exercise04.tex

\documentclass[11pt]{article}
% % \def\hidesolutions{}
\input{header}
\begin{document}
\exsheet{4}{3}{October} % parameters are the number of the session and the day




\begin{exercise}
    Compute the line integral of the vector field $\vec{F}$ along the curve $\gamma$, where 
    \begin{align*}
        \vec F : \bbR^2 \to \bbR^2, \quad (x_1,x_2) \mapsto ( x_2, 0 ), \qquad \gamma : [0,2\pi]  \to \bbR^2, \quad t \mapsto ( \cos(t), \sin(t) )
    \end{align*}
    Compute the line integral of the vector field $\vec{G}$ along the curve $\delta$, where 
    \begin{align*}
        \vec G : \bbR^3 \to \bbR^3, \quad (x_1,x_2,x_3) \mapsto ( 2, 3, -1 ), \qquad \delta : [0,4]  \to \bbR^2, \quad t \mapsto  ( t^2, \cos(t), e^t )
    \end{align*}
\end{exercise}
\begin{solution}
    \begin{align*}
        \int_{\Gamma} \vec{F}(x_1,x_2) d\ell 
        &= 
        \int_{0}^{2\pi} (\vec{F}\circ \gamma)(t)\cdot \dot{\gamma}(t)dt
        \\&= 
        \int_{0}^{2\pi} \begin{pmatrix}\sin t\\ 0\end{pmatrix}\cdot \begin{pmatrix} -\sin(t)\\ \cos(t)\end{pmatrix}dt
        \\&= 
        \int_{0}^{2\pi}  -\sin^2(t)dt
        \\&= 
        \int_{0}^{2\pi}  -\frac{1}{2}+\frac{1}{2}\cos(2t)dt
        = 
        \left[-\frac{1}{2}t + \frac{1}{4}\sin(2t)\right]_{0}^{2\pi} = -\pi
    \end{align*}

    \begin{align*}
        \int_{\Gamma} \vec{G}(x_1,x_2,x_3) d\ell &= \int_{0}^{4} (\vec{G}\circ \delta)(t)\cdot \dot{\delta}(t)dt
        \\&= \int_{0}^{4}\begin{pmatrix} 2\\ 3\\ -1\end{pmatrix} \cdot  \begin{pmatrix}2t\\ -\sin(t) \\ e^t \end{pmatrix}dt
        \\&= \int_{0}^{4} 4t - 3\sin(t) - e^t dt
        \\&= \left[2t^2 + 3\cos(t) - e^{t} \right]_{0}^{4}
        \\&= 32 + 3\cos4-e^4 - \left(3 - 1\right) = 30+3\cos4 -e^ 4
    \end{align*}
        % TODO: we need of those cosine half angle formulas to turn the square of sin into a cos 2t      
        % TODO: the second one is easy
\end{solution}


\begin{exercise}
    Compute the line integral of the vector field $\vec{F}$ along the curve $\gamma$, where 
    \begin{align*}
        \vec F : \bbR^2 \to \bbR^2, \quad (x_1,x_2) \mapsto ( e^{ x_1x_2 } x_2, e^{ x_1 x_2 } x_1 )
    \end{align*}
    and 
    \begin{align*}
        \gamma : [0,1] \to \bbR^2, \quad t \mapsto \left( \arctan( \cos(\pi t)^2 - \sin(\pi t)^2 ), 1 + \sqrt[2]{1+\arctan(t)} \right)
    \end{align*}
\end{exercise}
\begin{solution}
    The vector field is the gradient of $e^{ x_1 x_2 }$, and so we can use the formula for vector fields that admit a potential.
    We find that
    \begin{align*}
        \int_\Gamma \vec F d\ell = \int_\Gamma \grad f d\ell = f(\gamma(1)) - f(\gamma(0))
    \end{align*}
    Now we only need to compute the function values of $f$. Explicitly,
    \begin{align*}
        \gamma(0) 
		&= 
		\left( \arctan( \cos(\pi 0)^2 - \sin(\pi 0)^2 ), 1 + \sqrt[2]{1+\arctan(0)} \right)
        \\&
		=
        \left( \arctan( 1 ), 1 + \sqrt[2]{1} \right) 
        =
        \left( \pi/4, 2 \right)
    \end{align*}
    \begin{align*}
        \gamma(1) 
		&= 
		\left( \arctan( \cos(\pi 1)^2 - \sin(\pi 1)^2 ), 1 + \sqrt[2]{1+\arctan(1)} \right)
        \\&
		=
        \left( \arctan( 1 ), 1 + \sqrt[2]{1+\pi/4} \right)
        =
        \left( \pi/4, 1 + \sqrt[2]{1+\pi/4} \right)
    \end{align*}
    We then compute
    \begin{align*}
        f(\gamma(1)) - f(\gamma(0)) 
        =
        \exp\left( \frac \pi 4 ( 1 + \sqrt[2]{1+\pi/4} ) \right) - \exp\left( \pi/2 \right).
    \end{align*}
\end{solution}

