Week 9: Preparation

Week 9: Preparation#

Reading Material#

Key Concepts#

Long Day will cover:

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

  • Normal vector of a surface

  • Line integral and surface integral

  • The general Jacobian “function”

  • Vector fields and gradient vector fields

  • Flow curves

  • Tangental line integral: Integration along a curve

Short Day will cover:

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

  • Circulation

  • Orientation of parametrized curves and surfaces

  • 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.

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

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