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Chapter 1: Problem 3

Compute the curvature of the ellipse $$ x=a \cos t, \quad y=b \sin t, \quad t \in[0,2 \pi], a \neq b, $$ and show that it has exactly four vertices, namely, the points \((a, 0)\), \((-a, 0),(0, b),(0,-b)\).

Short Answer

Expert verified
The curvature is \( \kappa = \frac{ab}{(a^2 \sin^2(t) + b^2 \cos^2(t))^{3/2}} \), and the ellipse has four vertices: (a, 0), (-a, 0), (0, b), (0, -b).

Step by step solution

01

Parametrize the Ellipse

Given the parametric equations of the ellipse: \[ x = a \cos(t), \quad y = b \sin(t) \].
02

Compute First Derivatives

Compute the first derivatives with respect to \(t\): \[ \frac{dx}{dt} = -a \sin(t), \quad \frac{dy}{dt} = b \cos(t) \].
03

Compute Second Derivatives

Compute the second derivatives with respect to \(t\): \[ \frac{d^2x}{dt^2} = -a \cos(t), \quad \frac{d^2y}{dt^2} = -b \sin(t) \].
04

Apply the Curvature Formula

The formula for the curvature \(\kappa\) of a parametric curve \((x(t), y(t))\) is: \[ \kappa = \frac{|x' y'' - y' x''|}{((x')^2 + (y')^2)^{3/2}} \]. Substituting the derivatives: \[ \kappa = \frac{|(-a \sin(t))(-b \sin(t)) - (b \cos(t))(-a \cos(t))|}{((-a \sin(t))^2 + (b \cos(t))^2)^{3/2}} \].
05

Simplify Numerator

Simplify the numerator: \[ |ab \sin^2(t) + ab \cos^2(t)| = |ab(\sin^2(t) + \cos^2(t))| = |ab| \].
06

Simplify Denominator

Simplify the denominator: \[ ((-a \sin(t))^2 + (b \cos(t))^2)^{3/2} = (a^2 \sin^2(t) + b^2 \cos^2(t))^{3/2} \].
07

Calculate Curvature

Combine the expressions to find the curvature: \[ \kappa = \frac{ab}{(a^2 \sin^2(t) + b^2 \cos^2(t))^{3/2}} \].
08

Determine Vertices

Vertices of the ellipse occur where the curvature reaches maximum or minimum. Check values of \[ t = 0, \tfrac{\pi}{2}, \pi, \frac{3\pi}{2} \]. These correspond to the points: \[ t = 0 \rightarrow (a, 0) \ t = \frac{\pi}{2} \rightarrow (0, b) \ t = \pi \rightarrow (-a, 0) \ t = \frac{3\pi}{2} \rightarrow (0, -b) \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametric Equations
Parametric equations are a way to represent curves by expressing the coordinates of the points as functions of a parameter, typically denoted as \( t \). For the ellipse problem at hand, the parametric equations are given as \( x = a \cos(t) \) and \( y = b \sin(t) \).
These equations tell us how the x and y coordinates change as \( t \) varies from 0 to \( 2\pi \). It is essential to understand that \( a \) and \( b \) are constants representing the semi-major and semi-minor axes of the ellipse.

    Using parametric equations makes it easier to handle the geometry and calculus of curves, as they provide a direct link between the coordinate points and a single parameter.
    In this ellipse example, \( a \) and \( b \) must not be equal, ensuring the ellipse is not a perfect circle.
    Parametric forms are popular because many types of curves, including ellipses, spirals, and parabolas, become easier to work with.
    Rather than dealing with one equation in Cartesian coordinates, we deal with two linked equations in terms of \( t \).
Calculus
Calculus is used to analyze the properties of curves, including finding the curvature. To find the curvature \( \kappa \) of a parametric curve \( (x(t), y(t)) \), we use derivatives.

