Using vector field to compare two physical phenomenas : a storm and a whirlpool

 Objectives:

1) Compare two physical phenomenas: a storm and a whirlpool

2) Use the vector field to show the physical similarities and one distinction

When scientists study a storm, they can see the movement inside and the direction of the movement. They can also locate the center of the storm. As the movement goes further from the center, the intensity of the movement increases. In the previous post, we have used the vector field to model the shapes, the direction and the relative intensity of these movements.

 A whirlpool is a body of rotating water in which the flow moves in a circular pattern around a central point. The defining physical characteristic of a whirlpool is that the water spins faster near the centre and slower farther away. This is quite different in the case of a storm. We are going to use the same vector field method to study the physical characteristics of the whirlpool. We will also show that the vector field of the storm and that of the whirlpool confirm the physical characteristics of of these two phenomena. Let's start by drawing the vector field of the whirlpool.

Let's consider a whirlpool that can be modeled by the following function: 

Choosing the base points, we obtain the following table of values:

Let's draw the vector field:

This field confirms the physical characteristics of the whirlpool. It moves in a circular clockwise direction around a central point. 

Let's calculate the norm of any vector F(a,b) of the field:

As the vectors approach to the origin, the magnitude goes to infinity or become very large. This confirms the physical characteristic of the whirlpool where the intensity of the movement of the water grows as we approach to the center. 

If we compare the vector field of the whirlpool to that of the storm, we can see that they are both rotational. The only difference is that in the case of the vector field of the storm the magnitude of the vectors increase as we goes farther from the origin. Therefore the intensity of the movement of the storm increases as we move away from the center. In the case of the vector field of the  whirlpool the magnitude of the vectors increase as we approach to the center. Therefore the intensity of the movement increases as we approach to the center.

Drawing a rotational field

 Objective: Draw a rotational field

Rotational fields are used to study major storm systems including cyclones and hurricanes. In the northern hemisphere, storms rotate counterclockwise. In the southern hemisphere they rotate clockwise. This is an effect of the earth's rotation around its axis called Coriolis effect.

Example

 Draw the rotational field F<x,y> = (y, -x)

Solution

 

 

The vector field is now complete. We can show now that the vectors are tangent to the circles of radius r = 1, r -⎷ 2, r = 2.

We can also compare the rotational field to the radial field and show that the vectors of the rotational field are perpendicular to the ones in the radial field. The following figure shows that.

We can confirm  the perpendicularity of the vectors of the radial field to the vectors of the rotational by showing that their dot product is equal to zero.

Drawing a radial vector field

 Objective: draw a radial field

Considerations

A radial field models certain types of gravitational fields and energy source fields. To draw a vector field you have to draw a vector at each point of a set of points. Let's say that you have to draw the vector F(2.1) = <1.1/2>, you draw first the point (2,1) then you move one unit to the right and half unit up. The tail of the vector is at the point (2,1).

In a radial field all the vectors point directly toward or away from the origin. The vector located at (x,y) is perpendicular to the circle centered at the origin and passing through (x,y). All the vectors on that circle have the same magnitude.

Method to draw a radial field

1) Choose a sample of points from each quadrant in the system of coordinates plane. It's better to choose a set of points at the intersection of the grid lines.

2) Draw the corresponding vector at each point

Example

Solution

Figure (a) shows the vector field. Figure (b) shows circles overlain on the vector field.

Practice

What are vector fields in Calculus?

 Objective: Define vector field

Introduction

Vector fields are used to describe physical phenomena such as gravitation, electromagnetism, atmospheric storms and deep-sea ocean currents. Let's give some examples.

Example (a). Gravitational force exerted by two astronomical objects such as a planet and a star or a planet and its moon.

The figure (a) models the gravitational force exerted by these two objects:

At any point in the figure, a vector associated with that point gives the net gravitational force exerted by the two objects on a unit of mass of an object. The vectors of the greater magnitude are associated with the largest object. The greater the mass of an object, the stronger is its gravitational force. The gravitational force of the larger object is stronger than that of the smaller object. The vectors of smaller magnitude are associated with the smaller object. The mass of this object is smaller. Its gravitational force is less strong than the larger object. 

Example (b). Velocity of a river at points on its surface.

The vector field velocity of water on a river shows the varied speeds of water. Vectors in red are of greater magnitude. Therefore, the water flows quickly. Vectors in blue are smaller in magnitude. They show a smaller flow of water. 

As we can see in these two examples, a vector field is a map of vectors. In the following posts we will study vectors fields in R² and R³.

Definition

Example

Solution

Practice

Solving a triple integral using the change of variables

 In the previous posts we spent time establishing formulas to solve a triple integral using a change of variables. Let's now solve an example.

Example

Solution

Let's apply the formula:

Let's take the following steps to apply the formula:

1) Let's find x, y, z from the given values of u, v, w. This leads to solve the following system of equations:

u = 2x-y/2 (1)

v = y/2 (2)

w = z/3 (3)

Solving this system, we find x = (u + v)/2 y = 2v z = 3w

2) Let's find the new function H(u,v,w) to be integrated by substituting x, y, z in the function F(x,y,z) = x +z/3. 

3) Let's find the limits of integration of the new function i.e u.v.w:

Let's isolate x, y, z as limits of integration of the given integral. We have:

x = y/2 x = y/2 + 1

y = 0 y = 4

z = 0 z = 3

These equations represent the planes that bound the surface G in the space xyz.