\begin{exercise}
    What is the length of the graph of the function $g(x) = x^{\frac 3 2}$ over the interval $[0,2]$?
    Simplify as much as reasonable.
\end{exercise}
\begin{solution}
    We can express the graph as a curve 
    \begin{align*}
        \gamma : [0,2] \to \mathbb R^2, \quad t \mapsto ( t, t^{\frac 3 2} ) 
    \end{align*}
    As mentioned in the lecture, the integral of $1$ along that curve gives the length.
	Hence we find:
    \begin{align*}
        \int_\Gamma 1 \;dl = \int_0^2 \left| \left( 1, \frac 3 2 t^{\frac 1 2} \right) \right| \;dt = \int_0^2 \sqrt{ 1 + \frac{9t}{4} } \;dt
    \end{align*}
    We integrate the last expression. The integrand is a derivative, up to some numerical factor:
    \begin{align*}
        \int_0^2 \sqrt{ 1 + \frac 9 4 t } \;dt 
        &= 
        \frac 2 3 \cdot \frac 4 9 \int_0^2 \frac 3 2 \cdot \frac 9 4 \left( 1 + \frac 9 4 t \right)^{\frac 1 2} \;dt 
        \\&= 
        \frac 2 3 \cdot \frac 4 9 \int_0^2 \partial_t \left( 1 + \frac 9 4 t \right)^{\frac 3 2} \;dt 
        \\&= 
        \frac 2 3 \cdot \frac 4 9 \left[ \left( 1 + \frac 9 4 t \right)^{\frac 3 2} \right]_0^2
        \\&= 
        \frac 2 3 \cdot \frac 4 9 \left( \left( 1 + \frac 9 2 \right)^{\frac 3 2} - 1 \right)
        \\&= 
        \frac{8}{27} \left( \left( \frac {11} 2 \right)^{\frac 3 2} - 1 \right)
        .
    \end{align*}
    (You can compute the integral via \textit{Intégration par changement de variable}, 
	which is the same calculation but in a different formalism.)
    % \begin{align*}
    %     \int_0^2 \sqrt{1 + \left[\frac{\;d}{\;dx} (x^2\sin x)\right]^2}dx &= \int_0^2 \sqrt{1 + (2x\sin x + x^2\cos x)^2}\;dx\\
    % &=\int_0^2 \sqrt{1 + 4x^2\sin^2 x + x^4\cos^2 x + 4x^3\cos x\sin x}\;dx
    % \end{align*}
\end{solution}