    First, we find the first derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For the ellipse, these are \( \frac{dx}{dt} = -a \sin(t) \) and \( \frac{dy}{dt} = b \cos(t) \).
    Next, we find the second derivatives \( \frac{d^2x}{dt^2} \) and \( \frac{d^2y}{dt^2} \). For this example, these derivatives are \( -a \cos(t) \) and \( -b \sin(t) \) respectively.
    The curvature formula for a parametric curve is:
    \( \kappa = \frac{|x'y'' - y'x''|}{((x')^2 + (y')^2)^{3/2}} \)
    By substituting the values we're able to define the exact curvature expression for the given ellipse.
Geometry of Curves
The geometry of curves concerns understanding the shapes and properties of various curves. For ellipses, it involves studying how the points are arranged and how they differ from other curves like circles or parabolas.
The vertices of the ellipse are special points where the curvature is maximum or minimum. These points provide valuable insights into the shape and size of the ellipse.

    For the ellipse with parametric equations \( x = a \cos(t) \) and \( y = b \sin(t) \), the critical points (vertices) correspond to the angles \( t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \).
    At these points, the curvature is assessed, revealing the vertices at \( (a, 0), (-a, 0), (0, b), (0, -b) \).
    Understanding these points helps us see how the ellipse stretches and deforms compared to a circle.
    This geometry is fundamental in practice, as ellipses model numerous real-world phenomena, from planetary orbits to optical lenses.

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Most popular questions from this chapter

If a closed plane curve \(C\) is contained inside a disk of radius \(r\), prove that there exists a point \(p \in C\) such that the curvature \(k\) of \(C\) at \(p\) satisfies \(|k| \geq 1 / r .\)

