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Chapter 14: Problem 10

Sketch the oriented quarter circle from (1,0) to (0,1) and supply a parameterization for the curve. Draw the unit normal vector at several points on the curve.

Short Answer

Expert verified
Question: Sketch the oriented quarter-circle from (1,0) to (0,1), provide its parameterization, and draw the unit normal vector at several points on the curve. Answer: The oriented quarter-circle starts at point (1,0) and ends at point (0,1) in the counter-clockwise direction. Its parameterization is given by x(t) = cos(t) and y(t) = sin(t), where t varies from 0 to 蟺/2. The unit normal vector N(t) = [cos(t), sin(t)]. At t = 0, N(0) = [1, 0]; at t = 蟺/4, N(蟺/4) = [1/鈭�2, 1/鈭�2]; and at t = 蟺/2, N(蟺/2) = [0, 1].

Step by step solution

01

Sketch the oriented quarter circle from (1,0) to (0,1)

The given curve is an oriented quarter circle, which indicates that it is a section of a circle that goes from point (1,0) to point (0,1). Since the circle is located at the first quadrant, we draw the quarter circle counter-clockwise.
02

Parameterize the curve

To parameterize the curve from (1,0) to (0,1), we will use the polar coordinates. The radius of the quarter circle is 1. We will use the parameter t, where t varies from 0 to 蟺/2. So, the parameterized curve is given by: x(t) = cos(t) y(t) = sin(t)
03

Draw the unit normal vector at several points on the curve

To find the unit normal vector, we first need to find the tangent vector to the curve. The tangent vector T(t) is the derivative of the position vector: dx/dt = -sin(t) dy/dt = cos(t) T(t) = [-sin(t), cos(t)] Now, we need to find the unit normal vector N(t) by rotating the tangent vector by 90 degrees in the counter-clockwise direction: N(t) = [cos(t), sin(t)] At different points along the curve, we can find N(t) using different values of t. For example, let's find the normal vector at three specific points (t = 0, t = 蟺/4, and t = 蟺/2): 1. t = 0: N(0) = [cos(0), sin(0)] = [1, 0] 2. t = 蟺/4: N(蟺/4) = [cos(蟺/4), sin(蟺/4)] = [1/鈭�2, 1/鈭�2] 3. t = 蟺/2: N(蟺/2) = [cos(蟺/2), sin(蟺/2)] = [0, 1] Now, draw these normal vectors at their corresponding points on the curve. Make sure they are perpendicular to the tangent vectors and have a magnitude of 1.

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Key Concepts

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

Parametric Curves
Understanding parametric curves is key for many topics in calculus. A parametric curve is defined using one or more parameters. The main advantage of parametric equations is that they allow us to describe more complex curves that cannot be represented as a single function in the Cartesian coordinate system. In the previous exercise, the curve was a quarter-circle on a plane.The mathematical representation of a parametric curve involves expressing the coordinates as functions of a parameter, often denoted as \( t \). For instance, the parameterization of the quarter-circle from (1,0) to (0,1) can be expressed as:
  • \( x(t) = \cos(t) \)
  • \( y(t) = \sin(t) \)
This representation was achieved by considering \( t \) to range from \( 0 \) to \( \pi/2 \), describing the curve as it moves counter-clockwise from the x-axis to the y-axis.
Tangent and Normal Vectors
Tangent and normal vectors are essential for understanding the orientation and geometry of curves. The tangent vector of a parametric curve gives the direction of the curve at any point, while the normal vector is perpendicular to the tangent vector.

Finding Tangent Vectors

The tangent vector is found by taking the derivative of the position vector with respect to the parameter \( t \). For our quarter-circle:
  • \( \frac{dx}{dt} = -\sin(t) \)
  • \( \frac{dy}{dt} = \cos(t) \)
Therefore, the tangent vector \( T(t) \) is \([-\sin(t), \cos(t)]\).

