exercise01.tex

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\exsheet{1}{12}{September} % parameters are the number of the session and the day


\begin{exercise}
	Compute the norms (that is, lengths), the $3$ scalar products and $6$ vector products of the following vectors. 
    Find the volume of the parallelepiped spanned by these vectors.
    \[
    \vec{a} = \left(\begin{array}{c} 1 \\ 2 \\ -1 \end{array}\right)
    ,
    \vec{b} = \left(\begin{array}{c} 2 \\ -1 \\ 0 \end{array}\right)
    ,
    \vec{c} = \left(\begin{array}{c} 1 \\ 1 \\ 3 \end{array}\right)
    \]
\end{exercise}

\begin{solution}
    \begin{gather*}
        \vec{a} = \sqrt{6}, \quad \vec{b} = \sqrt{5}, \quad \vec{c} = \sqrt{11}
        \\
        \vec{a} \cdot \vec{b} = 0,
        \quad 
        \vec{a} \cdot \vec{c} = 0,
        \quad 
        \vec{b} \cdot \vec{c} = 1,
        \\ 
        \vec{a} \times \vec{b} = - \vec{b} \times \vec{a} = \left(\begin{array}{c} -1 \\ -2 \\ -5 \end{array}\right),
        \quad 
        \vec{a} \times \vec{c} = - \vec{c} \times \vec{a} = \left(\begin{array}{c} 7 \\ -4 \\ -1 \end{array}\right),
        \quad 
        \vec{b} \times \vec{c} = - \vec{c} \times \vec{b} = \left(\begin{array}{c} -3 \\ -6 \\ 3 \end{array}\right),
    \end{gather*}
    The volume of the parallelepiped spanned by the three vectors is $18$.
    
    
\end{solution}


\begin{exercise}
	Consider the function 
    \[
        f(x,y) = \sin\left( 2\pi x^2 + 2\pi y^2 \right)
    \]
    Find the first derivatives in $x$ and $y$.
    Sketch the level sets of $f$ for the values $0$ and $1$.
\end{exercise}

\begin{solution}
    We easily compute 
    \[
        \partial_x f(x,y) = \cos\left( 2\pi x^2 + 2\pi y^2 \right) \cdot 4\pi \cdot x,
        \quad 
        \partial_y f(x,y) = \cos\left( 2\pi x^2 + 2\pi y^2 \right) \cdot 4\pi \cdot y.
    \]
    The level sets are concentric circles centered at zero. One approach to understanding works as follows.
    First, we rewrite 
    \[
        f(x,y) = \sin\left( 2\pi ( x^2 + y^2 ) \right)
    \]
    We remember 
    that 
    \[
        \sin( 2\pi t ) = 0 \text{ for the positive values } t = 0, \frac 1 2, 1, \frac 3 2, \dots
    \] 
    and 
    that 
    \[
        \sin( 2\pi t ) = 1 \text{ for the positive values } t = \frac 1 4, \frac 5 4, \frac 9 4, \dots.
    \]
    Hence, whenever $x^2 + y^2$ hits one these values listed above, then we are on the level of $0$ or $1$.
    Specifically, $f(x,y) = 0$ if $(x,y)$ lies on a circle centered at zero and with radius
    \[
        0, \sqrt{\frac 1 2}, \sqrt{1}, \sqrt{\frac 3 2}, \sqrt{2}, \dots, 
    \]
    Similarly, $f(x,y) = 1$ if $(x,y)$ lies on a circle centered at zero and with radius
    \[
        0, \sqrt{\frac 1 4}, \sqrt{\frac 5 4}, \sqrt{\frac 9 2}, \sqrt{\frac {13} 4}, \dots. 
    \]
\end{solution}





\begin{exercise}
	Sketch the following vector field:
    \[
        f(x,y) = \left(\begin{array}{c} \sin(x) \\ sin(y) \end{array}\right)
    \]
\end{exercise}

\begin{solution}
    Please do this in the exercise session.
\end{solution}



\begin{exercise}
	Compute the following integrals:
    \[
        \int_0^1 \int_0^1 x^2 e^y \; dx dy,
        \quad 
        \int_0^1 x^2 \cos(x) \; dx.
    \]
\end{exercise}

\begin{solution}
    For the first integral, we can either split up the integral into two factors, or we compute one after the other.
    We find 
    \[
        \int_0^1 \int_0^1 x^2 e^y \; dx dy
        =
        \int_0^1 x^2 \; dx 
        \cdot 
        \int_0^1 e^y \; dy
        =
        \frac 1 3 \cdot ( e - 1).
    \]
    For the second integral, we use integration by parts. 
    \begin{align*}
        \int_0^1 x^2 \cos(x) \; dx
        &=
        x^2 \sin(x) |_0^1
        -
        \int_0^1 2x \sin(x) \; dx
        \\&=
        x^2 \sin(x) |_0^1
        -
        \left( 2x (-\cos(x)) |_0^1 - \int_0^1 2 (-\cos(x)) \; dx \right)
        \\&=
        x^2 \sin(x) |_0^1
        -
        2x (-\cos(x)) |_0^1 
        +
        \int_0^1 2 (-\cos(x)) \; dx
        \\&=
        x^2 \sin(x) |_0^1
        -
        2x (-\cos(x)) |_0^1 
        -
        2 \int_0^1 \cos(x) \; dx
        \\&=
        x^2 \sin(x) |_0^1
        -
        2x (-\cos(x)) |_0^1 
        -
        2 \left( \sin(1) - \sin(0) \right)
        \\&=
        \sin(1)
        -
        2 (-\cos(1)) 
        -
        2 \sin(1)
        \\&=
        2 \cos(1)
        - 
        \sin(1)
        .
    \end{align*}
\end{solution}




\end{document}