\begin{exercise}
    Take a look at the functions 
    \begin{align*}
        f(x_1,x_2) = 1, \quad g(x_1,x_2) = x_2.
    \end{align*}
    We have the following curves:
    \begin{align*}
        &\alpha : [0,1] \to \bbR^2, \quad t \mapsto (t,t),
        \\
        &\beta  : [0,2] \to \bbR^2, \quad t \mapsto \begin{cases} (t,0) & 0 \leq t < 1, \\ (0,t-1) & 1 \leq t \leq 2, \end{cases}
        \\
        &\gamma : [0,1] \to \bbR^2, \quad t \mapsto (t^2,t).
%         \\
%         &\delta : [0,1] \to \bbR^2, \quad t \mapsto (1-t^2,1-t).
    \end{align*}
    Compute:
    \begin{align*}
        \int_A f \;d\ell, \quad \int_B f \;d\ell, \quad 
        \int_A g \;d\ell, \quad \int_B g \;d\ell, \quad 
        \int_\Gamma g \;d\ell
%         \quad \int_\Delta g \;d\ell
        .
    \end{align*}
\end{exercise}
\begin{solution}  
     We compute these integrals as follows:
     \begin{align*}
        \int_A f \;d\ell
        =
        \int_0^1 \sqrt{2} \;dt = \left[\sqrt{2}t\right]_0^1 = \sqrt{2}
        ,
        \quad 
        \int_B f \;d\ell
        = 
        \int_0^2 \;dt = \left[t \right]_0^2  = 2
        %
    \end{align*}  
    \begin{align*}
        \int_A g \;d\ell
        &= \int_0^1 t \sqrt{2} \;dt = \left[\frac{\sqrt{2}}{2}t^2 \right]_0^1 = \frac{\sqrt{2}}{2}
        %
    \end{align*}
    \begin{align*}
        \int_B g \;d\ell
        &=
        \int_0^1 (0) \;dt + \int_1^2 (t-1) \;dt  
        = 
        \int_1^2 (t-1) \;dt 
        = 
        \left[ \frac{1}{2} (t-1)^2 \right]_1^2 = \frac{1}{2}
        %
    \end{align*}
    For the interal of $g$ over the third curve, we proceed as follows:
    \begin{align*}
        \int_\Gamma g \;d\ell
        &=
        \int_0^1 t \sqrt{ 4t^2 + 1 } \;dt
        \\&=
        \int_0^1 t \left( 4t^2 + 1 \right)^{\frac 1 2} \;dt
        \\&=
        \frac 2 3 \cdot \frac 1 8 \cdot 
        \int_0^1 \frac 3 2 \cdot 8t \left( 4t^2 + 1 \right)^{\frac 1 2} \;dt
        \\&=
        \frac 2 3 \cdot \frac 1 8 \cdot 
        \int_0^1 \partial_t \left( 4t^2 + 1 \right)^{\frac 3 2} \;dt
        =
        \frac 2 3 \cdot \frac 1 8 \cdot 
        \left( 5^{\frac 3 2} - 1 \right)
        = 
        \frac{ 5^{\frac 3 2} - 1 }{12}
        .
    \end{align*}
%     For the integral of $g$ over the last curve, we take a sharp look and integrate by parts:
%     \begin{align*}
%         \int_\Delta g \;d\ell
%         &=
%         \int_0^1 t^2 \left( 1 + 4t^2 \right)^{\frac 1 2} \;dt
%         =
%         \int_0^1 t^2 \left( 2t + 1 \right)^{\frac 1 2} \;dt
%         \\&=
%         \int_0^1 t^2 \cdot \partial_t\left( \frac 4 3 \left( 1 + 2t \right)^{\frac 3 2} \right) \;dt
%         \\&=
%         \frac 4 3 \left[ t^2 \left( 2t + 1 \right)^{\frac 3 2} \right]_0^1
%         -
%         \frac 4 3 \int_0^1 2t \left( 2t + 1 \right)^{\frac 3 2} \;dt
%         =
%         3^{\frac 3 2}
%         -
%         \frac 4 3 \int_0^1 2t \left( 2t + 1 \right)^{\frac 3 2} \;dt
%         .
%     \end{align*}
%     We integrate by parts once more:
%     \begin{align*}
%         \int_0^1 2t \cdot \partial_t\left( \frac 4 5 \left( 2t + 1 \right)^{\frac 5 2} \right) \;dt
%         &=
%         \left[ 2t \cdot \frac 4 5 \left( 2t + 1 \right)^{\frac 5 2} \right]_0^1
%         -
%         \int_0^1 2 \cdot \frac 4 5 \left( 2t + 1 \right)^{\frac 5 2} \;dt
%         \\&=
%         \frac 8 5 \cdot 3^{\frac 5 2}
%         -
%         \int_0^1 2 \cdot \frac 4 5 \left( 2t + 1 \right)^{\frac 5 2} \;dt
%         \\&=
%         \frac 8 5 \cdot 3^{\frac 5 2}
%         -
%         \frac 8 5 
%         \int_0^1 \left( 2t + 1 \right)^{\frac 5 2} \;dt
%         ,
%     \end{align*}
%     and then we calculate, using the substitution $u = 2t + 1$, which means $du = 2 dt$
%     \begin{align*}
%         \int_0^1 \left( 2t + 1 \right)^{\frac 5 2} \;dt
%         &=
%         \frac 1 2 
%         \int_1^3 \left( u \right)^{\frac 5 2} \;du
%         =
%         \frac 1 2 
%         \left[ \frac 2 7 u^{\frac 7 2}\right]_1^3
%         =
%         \frac 1 7 
%         \left( 3^{\frac 7 2} - 1 \right)
%         .
%     \end{align*}
%     In combination,
%     \begin{align}
%         \int_\Delta g \;d\ell 
%         &= 
%         3^{\frac 3 2} - \frac 4 3 \cdot \frac 8 5 \cdot 3^{\frac 5 2} + \frac 4 3 \cdot \frac 8 5 \cdot \frac 1 7 \left( 3^{\frac 7 2} - 1 \right)
%         \\&=
%         3^{\frac 3 2} - \frac{32}{15} 3^{\frac 5 2} + \frac{32}{105} \left( 3^{\frac 7 2} - 1 \right)
%         .
%     \end{align}
    Note that the curves all have the same starting and end points, $(0,0)$ and $(1,1)$, 
    but the curve integrals of the function may (generally) differ from curve to curve.  
\end{solution}





