Week 9: Preparation

Week 9: Preparation#

Reading Material#

  • Review: In Chapter 3, re-read Section 3.1 from the headline “Vector-valued functions of one variable”.

  • Reading: In Chapter 7, read the parts in Section 7.1 that relate to curves (especially Definition 7.1.1 and Lemma 7.1.1), and read Section 7.2 through the headline “The Line Integral”, and Sections 7.3 through 7.4.

  • Python: Demo 9, the sections on curve integrals, and Demo 10.

Note that surface parametrizations and surface integrals (as well as the topics of flux and Gauss’ Divergence Theorem) are not part of the syllabus. Skip these parts when you meet them in the textbook and the Python demos. They are available in case they are needed for certain students in the project period.

Key Concepts#

Long Day will cover:

  • Parametrizations of curves in \(\mathbb{R}^n\)

  • Curve length

  • The line integral

  • The universal Jacobian (function)

  • Vector fields and gradient vector fields

  • The tangental line integral: Integration of vector field along a curve

Short Day will cover:

  • The antiderivative problem in \(\mathbb{R}^n\)

  • The stair-line method

  • Flow curves

Note

At DTU (but not much elsewhere), the line integral of a vector field along a curve is often called the tangential line integral.

Preparatory Exercises#

I: Parametric Representation and Curve Length of a Circle#

Let \(\mathcal{C}\) denote a circle in \(\mathbb{R}^2\) given by the equation

\[\begin{equation*} (x-1)^2 + y^2 = 4. \end{equation*}\]

Question a#

State the center and radius of \(\mathcal{C}\).

Answer

The center is \((1,0)\) and the radius is \(2\).

Question b#

Choose a parametric representation \(\pmb{r}(t)\) for \(\mathcal{C}\) with \(t \in [0, 2\pi]\).

Answer

A natural parametric representation of this shape is:

\[\begin{equation*} \pmb{r}(t) = \bigl(1+2\cos(t),\,2\sin(t)\bigr), \quad t\in [0,2\pi]. \end{equation*}\]

Question c#

We know that its curve length is \(2\pi r\), so \(4 \pi\), as this is the well-known circumference formula for a circle. We here want to rediscover this value using the general formula for curve length. First, determine the Jacobian function, meaning the norm of \(\pmb{r}'(t)\), and calculate the curve length of \(\mathcal{C}\) using a fitting formula from the textbook.

Answer

Calculate the derivative:

\[\begin{equation*} \pmb{r}'(t)=(-2\sin(t),\,2\cos(t)). \end{equation*}\]

The norm is

\[\begin{equation*} \Vert \pmb{r}'(t)\Vert = \sqrt{4\sin^2(t)+4\cos^2(t)} = \sqrt{4}=2. \end{equation*}\]

The curve length \(L\) is then

\[\begin{equation*} L=\int_0^{2\pi} 2\,\mathrm dt = 4\pi. \end{equation*}\]

II: Line Integral of Scalar Function#

Let \(f(x,y)=x^2+y^2\), and let \(\mathcal{C}\) be the same circle as in exercise I: Parametric Representation and Curve Length of a Circle with the parametric representation

\[\begin{equation*} \pmb{r}(t)=\bigl(1+2\cos(t),\,2\sin(t)\bigr),\quad t\in[0,2\pi]. \end{equation*}\]

Question a#

Find the expression for \(f(\pmb{r}(t))\).

Answer

Substitute in the parametric representation:

\[\begin{equation*} f(\pmb{r}(t)) = (1+2\cos(t))^2+(2\sin(t))^2. \end{equation*}\]

Expand and simplify, first the first term:

\[\begin{equation*} (1+2\cos(t))^2 = 1 + 4\cos(t)+4\cos^2(t), \end{equation*}\]

and then the second term:

\[\begin{equation*} (2\sin(t))^2= 4\sin^2(t). \end{equation*}\]

Hence we have:

\[\begin{equation*} f(\pmb{r}(t)) = 1+4\cos(t)+4\cos^2(t)+4\sin^2(t), \end{equation*}\]

which simplifies, since \(4\cos^2(t)+4\sin^2(t)=4\), to:

\[\begin{equation*} f(\pmb{r}(t)) = 5+4\cos(t). \end{equation*}\]

Question b#

Calculate the line integral

\[\begin{equation*} \int_{\mathcal{C}} f(x,y)\,\mathrm ds. \end{equation*}\]

Answer

The line integral becomes:

\[\begin{equation*} \int_{\mathcal{C}} f(x,y)\,\mathrm ds = \int_0^{2\pi} \Bigl(5+4\cos(t)\Bigr)\Vert \pmb{r}'(t)\Vert\,\mathrm dt. \end{equation*}\]

Since \(\Vert \pmb{r}'(t)\Vert = 2\), we get:

\[\begin{equation*} \int_0^{2\pi} \Bigl(5+4\cos(t)\Bigr) 2\,\mathrm dt = 2\int_0^{2\pi}\Bigl(5+4\cos(t)\Bigr) \mathrm dt. \end{equation*}\]

We carry out the integration:

\[\begin{equation*} 2\left[5t + 4\sin(t)\right]_0^{2\pi} = 2\left(5\cdot2\pi + 4\cdot(\sin(2\pi)-\sin(0))\right)=2\cdot10\pi = 20\pi. \end{equation*}\]

III: Determination of a Gradient Field#

Consider the vector field \(\pmb{V}: \mathbb{R}^2 \to \mathbb{R}^2\), \(\pmb{V}(x,y)=(2xy,\,x^2)\).

Question a#

Can you guess a function \(f: \mathbb{R}^2 \to \mathbb{R}\) such that \(\nabla f = \pmb{V}\)?

Question b#

Maybe you cannot easily guess whether an \(f\) exists such that \(\nabla f = \pmb{V}\). So, first, find a way to investigate whether \(\pmb{V}\) is a gradient field. Then, if it is, find an antiderivative \(f(x,y)\), such that \(\nabla f = \pmb{V}\).

Answer

For \(\pmb{V}\) to be a gradient field, it must hold that

\[\begin{equation*} \frac{\partial V_1}{\partial y} = \frac{\partial V_2}{\partial x}. \end{equation*}\]

Here we have

\[\begin{equation*} \frac{\partial V_1}{\partial y} = \frac{\partial (2xy)}{\partial y} = 2x, \end{equation*}\]

and

\[\begin{equation*} \frac{\partial V_2}{\partial x} = \frac{\partial (x^2)}{\partial x} = 2x. \end{equation*}\]

Since these are equal, then \(\pmb{V}\) is a gradient field.

To find an antiderivative \(f(x,y)\), we integrate the first component with respect to \(x\):

\[\begin{equation*} f(x,y)=\int 2xy\,dx = x^2y + g(y), \end{equation*}\]

where \(g(y)\) is a function of \(y\) only.

We then differentiate \(f(x,y)\) with respect to \(y\):

\[\begin{equation*} \frac{\partial f}{\partial y}=x^2 + g'(y). \end{equation*}\]

But we need \(\frac{\partial f}{\partial y}= x^2\) (corresponding to \(V_2\)), which implies that \(g'(y)=0\). Hence, \(g(y)=C\) for any constant \(C\).

So, any antiderivative is:

\[\begin{equation*} f(x,y)= x^2y + C, \end{equation*}\]

and we can for example choose \(C=0\) to get the rather simple antiderivative: \(f(x,y)= x^2y\).

Contents