### Holmes: Not so fast, Watson – you are forgetting Newton’s Law of Cooling!

###### Watson: Huh?

###### Holmes: Elementary, my dear Watson. Moore was found standing over McHam at 10:06 p.m., at which time the coroner recorded a body temperature of $77.9^{\circ} F$ and noted that the room thermostat was set to $72^{\circ} F$. At 11:06 p.m. the coroner took another reading and recorded a body temperature of $75.6^{\circ} F$. Since McHam’ s normal temperature was $98.6^{\circ} F$, and since Moore was on stage between 6:00 p.m. and 8:00 p.m., Moore is obviously innocent. Ask any calculus student to figure it out for you.

### How did Holmes know that Moore was innocent? Explain.

By Newton's Law of Cooling we know that the rate at which an object's temperture changes is proportional to the difference between the object's temperture and the ambient temperture. This tells us that $\frac{du}{dt} = k(u(t) - A)$. First, we have to find the general solution for the ODE. We have:

$$
\frac{1}{u(t)-A} du = k dt \\
\int \frac{1}{u(t)-A} du = \int k dt \\
\ln(u(t) - A) = kt + C \\
e^{\ln(u(t) - A)} = e^{kt + C} \\
u(t) - A = Ce^{kt} \\
u(t) = Ce^{kt} + A
$$

Here are some other given information:
* The room temperature is $A = 72^{\circ} F$.
* Moore was standing over McHam's body at 10:06 PM and the temperature recorded gives us $u(0) = 77.9^{\circ} F$.
* Checking the temperature again an hour later gives us $u(1) = 75.6^{\circ} F$.
* McHam's body temperature pior to death is $u(t) = 98.6^{\circ} F$ where we are trying to find $t$.

We first have to find $C$ of $u(t) = Ce^{kt} + A$ when $u(0) = 77.9$ so we have:

$$
77.9 = Ce^{k(0)} + 72 \\
C = 5.9
$$

Now that we have $C$ we have the updated general solution of $u(t) = 5.9e^{kt} + 72$. Next we find $k$:

$$
75.6 = 5.9e^{k(1)} + 72 \\
3.6 = 5.9e^{k} \\
\frac{3.6}{5.9} = e^{k} \\
\ln \bigg( \frac{3.6}{5.9} \bigg) = \ln (e^{k}) \\
k = \ln \bigg( \frac{3.6}{5.9} \bigg) = -0.494
$$

Now that we have $u(t) = 5.9e^{(-0.494)t} + 72$ we can solve for $t$:
$$
98.6 = 5.9e^{(-0.494)t} + 72 \\
26.6 = 5.9e^{(-0.494)t} \\
\frac{26.6}{5.9} = e^{-0.494t} \\
\ln \bigg( \frac{26.6}{5.9} \bigg) = \ln (e^{-0.494t}) \\
\ln \bigg( \frac{26.6}{5.9} \bigg) = -0.494t \\
t = \frac{\ln \big( \frac{26.6}{5.9} \big)}{-0.494} = -3.048
$$

Note that $0.048 * 60 = 2.88$ minutes. This tells us that the murder happened about 3 hours and 3 minutes before the first temperature was recorded. The time of death was approximately 7:03 PM. Therefore Moore is innocent since he was on stage from 6:00 PM to 8:00 PM.

--------------------------------------------------

# Problem 4: Time of Death

### It was hot for this London fall day – $70^{\circ} F$. Holmes arrived at Baker Street Annex to find the inspector hunched over the body. “It is important that we determine the exact time of death, sir, for in that way we may immediately determine the motive,” said the inspector. Not wishing to pursue the unpursuable non sequitur , Holmes took out his thermometer and after a few moments of discrete investigation, announced, “I say! $94.6^{\circ} F$. And it is presently noon.” With that he departed into the London fog, to return to the body at the same spot in one hour. After performing another investigation Holmes declared, “$93.4^{\circ} F$ at 1 o’clock.” And then silence. . . “Inspector, the murder occurred at exactly 8:58.51204 o’clock a.m. Good day to you, sir!” Later in their chambers Watson asked, “I say! How did you do that, Holmes?” “Elementary, my dear Watson. A simple application of a law of Newton,” said Holmes. “Here, here Holmes, you don’t mean to say he fell out of a window?” snapped Watson, sensing the gravity of the issue. Remaining cool Holmes began at the beginning, where he always began with Watson, "You see ..."

**1. Use a mathematical model to either refute or corroborate Sherlock Holmes’ conclusion. Be sure you state your assumptions and defend the use of your model.**

Just like problem 3 we'll use Newton's Law of Cooling formula: $u(t) = Ce^{kt} + A$. Based on the information given we know:

* The ambient temperature is $A = 70^{\circ} F$
* After the first temperature reading, we have $u(0) = 94.6^{\circ} F$
* After second temperature reading, we have $u(1) = 93.4^{\circ} F$
* The normal body temperature is $u(t) = 98.6^{\circ} F$ where we are trying to find $t$.