Calculating Normal Vectors

To determine the normal vector, we rotate the tangent vector by 90 degrees. This involves swapping the components and changing the sign of the first component, giving us \( N(t) = [\cos(t), \sin(t)] \). The normal vector helps in understanding how the curve is curving at any given point, offering insights into its geometry and structure.
Polar Coordinates
Polar coordinates provide a different way of describing points on a plane. They are particularly useful for dealing with circular and rotational symmetries. Each point in the polar coordinate system is represented by a radius and an angle.In our exercise, the quarter-circle can naturally be described using polar coordinates where the radius is constant, and the angle increases linearly from \( 0 \) to \( \pi/2 \). This makes parameterizing curves like circles much more straightforward.In this system, any point \( (r, \theta) \) is translated into Cartesian coordinates through:
  • \( x = r \cos(\theta) \)
  • \( y = r \sin(\theta) \)
For our specific curve, the radius \( r \) is 1, simplifying our equations to \( x = \cos(t) \) and \( y = \sin(t) \). This natural synthesis of polar and parametric forms allows for a seamless representation of a variety of curves.

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Most popular questions from this chapter

The Navier-Stokes equation is the fundamental equation of fluid dynamics that models the flow in everything from bathtubs to oceans. In one of its many forms (incompressible, viscous flow), the equation is $$\rho\left(\frac{\partial \mathbf{V}}{\partial t}+(\mathbf{V} \cdot \nabla) \mathbf{V}\right)=-\nabla p+\mu(\nabla \cdot \nabla) \mathbf{V}.$$ In this notation, \(\mathbf{V}=\langle u, v, w\rangle\) is the three-dimensional velocity field, \(p\) is the (scalar) pressure, \(\rho\) is the constant density of the fluid, and \(\mu\) is the constant viscosity. Write out the three component equations of this vector equation. (See Exercise 40 for an interpretation of the operations.)

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Consider the potential function \(\varphi(x, y, z)=G(\rho),\) where \(G\) is any twice differentiable function and \(\rho=\sqrt{x^{2}+y^{2}+z^{2}} ;\) therefore, \(G\) depends only on the distance from the origin. a. Show that the gradient vector field associated with \(\varphi\) is \(\mathbf{F}=\nabla \varphi=G^{\prime}(\rho) \frac{\mathbf{r}}{\rho},\) where \(\mathbf{r}=\langle x, y, z\rangle\) and \(\rho=|\mathbf{r}|\) b. Let \(S\) be the sphere of radius \(a\) centered at the origin and let \(D\) be the region enclosed by \(S\). Show that the flux of \(\mathbf{F}\) across \(S\) is $$\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=4 \pi a^{2} G^{\prime}(a) $$ c. Show that \(\nabla \cdot \mathbf{F}=\nabla \cdot \nabla \varphi=\frac{2 G^{\prime}(\rho)}{\rho}+G^{\prime \prime}(\rho)\) d. Use part (c) to show that the flux across \(S\) (as given in part (b)) is also obtained by the volume integral \(\iiint_{D} \nabla \cdot \mathbf{F} d V\). (Hint: use spherical coordinates and integrate by parts.)

Electric field due to a point charge The electric field in the \(x y\) -plane due to a point charge at (0,0) is a gradient field with a potential function \(V(x, y)=\frac{k}{\sqrt{x^{2}+y^{2}}},\) where \(k>0\) is a physical constant. a. Find the components of the electric field in the \(x\) - and \(y\) -directions, where \(\mathbf{E}(x, y)=-\nabla V(x, y).\) b. Show that the vectors of the electric field point in the radial direction (outward from the origin) and the radial component of \(\mathbf{E}\) can be expressed as \(E_{r}=k / r^{2},\) where \(r=\sqrt{x^{2}+y^{2}}.\) c. Show that the vector field is orthogonal to the equipotential curves at all points in the domain of \(V.\)
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