\begin{exercise}
    We want to verify that the curve integrals do not depend on the direction of the parameterization.
    Consider the curve parameterizations
    \begin{align}
        \gamma_+ : [0,1] \rightarrow \bbR^2, \quad t \mapsto ( t, 1 + \frac 1 2 t^2 ),
        \\
        \gamma_- : [0,1] \rightarrow \bbR^2, \quad s \mapsto ( 1-s, 1 + \frac 1 2 (1-s)^2 ),
    \end{align}
    \begin{itemize}
     \item Given the scalar field 
     \begin{align*}
        f(x,y) 
        = x ( y - \frac 1 2 x^2 )
        = x y - \frac 1 2 x^3
        ,
     \end{align*}
     show that 
     \begin{align}
        \int_{\gamma_+} f \;dl = \int_{\gamma_-} f \;dl
     \end{align}
     \item Compute the tangent and unit tangent vectors of each curve,
     and show that $\dot\gamma_+(t) = - \dot\gamma_-(1-t)$.
     Show that the curve integrals along $\gamma_+$ and $\gamma_-$ of the vector field
     \begin{align}
        F(x,y) = (-y,x)
     \end{align}
     are the negative of each other.
    \end{itemize}
\end{exercise}
\begin{solution}
    \begin{itemize}
		\item We compute the first integral directly:
		\begin{align*}
			\int_{\gamma_+} f \;dl
			&=
			\int_{0}^1 f( t, 1 + \frac 1 2 t^2 ) |\dot{\gamma}_+(t)|dt
			\\&
			=
			\int_{0}^1 t \sqrt{ 1 + t^2 } dt
			\\&
			=
			\frac 1 2 \int_{0}^1 2t \sqrt{ 1 + t^2 } dt
			.
		\end{align*}
		We substitute $u = 1 + t^2$, so that $du = 2 dt$. Thus
		\begin{align*}
			\frac 1 2 \int_{0}^1 2t \sqrt{ 1 + t^2 } dt
			&=
			\frac 1 2 \int_1^2 \sqrt{ u } \;du
			=
			\frac 1 2 \left[ \frac 2 3 u^{\frac 3 2} \right]_1^2 = \frac 1 3 \left( 2 \sqrt{2} - 1 \right)
			.
		\end{align*}
		The second integral can be computed directly as well, or with a change of variables. For the latter, we use
		\begin{align*}
			\int_{\gamma_-} f \;dl
			&=
			\int_{0}^1 f( 1 - s, 1 + \frac 1 2 (1 - s)^2 ) |\dot{\gamma}_-(s)|ds
			\\&
			=-\int_1^0 f(t, 1 + \frac 1 2 t^2) |\dot{\gamma}_-(1-t)|dt
			\\&
			= \int_0^1 t \sqrt{1 + (1 - (1 - t))^2} dt = \int_0^1 t \sqrt{1 + t^2} dt
			\\&
			= \int_{\gamma_+}f dl,
		\end{align*}
		where we used the substitution $t = 1 - s$ with $dt = -ds$.

