The problem has three parts. 2.1: A balloon is rising at 2 m/s and a boy is cycling at 5 m/s. When the boy passes under the balloon, it is 15 m above him. Find how fast the distance between them is increasing 3 seconds later. 2.2: Using the Intermediate Value Theorem (IVT), show that the equation $\sqrt[3]{x} = 1-x$ has at least one solution. 2.3: Verify that $f(x) = 2x^3 - 3x + 1$ satisfies the Mean Value Theorem (MVT) on the interval $[0, 2]$ and find all numbers $c$ that satisfy the conclusion of the MVT.

1. Problem Description

The problem has three parts.

2.1: A balloon is rising at 2 m/s and a boy is cycling at 5 m/s. When the boy passes under the balloon, it is 15 m above him. Find how fast the distance between them is increasing 3 seconds later.

2.2: Using the Intermediate Value Theorem (IVT), show that the equation $\sqrt[3]{x} = 1-x$ has at least one solution.

2.3: Verify that $f(x) = 2x^3 - 3x + 1$ satisfies the Mean Value Theorem (MVT) on the interval $[0, 2]$ and find all numbers $c$ that satisfy the conclusion of the MVT.

2. Solution Steps

2.1

Let the height of the balloon at time $t$ be $h(t)$, and the horizontal distance of the boy from the initial point be $d(t)$. We are given that the balloon's initial height is 15 m, and it rises at 2 m/s. The boy cycles at 5 m/s. Thus,

$h(t) = 15 + 2t$

$d(t) = 5t$

Let $s(t)$ be the distance between the boy and the balloon at time $t$.

$s(t) = \sqrt{h(t)^2 + d(t)^2} = \sqrt{(15+2t)^2 + (5t)^2} = \sqrt{225 + 60t + 4t^2 + 25t^2} = \sqrt{29t^2 + 60t + 225}$

We want to find the rate of change of $s(t)$ at $t=3$. So we want to find $s'(3)$.

$s'(t) = \frac{1}{2\sqrt{29t^2 + 60t + 225}} (58t + 60) = \frac{29t + 30}{\sqrt{29t^2 + 60t + 225}}$

$s'(3) = \frac{29(3) + 30}{\sqrt{29(3^2) + 60(3) + 225}} = \frac{87 + 30}{\sqrt{261 + 180 + 225}} = \frac{117}{\sqrt{666}} = \frac{117}{\sqrt{666}} \approx \frac{117}{25.81} \approx 4.53$

2.2

Let $f(x) = \sqrt[3]{x} - (1-x) = \sqrt[3]{x} + x - 1$. We want to show that $f(x) = 0$ has at least one solution.

$f(0) = \sqrt[3]{0} + 0 - 1 = -1$

$f(1) = \sqrt[3]{1} + 1 - 1 = 1$

Since $f(0) = -1 < 0$ and $f(1) = 1 > 0$, and $f(x)$ is continuous for all $x$, by the Intermediate Value Theorem, there exists a $c \in (0, 1)$ such that $f(c) = 0$. Therefore, $\sqrt[3]{x} = 1-x$ has at least one solution.

2.3

$f(x) = 2x^3 - 3x + 1$.

First, $f(x)$ is a polynomial and is continuous and differentiable on $[0, 2]$. Thus, $f(x)$ satisfies the conditions of the Mean Value Theorem on the interval $[0, 2]$.

$f(0) = 2(0)^3 - 3(0) + 1 = 1$

$f(2) = 2(2)^3 - 3(2) + 1 = 2(8) - 6 + 1 = 16 - 6 + 1 = 11$

$f'(x) = 6x^2 - 3$

By the Mean Value Theorem, there exists a $c \in (0, 2)$ such that

$f'(c) = \frac{f(2) - f(0)}{2 - 0}$

$6c^2 - 3 = \frac{11 - 1}{2} = \frac{10}{2} = 5$

$6c^2 = 8$

$c^2 = \frac{8}{6} = \frac{4}{3}$

$c = \pm \sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}} = \pm \frac{2\sqrt{3}}{3}$

Since $c \in (0, 2)$, we take the positive value.

$c = \frac{2\sqrt{3}}{3} \approx \frac{2(1.732)}{3} \approx \frac{3.464}{3} \approx 1.155$

Since $0 < \frac{2\sqrt{3}}{3} < 2$, $c = \frac{2\sqrt{3}}{3}$ is the value we are looking for.

3. Final Answer

2.1: The distance is increasing at $\frac{117}{\sqrt{666}} \approx 4.53$ m/s.

2.2: By the Intermediate Value Theorem, the equation has at least one solution.

2.3: $c = \frac{2\sqrt{3}}{3}$