a) Two people start at the same point. One walks east at 3 km/h, and the other walks northeast at 2 km/h. We need to find the rate at which the distance between them is changing after 20 minutes. b) We need to solve the inequality $ (\lceil x^2 \rceil)^2 - \lceil x^2 \rceil - 12 < 0$, where $\lceil x \rceil$ denotes the ceiling function.

1. Problem Description

a) Two people start at the same point. One walks east at 3 km/h, and the other walks northeast at 2 km/h. We need to find the rate at which the distance between them is changing after 20 minutes.

b) We need to solve the inequality $(\lceil x^2 \rceil)^2 - \lceil x^2 \rceil - 12 < 0$, where $\lceil x \rceil$ denotes the ceiling function.

2. Solution Steps

a)

Let $x$ be the distance traveled by the person walking east, and $y$ be the distance traveled by the person walking northeast.

The person walking east has a speed of 3 km/h. After 20 minutes (which is 1/3 of an hour), the distance traveled is $x = 3 \times \frac{1}{3} = 1$ km.

The person walking northeast has a speed of 2 km/h. After 20 minutes, the distance traveled is $y = 2 \times \frac{1}{3} = \frac{2}{3}$ km.

Let $z$ be the distance between the two people. We can use the law of cosines to relate $x$, $y$, and $z$. The angle between east and northeast is $45^{\circ}$, so

$z^2 = x^2 + y^2 - 2xy \cos(45^{\circ})$

$z^2 = x^2 + y^2 - 2xy (\frac{\sqrt{2}}{2}) = x^2 + y^2 - xy\sqrt{2}$

Differentiating with respect to time $t$, we get:

$2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} - \sqrt{2} (\frac{dx}{dt}y + x\frac{dy}{dt})$

At $t = \frac{1}{3}$ hour, $x = 1$, $y = \frac{2}{3}$, $\frac{dx}{dt} = 3$, $\frac{dy}{dt} = 2$.

$z^2 = (1)^2 + (\frac{2}{3})^2 - (1)(\frac{2}{3})\sqrt{2} = 1 + \frac{4}{9} - \frac{2\sqrt{2}}{3} = \frac{13}{9} - \frac{2\sqrt{2}}{3} = \frac{13 - 6\sqrt{2}}{9}$

$z = \sqrt{\frac{13 - 6\sqrt{2}}{9}} = \frac{\sqrt{13 - 6\sqrt{2}}}{3} \approx \frac{\sqrt{13 - 6(1.414)}}{3} \approx \frac{\sqrt{13 - 8.484}}{3} \approx \frac{\sqrt{4.516}}{3} \approx \frac{2.125}{3} \approx 0.708$

$2z \frac{dz}{dt} = 2(1)(3) + 2(\frac{2}{3})(2) - \sqrt{2}(3(\frac{2}{3}) + 1(2)) = 6 + \frac{8}{3} - \sqrt{2}(2+2) = 6 + \frac{8}{3} - 4\sqrt{2}$

$2z \frac{dz}{dt} = \frac{18 + 8}{3} - 4\sqrt{2} = \frac{26}{3} - 4\sqrt{2} = \frac{26 - 12\sqrt{2}}{3}$

$\frac{dz}{dt} = \frac{26 - 12\sqrt{2}}{6z} = \frac{26 - 12\sqrt{2}}{6 \frac{\sqrt{13 - 6\sqrt{2}}}{3}} = \frac{26 - 12\sqrt{2}}{2\sqrt{13 - 6\sqrt{2}}} = \frac{13 - 6\sqrt{2}}{\sqrt{13 - 6\sqrt{2}}} = \sqrt{13 - 6\sqrt{2}} \approx \sqrt{4.516} \approx 2.125$ km/h

b)

Let $y = \lceil x^2 \rceil$. The inequality becomes $y^2 - y - 12 < 0$.

$(y-4)(y+3) < 0$.

Since $y = \lceil x^2 \rceil$, $y$ must be an integer.

The inequality holds when $-3 < y < 4$. Since $y$ is the ceiling function, $y$ must be an integer greater than or equal to

0. So we need $0 \le y < 4$, which means $y = 0, 1, 2, 3$.

$\lceil x^2 \rceil = 0$ has no solutions since $\lceil x^2 \rceil \ge 0$ and $\lceil x^2 \rceil = 0$ only when $x=0$. But if $x=0$, then $\lceil x^2 \rceil = 0$ and the inequality becomes $-12 < 0$, which holds. Thus $x = 0$ is a solution.

$\lceil x^2 \rceil = 1 \implies 0 < x^2 \le 1 \implies -1 \le x \le 1$. So $-1 \le x \le 1$.

$\lceil x^2 \rceil = 2 \implies 1 < x^2 \le 2 \implies 1 < x \le \sqrt{2}$ or $-\sqrt{2} \le x < -1$.

$\lceil x^2 \rceil = 3 \implies 2 < x^2 \le 3 \implies \sqrt{2} < x \le \sqrt{3}$ or $-\sqrt{3} \le x < -\sqrt{2}$.

Combining the intervals we have $[-\sqrt{3}, -\sqrt{2}) \cup (-\sqrt{2}, -1) \cup [-1, 1] \cup (1, \sqrt{2}] \cup (\sqrt{2}, \sqrt{3}] = [-\sqrt{3}, \sqrt{3}]$.

But we need to exclude $x$ such that $x = \pm\sqrt{2}$ since $\lceil (\pm\sqrt{2})^2 \rceil = 2$, which means $\lceil x^2 \rceil = 2$ and $y=2$. In this case $(2-4)(2+3) = -2(5) = -10 < 0$, so $x = \pm \sqrt{2}$ is a valid solution.

We require $\lceil x^2 \rceil < 4$, so $x^2 \le 3$. This gives $-\sqrt{3} \le x \le \sqrt{3}$.

3. Final Answer

a) The distance between the people is changing at a rate of $\sqrt{13-6\sqrt{2}} \approx 2.125$ km/h after 20 minutes.

b) The solution to the inequality is $-\sqrt{3} \le x \le \sqrt{3}$.