        \item
		Given the two parameterizations as stated above, 
		\begin{align}
			\gamma_+ &: [0,1] \rightarrow \bbR^2, \quad t \mapsto ( t, 1 + \frac 1 2 t^2 ),
			\\
			\gamma_- &: [0,1] \rightarrow \bbR^2, \quad s \mapsto ( 1-s, 1 + \frac 1 2 (1-s)^2 ),
		\end{align}
		We compute the tangent vectors:
        \begin{align*}
            \dot\gamma_+(t) = \begin{pmatrix} 1 \\ t \end{pmatrix}, 
            \qquad 
            \dot\gamma_-(s) = \begin{pmatrix} -1 \\ s - 1 \end{pmatrix}.
        \end{align*}
		Their norms are 
		\begin{align*}
            |\dot \gamma_+(t)| = \sqrt{1 + t^2},
            \\
            |\dot \gamma_-(s)| = \sqrt{1 + (s-1)^2}.
        \end{align*}
        It is intuitive that $\dot\gamma_+(t) = - \dot\gamma_-(1-t)$, as we check easily:
        \begin{align*}
            \dot\gamma_-(1-t) 
            = 
            -
            \begin{pmatrix} -1 \\ 1-t - 1 \end{pmatrix}
            =
            \begin{pmatrix}  1 \\ t       \end{pmatrix}
            =
            \dot\gamma_+(t)
            .
        \end{align*}
        Finally, for the curve integral we find again,
        using the change of variables $s = 1 - t$, hence $ds = - dt$, 
        the following identity:
        \begin{align*}
            \int_{\gamma_+} F \;dl 
            &= 
            \int_0^1 F(\gamma_+(t)) \cdot \dot\gamma_+(t) \;dt
            \\&
            = 
            -
            \int_0^1 F(\gamma_-(1 - t)) \cdot \dot\gamma_-(1 - t) \;dt
            \\&
            = 
            \int_1^0 F(\gamma_-(s)) \cdot \dot\gamma_-(s) \;ds
            .
        \end{align*}
        We switch the bounds of integration into normal order and find 
        \begin{align*}
            \int_1^0 F(\gamma_-(s)) \cdot \dot\gamma_-(s) \;ds
            = 
            -
            \int_0^1 F(\gamma_-(s)) \cdot \dot\gamma_-(s) \;ds
            = 
            -\int_{\gamma_-} F \;dl
            .
        \end{align*}
    \end{itemize}
\end{solution}












\begin{exercise}
    Consider the open cube 
    \begin{align*}
        \Omega = (0,1)^2 = \left\{ (x,y) \in \mathbb R^2 : 0 < x,y < 1 \right\}. 
    \end{align*}
    \begin{itemize}
        \item Show that $\Omega$ is convex.
        \item Let $z \in \Omega$. Show that $\Omega$ is star-shaped with respect to $z$.
        \item Show that $\Omega$ is simply-connected.
    \end{itemize}
    For the last part, show that any two continuous curves with the same endpoints are homotopic relative to endpoints. 
\end{exercise}
\begin{solution}
    \item
	Graphically, it is clear that each segment between points of $\Omega$ is also contained in $\Omega$. 
	But we prove this formally. 
	
