First Principles AI

Solution

Given:

H(t)=100+20t,\qquad T(t)=50+10t+t^2

Set $H(t)=T(t)$ :

\begin{aligned} 100+20t&=50+10t+t^2\\ 0&=t^2-10t-50 \end{aligned}

Solve:

\begin{aligned} t&=\frac{10\pm\sqrt{(-10)^2-4(1)(-50)}}{2}\\ &=\frac{10\pm\sqrt{100+200}}{2}\\ &=\frac{10\pm\sqrt{300}}{2}\\ &=\frac{10\pm 10\sqrt{3}}{2}\\ &=5\pm 5\sqrt{3} \end{aligned}

Time cannot be negative, so:

t=5+5\sqrt{3}\ \text{months}

\boxed{t=5+5\sqrt{3}\ \text{months}}\approx \boxed{13.66\ \text{months}}

Substitute into $H(t)$ :

\begin{aligned} H(5+5\sqrt{3})&=100+20(5+5\sqrt{3})\\ &=100+100+100\sqrt{3}\\ &=200+100\sqrt{3} \end{aligned}

\boxed{\text{Height }=200+100\sqrt{3}\ \text{m}}\approx \boxed{373.2\ \text{m}}

Initially, $H(0)=100$ and $T(0)=50$ , so the building is taller.

Since $T(t)$ includes a $t^2$ term, the tree eventually grows faster and overtakes the building after the intersection time $t=5+5\sqrt{3}$ .