Week 9: Preparation

Week 9: Preparation#

Key Concepts#

After reading, you should be able to explain the following key concepts:

  • Parametric representations of curves and surfaces in \(\mathbb{R}^n\)

  • Curve length

  • The normal vector of a surface

  • The line integral and the surface integral

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

  • Vector fields and gradient vector fields

  • Flux

Note

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

This week, we will explore these key concepts in great detail. We expect you to have familiarized yourself with these topics before lectures.

Reading Material#

We recommend that you read the textbook. Watching YouTube videos on the week’s topics can be useful, but it should not replace proper preparation for the week’s program and is not recommended as a standalone approach.

Read and study the following:

Preparatory Exercises#

1: 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

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

2: Line Integral of Scalar Function#

Let \(f(x,y)=x^2+y^2\), and let \(\mathcal{C}\) be the same circle as in exercise 1: 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:

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

3: 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: \(f(x,y)= x^2y\).

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