	Suppose that $a = (a_1,a_2), b = (b_1,b_2) \in \Omega$ and $t \in [0,1]$. We want to show that $ta + (1-t)b \in \Omega$.
	We compute that 
	\begin{align*}
		ta + (1-t)b
		=
		\left( t a_1 + (1-t) b_1, t a_2 + (1-t) b_2 \right).
	\end{align*}
	This point is in $\Omega$ if each coordinate is in the interval $(0,1)$. We show that for the first coordinate,
	because the argument for the second coordinate is almost the same. 
	Clearly,
	\begin{align*}
		t a_1 + (1-t) b_1 > t \min(a_1,b_1) + (1-t) \min(a_1,b_1) = \min(a_1,b_1) > 0,
	\end{align*}
	\begin{align*}
		t a_2 + (1-t) b_2 < t \max(a_2,b_2) + (1-t) \max(a_2,b_2) = \max(a_2, b_2) < 1.
	\end{align*}
	We conclude that $\Omega$ is convex.
	\item 
	As already observed, the segment between any two points $z,a \in \Omega$ is already contained in $\Omega$. 
	If we fix any point $z \in \Omega$, then $\Omega$ is star-shaped. \textit{Note that we do not use any property except that $\Omega$ is convex. Indeed, every convex domain is also star-shaped (with respect to any point inside).}
    \item 
    Given two continuous curves $\gamma_0, \gamma_1 : [a,b] \rightarrow \Omega$ with $\gamma_0(a) = \gamma_1(a) =: g_a$ and $\gamma_0(b) = \gamma_1(b) =: g_b$, we define 
    \begin{align*}
        \gamma : [a,b] \times [0,1] \rightarrow \bbR^2
    \end{align*}
    by setting 
    \begin{align*}
        \gamma(t,s) = (1-s) \cdot \gamma_0(t) + s \cdot \gamma_1(t).
    \end{align*}
    For any $t \in [a,b]$ we have $\gamma_0(t), \gamma_1(t) \in \Omega$,
    and since $\Omega$ is convex, we have $(1-s) \cdot \gamma_0(t) + s \cdot \gamma_1(t) \in \Omega$ for all $s \in [0,1]$.
    Hence $\gamma$ defines a function 
    \begin{align*}
        \gamma : [a,b] \times [0,1] \rightarrow \Omega
    \end{align*}
    Furthermore, $\gamma$ is obviously continuous in both variables. 
    
    We verify the remaing conditions. For all $t \in [a,b]$ we have 
    \begin{align}
        \gamma(t,0) = (1-0) \cdot \gamma_0(t) + 0 \cdot \gamma_1(t) = \gamma_0(t)
    \end{align}
    and 
    \begin{align}
        \gamma(t,1) = (1-1) \cdot \gamma_0(t) + 1 \cdot \gamma_1(t) = \gamma_1(t)
        .
    \end{align}
    Additionally, for all $s \in [0,1]$, we have 
    \begin{align}
        \gamma(a,s) = (1-s) \cdot \gamma_0(a) + s \cdot \gamma_1(a) = (1-s) \cdot g_a + s \cdot g_a = g_a
    \end{align}
    and 
    \begin{align}
        \gamma(b,s) = (1-s) \cdot \gamma_0(b) + s \cdot \gamma_1(b) = (1-s) \cdot g_b + s \cdot g_b = g_b
        .
    \end{align}
    This shows that $\gamma$ satisfies the required axioms.
\end{solution}






































\end{document}