Westworld 1973 Problem #3

Calculus Level 3

In order to bring the manic robots under control, Delos' Chief Engineer orders to completely shut down power to the entire park. Unfortunately, this cut power to the Control Center's ventilation system and electric doorways.....effectively trapping them inside!

At the start of the shutdown, the Control Center was a comfortable $68^{\degree} F$ . After $t=60$ minutes, the ambient temperature rose to $98^{\degree} F$ , and the oxygen level had fallen to $17$ % volume (NOTE: There was 100% oxygen at the start of the shutdown with certain death occurring at 0% volume).

Knowing the ambient temperature can be modeled per $T(t) = 68 + \alpha \cdot \ln(t+1)$ (where $t =$ minutes), and the oxygen level can be modeled per $A(T) = 1 - \sqrt{\beta(T-68)}$ , the oxygen level rate can be computed according to:

$dA/dt = dA/dT \times dT/dt$

Of course, 'Westworld' wouldn't be the gripping sci-fi thriller it is if Delos personnel survived the chaos of the robots' mass-malfunction! The question remains: What was the oxygen level's depletion rate at the time the Control Center met its demise (as pictured above). The answer is of the form $k \times 10^{-4}$ , which you're required to enter ' $k$ '.

The answer is -2.148.

2 solutions

David Vreken
Mar 28, 2021

Since after $t = 60$ minutes the temperature rose to $98°F$ , $T(60) = 68 + \alpha \cdot \ln(60 + 1) = 98$ , which solves to $\alpha = \cfrac{30}{\ln(61)}$ , so $T(t) = 68 + \cfrac{30}{\ln(61)} \cdot \ln (t + 1)$ .

Substituting $T(t) = 68 + \cfrac{30}{\ln(61)} \cdot \ln (t + 1)$ into $A(T) = 1 - \sqrt{\beta(T - 68)}$ gives $A(t) = 1 - \sqrt{\beta \cdot \cfrac{30}{\ln(61)} \cdot \ln (t + 1)}$ .

Since after $t = 60$ minutes the oxygen level had fallen to $17 \%$ , $A(60) = 1 - \sqrt{\beta \cdot \cfrac{30}{\ln(61)} \cdot \ln (60 + 1)} = 0.17$ , which solves to $\beta = \cfrac{0.83^2}{30}$ , so $A(t) = 1 - \cfrac{0.83}{\sqrt{\ln 61}} \sqrt{\ln (t + 1)}$ .

The oxygen's rate of depletion is then $A'(t) = \cfrac{0.83}{2\sqrt{\ln 61} (t + 1) \sqrt{t + 1}}$ .

The Control Center met its demise at $A(t) = 1 - \cfrac{0.83}{\sqrt{\ln 61}} \sqrt{\ln (t + 1)} = 0$ , which solves to $t = e^{\frac{\ln 61}{0.83^2}} - 1$ .

At this time, the oxygen's rate of depletion was $A'(t) = \cfrac{0.83}{2\sqrt{\ln 61} (e^{\frac{\ln 61}{.83^2}} - 1 + 1) \sqrt{e^{\frac{\ln 61}{.83^2}} - 1 + 1}} = \cfrac{0.83^2}{2 (\ln 61) e^{\frac{\ln 61}{0.83^2}}} \approx \boxed{-2.146} \times 10^{-4}$ .

Nice alternative solution, David....employing direct substitution of T(t) into A(T)!

tom engelsman - 2 months, 2 weeks ago

Thanks! I don't know if you noticed but you had -2.148 instead of -2.146, but I think it was because you rounded the time to 388.94. Anyway, it was close enough for the Brilliant algorithms to mark mine as correct.

David Vreken - 2 months, 2 weeks ago

Yup, just under 6 1/2 hrs. to transform the Control Center into a sealed Mojave Desert for the hapless Chief Engineer and his team.....yikes! I'd rather take my chances with Yul Brynner's Gunslinger droid instead ;)

Probably the most arduous part was trying to pick simple functions that could simulate this scene as close to reality as I could get while still sparking interest in the younger Brilliant.org solvers. Stay tuned.....I'm planning on a few more 'Westworld' probs to add later!

tom engelsman - 2 months, 2 weeks ago

Tom Engelsman
Mar 28, 2021

Let us first tackle the coefficients $\alpha$ in $T(t)$ and $\beta$ in $A(T).$ We know that $T(60) = 98$ , which yields:

$68 + \alpha \cdot \ln(60+1) = 98 \Rightarrow \alpha = \large \frac{30}{\ln 61}$ $F$ ;

and we are given that $A(98) = 0.17$ , or:

$0.17 = 1 - \sqrt{\beta(98-68)} \Rightarrow \beta = \large \frac{(1-.17)^2}{30} = \large \frac{0.83^2}{30}$ $F^{-1}$ .

Now, the Control Center's temperature at $0$ % oxygen computes to: $0 = 1-\sqrt{(\frac{0.83^2}{30})(T-68)} \Rightarrow T = \frac{30}{0.82^2} + 68 = 111.54^{\degree}$ F. Likewise, the time it takes to reach this zero-oxygen level is: $111.54 = 68 + \frac{30}{\ln 61} \cdot \ln (t+1) \Rightarrow t = e^{(\frac{\ln 61}{30}) (111.54-68)} - 1 = 388.94$ minutes.

Calculating the respective derivatives gives:

$\large{\frac{dA}{dT}}|_{T=111.54} = -\frac{1}{2}\sqrt{\frac{\beta}{T-68}} = -\frac{1}{2}\sqrt{\frac{0.83^2/30}{111.54-68}} = -0.0115$ ;

$\large{\frac{dT}{dt}}|_{t=388.94} = \frac{\alpha}{t+1} = \frac{30/ \ln 61}{388.94+1} = 0.0187$

which gives us an ultimate oxygen depletion rate of:

$\frac{dA}{dt} = (-1.15 \times 10^{-2})(1.87 \times 10^{-2}) = \boxed{-2.148 \times 10^{-4}}$ , or approximately $-0.02\%$ per minute.

Westworld 1973 Problem #3

The answer is -2.148.

2 solutions

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