First, we have to find $C$ when $u(0) = 94.6^{\circ} F$:

$$
94.6 = Ce^{k(0)} + 70 \\
C = 24.6
$$

Now we have the updated formula of $u(t) = 24.6e^{kt} +70$. Next we'll find $k$ when $u(1) = 93.4^{\circ} F$:

$$
93.4 = 24.6e^{k(1)} + 70 \\
23.4 = 24.6e^{k} \\
\frac{23.4}{24.6} = e^{k} \\
\ln \bigg( \frac{23.4}{24.6} \bigg) = \ln (e^{k}) \\
k = \ln \bigg( \frac{23.4}{24.6} \bigg) = -0.05
$$

This gives us $u(t) = 24.6e^{-0.05t} + 70$. Next to find $t$:

$$
98.6 = 24.6e^{(-0.05)t} + 70 \\
28.6 = 24.6e^{(-0.05)t} \\
\frac{28.6}{24.6} = e^{-0.05t} \\
\ln \bigg( \frac{28.6}{24.6} \bigg) = \ln (e^{-0.05t}) \\
\ln \bigg( \frac{28.6}{24.6} \bigg) = -0.05t \\
t = \frac{\ln \big( \frac{28.6}{24.6} \big)}{-0.05} = -3.013
$$

Note that $0.013 * 60 = .78$ minutes. This tells us the murder happened about 3 hours and 1 minute before the first temperature reading. The time of death was approximately $8:59 AM$ and it corroborates Sherlock Holmes’ conclusion.

**2. Suppose the forensic folks confirm that the victim had viral evidence of fever with a body temperature of $103.2^{\circ} F$ at the time of death. How does this change the evidence you offer in (1)?**

We would still use the formula that we found in the first half of (1), $u(t) = 24.6e^{-0.05t} +70$. The thing that would change is that $u(t) = 103.2$ and we have to find a new $t$:

$$
103.2 = 24.6e^{(-0.05)t} + 70 \\
33.2 = 24.6e^{(-0.05)t} \\
\frac{33.2}{24.6} = e^{-0.05t} \\
\ln \bigg( \frac{33.2}{24.6} \bigg) = \ln (e^{-0.05t}) \\
\ln \bigg( \frac{33.2}{24.6} \bigg) = -0.05t \\
t = \frac{\ln \big( \frac{33.2}{24.6} \big)}{-0.05} = -5.996
$$

Note that $0.996 * 60 = 59.76$ minutes. This tells us that the murder happened about 6 hours before the first temperature reading. The time of death was approximately $6:00$ AM. About 3 hours difference from the time of death in (1).

**3. What assumption are you making about the environmental temperature during the period from death until discovery? Drop this simplifying assumption and offer up a more realistic model. Then address (1) again in this context. How does this change the evidence you offer in (1)? How does Holmes predictive abilities look now?**

| Time | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| ----- | ---- | --- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- |
| Temp. | 60.8 | 59 | 57.2 | 55.4 | 55.4 | 53.6 | 55.4 | 57.2 | 59.0 | 62.6 | 66.2 | 68.0 |

| Time | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 |
| ----- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |
| Temp. | 69.8 | 71.6 | 73.4 | 73.4 | 73.4 | 69.8 | 69.8 | 68.0 | 66.2 | 62.6 | 59.0 | 57.2| 55.4 |

The assumption that we were making for parts (1) and (2) was that the ambient temperature is consistent. The table above shows us that is not the case as the weather fluctuates throughout the whole day. So a more realistic model of Newton's Law of Cooling would be $u'(t) = k(u(t) - A(t))$. First, we plotted all the points from the table onto Desmos. We noticed that it looks like a sinusoidal wave. Through a bit of research we found our $A(t) = 9.9cos(\frac{\pi}{10}t + \frac{\pi}{2}) +63.5$. It's pretty close but after discussing it with Nandita, she gave us a more accurate function: $A(t) = 64.0175 + 9.42834cos(0.294373t+1.87252)$ with the help of sagemath. Due to the table, we have new initial conditions as well. Where $u(12) = 94.6^{\circ} F$ and $u(13) = 93.5^{\circ} F$. With the help of sagemath again we have that $k = -0.0522305$


Now that we have our $k$, we have two cases for finding the time of death.
Case 1: Plugging in $u(t) = 98.6$ we get:


$t = 9.42$ and multiplying it by 60 we get that the time of death was approximately $9:25 AM$

Case 2: When the victim had a fever at the time of death we plug in $u(t) = 103.2$ to get $t = 7.23$ This tells us that the time of death was approximately $7:14 AM$.

With our more realistic model Sherlock Holmes was way off, at least 30 minutes for both cases. Therefore his predictive abilities are not that accurate in these type of situations when time makes or breaks someone's alibi.

--------------------------------------