Green's theorem in the plane is a basic fact of calculus and can be stated as follows. Let a simple closed plane curve be given by \(\alpha(t)=\) \((x(t), y(t)), t \in[a, b]\). Assume that \(\alpha\) is positively oriented, let \(C\) be its trace, and let \(R\) be the interior of \(C\). Let \(p=p(x, y), q=q(x, y)\) be real functions with continuous partial derivatives \(p_{x}, p_{y}, q_{x}, q_{y} .\) Then $$ \iint_{R}\left(q_{x}-p_{y}\right) d x d y=\int_{C}\left(p \frac{d x}{d t}+q \frac{d y}{d t}\right) d t, $$ where in the second integral it is understood that the functions \(p\) and \(q\) are restricted to \(\alpha\) and the integral is taken between the limits \(t=a\) and \(t=b .\) In parts a and \(\mathrm{b}\) below we propose to derive, from Green's theorem, a formula for the area of \(R\) and the formula for the change of variables in double integrals (cf. Eqs. (1) and (7) in the text). a. Set \(q=x\) and \(p=-y\) in Eq. (9) and conclude that $$ A(R)=\iint_{R} d x d y=\frac{1}{2} \int_{a}^{b}\left(x(t) \frac{d y}{d t}-y(t) \frac{d x}{d t}\right) d t $$ b. Let \(f(x, y)\) be a real function with continuous partial derivatives and \(T: R^{2} \rightarrow R^{2}\) be a transformation of coordinates given by the functions \(x=x(u, v), y=y(u, v)\), which also admit continuous partial derivatives. Choose in Eq. (9) \(p=0\) and \(q\) so that \(q_{x}=f\). Apply successively Green's theorem, the map \(T\), and Green's theorem again to obtain $$ \begin{aligned} &\iint_{R} f(x, y) d x d y \\ &\quad=\int_{C} q d y=\int_{T-1(C)}(q \circ T)\left(y_{u} u^{\prime}(t)+y_{v} v^{\prime}(t)\right) d t \\ &\quad=\iint_{T-1}\left\\{\frac{\partial}{\partial u}\left((q \circ T) y_{v}\right)-\frac{\partial}{\partial v}\left((q \circ T) y_{u}\right)\right\\} d u d v \end{aligned} $$ Show that $$ \begin{gathered} \frac{\partial}{\partial u}\left(q(x(u, v), y(u, v)) y_{v}\right)-\frac{\partial}{\partial v}\left(q(x(u, v), y(u, v)) y_{u}\right) \\\ \quad=f(x(u, v), y(u, v))\left(x_{u} y_{v}-x_{v} y_{u}\right)=f \frac{\partial(x, y)}{\partial(u, v)} \end{gathered} $$ Put that together with the above and obtain the transformation formula for double integrals: $$ \iint_{R} f(x, y) d x d y=\iint_{T^{-1}(R)} f(x(u, v), y(u, v)) \frac{\partial(x, y)}{\partial(u, v)} d u d v . $$15. Green's theorem in the plane is a basic fact of calculus and can be stated as follows. Let a simple closed plane curve be given by \(\alpha(t)=\) \((x(t), y(t)), t \in[a, b]\). Assume that \(\alpha\) is positively oriented, let \(C\) be its trace, and let \(R\) be the interior of \(C\). Let \(p=p(x, y), q=q(x, y)\) be real functions with continuous partial derivatives \(p_{x}, p_{y}, q_{x}, q_{y} .\) Then $$ \iint_{R}\left(q_{x}-p_{y}\right) d x d y=\int_{C}\left(p \frac{d x}{d t}+q \frac{d y}{d t}\right) d t, $$ where in the second integral it is understood that the functions \(p\) and \(q\) are restricted to \(\alpha\) and the integral is taken between the limits \(t=a\) and \(t=b .\) In parts a and \(\mathrm{b}\) below we propose to derive, from Green's theorem, a formula for the area of \(R\) and the formula for the change of variables in double integrals (cf. Eqs. (1) and (7) in the text). a. Set \(q=x\) and \(p=-y\) in Eq. (9) and conclude that $$ A(R)=\iint_{R} d x d y=\frac{1}{2} \int_{a}^{b}\left(x(t) \frac{d y}{d t}-y(t) \frac{d x}{d t}\right) d t $$ b. Let \(f(x, y)\) be a real function with continuous partial derivatives and \(T: R^{2} \rightarrow R^{2}\) be a transformation of coordinates given by the functions \(x=x(u, v), y=y(u, v)\), which also admit continuous partial derivatives. Choose in Eq. (9) \(p=0\) and \(q\) so that \(q_{x}=f\). Apply successively Green's theorem, the map \(T\), and Green's theorem again to obtain $$ \begin{aligned} &\iint_{R} f(x, y) d x d y \\ &\quad=\int_{C} q d y=\int_{T-1(C)}(q \circ T)\left(y_{u} u^{\prime}(t)+y_{v} v^{\prime}(t)\right) d t \\ &\quad=\iint_{T-1}\left\\{\frac{\partial}{\partial u}\left((q \circ T) y_{v}\right)-\frac{\partial}{\partial v}\left((q \circ T) y_{u}\right)\right\\} d u d v \end{aligned} $$ Show that $$ \begin{gathered} \frac{\partial}{\partial u}\left(q(x(u, v), y(u, v)) y_{v}\right)-\frac{\partial}{\partial v}\left(q(x(u, v), y(u, v)) y_{u}\right) \\\ \quad=f(x(u, v), y(u, v))\left(x_{u} y_{v}-x_{v} y_{u}\right)=f \frac{\partial(x, y)}{\partial(u, v)} \end{gathered} $$ Put that together with the above and obtain the transformation formula for double integrals: $$ \iint_{R} f(x, y) d x d y=\iint_{T^{-1}(R)} f(x(u, v), y(u, v)) \frac{\partial(x, y)}{\partial(u, v)} d u d v . $$

Let \(\alpha(s), s \in[0, l]\) be a closed convex plane curve positively oriented. The curve $$ \beta(s)=\alpha(s)-r n(s), $$ where \(r\) is a positive constant and \(n\) is the normal vector, is called a parallel curve to \(\alpha\) (Fig. 1-37). Show that a. Length of \(\beta=\) length of \(\alpha+2 \pi r\). b. \(A(\beta)=A(\alpha)+r l+\pi r^{2} .\) c. \(k_{\beta}(s)=k_{\alpha}(s) /\left(1+r k_{\alpha}(s)\right)\). For (a)-(c), \(A\) ( ) denotes the area bounded by the corresponding curve, and \(k_{\alpha}, k_{\beta}\) are the curvatures of \(\alpha\) and \(\beta\), respectively.

A parametrized curve \(\alpha(t)\) has the property that its second derivative \(\alpha^{\prime \prime}(t)\) is identically zero. What can be said about \(\alpha\) ?

Find a parametrized curve \(\alpha(t)\) whose trace is the circle \(x^{2}+y^{2}=1\) such that \(\alpha(t)\) runs clockwise around the circle with \(\alpha(0)=(0,1)\).
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