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## Instructions:

Measure theory is the branch of mathematics that provides a rigorous framework for the study of integration, probability, and other related areas of analysis. The basic idea of measure theory is to extend the concept of length, area, and volume to more abstract spaces.

In measure theory, a measure is a function that assigns a non-negative number to a set, which represents the size or magnitude of that set. The most important measure is the Lebesgue measure, which is used to define the Lebesgue integral.

Lp Spaces:

Lp spaces are a family of function spaces that are used in functional analysis and in probability theory. The Lp space consists of all functions for which the pth power of the absolute value is Lebesgue integrable.

A cycloid is the plane curve traced out by a point on the circumference of a circle as it rolls without slipping along a straight line. Show that, if the straight line is the $x$-axis and the circle has radius $a>0$, the cycloid can be parametrized as

$$

\gamma(t)=a(t-\sin t, 1-\cos t)

$$

To show that $\gamma(t)=a(t-\sin t, 1-\cos t)$ parametrizes the cycloid, we need to show that the curve traced out by this parametrization is the same as the curve traced out by a point on the circumference of a circle of radius $a$ as it rolls without slipping along the $x$-axis.

Let $P$ be a point on the circumference of the circle of radius $a$ centered at the origin, and let $Q$ be the point on the $x$-axis directly below $P$, as shown in the figure below.

Let $O$ be the center of the circle, and let $t$ be the angle $\angle AOP$, where $A$ is the point on the $x$-axis directly to the left of $P$. Then the angle $\angle AOQ$ is also $t$, since $OA$ and $OQ$ are radii of the circle.

As the circle rolls without slipping along the $x$-axis, the point $P$ moves along a distance equal to the arc length $s$ from $A$ to $P$. This arc length is equal to the length of the circumference of the circle times the angle $t$ in radians, or $s=at$. Meanwhile, $Q$ moves a distance equal to the horizontal component of $s$, which is $s-\sin t$. Thus, the position of $Q$ at time $t$ is $(s-\sin t, 0)=(at-a\sin t, 0)$. Finally, $P$ and $Q$ are vertically separated by the vertical component of $s$, which is $a(1-\cos t)$. Thus, the position of $P$ at time $t$ is $(at-a\sin t, a(1-\cos t))$.

This shows that the curve traced out by the point $P$ as the circle rolls without slipping along the $x$-axis is given by the parametrization $\gamma(t)=a(t-\sin t, 1-\cos t)$, as desired.

Find the Cartesian equations of the following parametrized curves:

(i) $\gamma(t)=(1+\cos t, \sin t(1+\cos t))$.

(ii) $\gamma(t)=\left(t^2+t^3, t^3+t^4\right)$

(i) To find the Cartesian equations of $\gamma(t)=(1+\cos t, \sin t(1+\cos t))$, we can eliminate the parameter $t$ by solving for $\cos t$ and $\sin t$ in terms of $x$ and $y$.

From the first coordinate, we have $x=1+\cos t$, so $\cos t=x-1$.

From the second coordinate, we have $y=\sin t(1+\cos t)$, so $\sin t=\frac{y}{1+\cos t}=\frac{y}{x}$. Substituting for $\cos t$ and $\sin t$ in terms of $x$ and $y$, we get

$\begin{aligned} & x=1+\cos t=1+(x-1)=x, \ & y=\sin t(1+\cos t)=\frac{y}{x} \cdot x(x-1)=y(x-1) .\end{aligned}$

Thus, the Cartesian equations of the curve are $x=1+\cos t$ and $y=\sin t(1+\cos t)$, or equivalently,

$x=1+\frac{x^2-y^2}{2 x}, \quad y=\frac{y}{x} \cdot\left(1+\frac{x^2-y^2}{2 x}\right)$.

Simplifying, we get

$x^2-2 x+y^2=1$,

which is the equation of a circle centered at $(1,0)$ with radius $1$.

(ii) To find the Cartesian equations of $\gamma(t)=(t^2+t^3, t^3+t^4)$, we can eliminate the parameter $t$ by solving for $t^2$ and $t^3$ in terms of $x$ and $y$.

From the first coordinate, we have $x=t^2+t^3=t^2(1+t)$, so $t^2=\frac{x}{1+t}$.

From the second coordinate, we have $y=t^3+t^4=t^3(1+t)=\left(\frac{x}{1+t}\right)^{\frac{3}{2}}(1+t)$. Substituting for $t^2$ and $t^3$ in terms of $x$ and $y$, we get

$\begin{aligned} t^2 & =\frac{x}{1+t} \ t^3 & =\frac{y}{(1+t)^{\frac{3}{2}}}\end{aligned}$

Solving for $t$ in terms of $x$ and $y$, we find

$t=\frac{\sqrt{x^2+4 y}-x}{2}$

Substituting this expression for $t$ into the equation for $x$, we get

$\frac{x}{1+\frac{\sqrt{x^2+4 y}-x}{2}}=\frac{x^2+2 \sqrt{x^2+4 y}-2 x}{\sqrt{x^2+4 y}}=\frac{\left(x^2+2 \sqrt{x^2+4 y}+4 y\right)-4 y}{\sqrt{x^2+4 y}}=\sqrt{x^2+4 y}$.

Calculate the length of the part of the curve

$$

\gamma(t)=(\sinh t-t, 3-\cosh t)

$$

cut off by the $x$-axis.

To find the length of the part of the curve $\gamma(t)$ cut off by the $x$-axis, we need to find the $t$-values where $\gamma(t)$ intersects the $x$-axis, and then integrate the magnitude of the derivative of $\gamma(t)$ over the interval of $t$ values corresponding to that part of the curve.

The $x$-coordinate of a point on the $x$-axis is zero, so we need to solve the equation $\sinh t – t = 0$ to find the $t$-values where $\gamma(t)$ intersects the $x$-axis. This equation cannot be solved analytically, so we will use numerical methods to find an approximation.

One method is to use the bisection method: we start with an interval that contains a root of the equation (for example, $t \in [1,2]$), and we bisect the interval repeatedly, discarding the half that does not contain a root. After several iterations, we get an interval that contains the root with the desired accuracy.

Using this method, we find that the equation $\sinh t – t = 0$ has a root in the interval $[1.5,1.6]$, which we will denote by $t_0$. To find the length of the part of the curve cut off by the $x$-axis, we need to integrate the magnitude of the derivative of $\gamma(t)$ over the interval $[0,t_0]$.

The derivative of $\gamma(t)$ is

$\gamma^{\prime}(t)=(\cosh t-1, \sinh t)$

and its magnitude is

$\left|\gamma^{\prime}(t)\right|=\sqrt{(\cosh t-1)^2+(\sinh t)^2}=\sqrt{\cosh ^2 t-2 \cosh t+2}$

To integrate this expression over the interval $[0,t_0]$, we make the substitution $u = \sinh t$, so that

$\int_0^{t_0}\left|\gamma^{\prime}(t)\right| d t=\int_0^{\sinh t_0} \sqrt{\cosh ^2\left(\sinh ^{-1} u\right)-2 \cosh \left(\sinh ^{-1} u\right)+2} d u$.

Using the identity $\cosh^2 x – \sinh^2 x = 1$, we can simplify the expression under the square root to $2 – 2 \sinh(\sinh^{-1} u) = 2(1 – u^2)$, so that

$\int_0^{t_0}\left|\gamma^{\prime}(t)\right| d t=\int_0^{\sinh t_0} 2 \sqrt{1-u^2} d u=\pi$

Therefore, the length of the part of the curve $\gamma(t)$ cut off by the $x$-axis is $\boxed{\pi}$.