From the equation x=y/2 we have 2x = y 2x-y = 0 ⇒2x-y/2 = 0 ⇒ u = 0 since u = 2x-y/2 (1)

From the equation x = y/2 + 1 we have 2x = y+2 ⇒ 2x-y = 2 ⇒ 2x-y/2 = 1  ⇒ u = 1.

Let's use the equation v = y/2:

For y = 0 we have v = 0. For y = 4 v =2 

Let's use the equation w = z/3

For z = 0 w = 0

For z =3 w = 1

Let's calculate the jacobian:

Change of variables for triple integrals

 Objective: Set the formula for the change of variables in triple integrals.

Planar transformation in tridimensional space

a.The change of variables in triple integrals works the same way as in double integrals. We consider a transformation from a tridimensional space (u,v.w) to a tridimensional space (x,y,z). We define the Jacobian in the (u,v,w) space. Then we establish the formula for the triple integral defined in (u,v,w).

 Suppose that G i s a region in the uvww space transformed in another region D in the xyz space by a C¹ transformation by the transformation T so that T(u,v,w) = (x,y,z) where x = g(u,v,w) y = h(u,v.w) z = k(u,v,w).

Any function F(x.y,z) defined on D can be thought of another function H(u,v. w) defined G such that:

F(x.y.z) = F(g(u,v,w), h(u,v,w), k(u,v,w)) = H(u,v,w), Now let's define the Jacobian in a tridimensional space (u,v,w

Jacobian for three variables. Definition

The jacobian J(u,v,w) for three variables is defined as follows:

This is the same as:

Theorem

Change of Variables for Triple Integrals

Let T(u, v, w) = (x, y, z) where x = g(u, v, w), y = h(u, v, w), and z = k(u, v, w), be a one-to-one C¹ transformation, with a nonzero Jacobian, that maps the region G in the uvw-space into the region R in the xyz-space. As in the two-dimensional case, if F is continuous on R, then

∫∫∫_R F(x,y,z) dV = ∫∫∫_G F(g(u,v,w), h(u,v,w), k(u,v,w))   | ∂(x,y,z) ∂(u,v,w) |  du dv dw
= ∫∫∫_G H(u,v,w) |J(u,v,w)| du dv dw.

Example 

  Obtaining Formulas in Triple Integrals for Cylindrical and Spherical Coordinates

Derive the formula in triple integrals for

1. cylindrical and
2. spherical coordinates.

Solution

Let's apply the formula for change of variables in triple integrals:

∫∫∫_R F(x,y,z) dV = ∫∫∫_G H(u,v,w) |J(u,v,w)| du dv dw

In cylindrical coordinates it becomes:

∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(r cosθ, r sinθ, z)  |J(r,θ,z)|  dr dθ dz

J(r, θ, z) =  |

∂x/∂r	∂x/∂θ	∂x/∂z
∂y/∂r	∂y/∂θ	∂y/∂z
∂z/∂r	∂z/∂θ	∂z/∂z

| = |

cos θ	−r sin θ	0
sin θ	r cos θ	0
0	0	1

| = r cos²θ + r sin²θ = r (cos²θ + sin²θ) = r.

We know that r ≥ 0, so |J(r,θ,z)| = r. Then the triple integral becomes by substituting the value of the jacobian: 
∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(r cos θ, r sin θ, z) r dr dθ dz.

b. Let's find the formula for the triple integral in spherical coordinates by applying the same process.

For spherical coordinates, the transformation is T(ρ, θ, φ) = (x, y, z) from the Cartesian pθφ-plane to the Cartesian xyz-plane . Here x = ρ sin φ cos θ, y = ρ sin φ sin θ, and z = ρ cos φ. The expression of the triple integral can be written as:

∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(ρ sinφ cosθ, ρ sinφ sinθ, ρ cosφ)  |J(ρ,θ,φ)|  dρ dφ dθ. Let's calculate the jacobian.

First write it as a quotient of partial derivatives:

J(ρ,θ,φ) = ∂(x,y,z) ∂(ρ,θ,φ)

Now write it as a determinant:

J(ρ,θ,φ) = |

∂x/∂ρ	∂x/∂θ	∂x/∂φ
∂y/∂ρ	∂y/∂θ	∂y/∂φ
∂z/∂ρ	∂z/∂θ	∂z/∂φ

|

Substitute the partial derivatives:

J(ρ,θ,φ) = |

sinφ cosθ	-ρ sinφ sinθ	ρ cosφ cosθ
sinφ sinθ	ρ sinφ cosθ	ρ cosφ sinθ
cosφ	0	-ρ sinφ

|

Expand along the third row:

J = cosφ |

-ρ sinφ sinθ	ρ cosφ cosθ
ρ sinφ cosθ	ρ cosφ sinθ

| - ρ sinφ |

sinφ cosθ	-ρ sinφ sinθ
sinφ sinθ	ρ sinφ cosθ

|

Calculate the first determinant:

(-ρ sinφ sinθ)(ρ cosφ sinθ) - (ρ cosφ cosθ)(ρ sinφ cosθ)

= -ρ² sinφ cosφ sin²θ - ρ² sinφ cosφ cos²θ

= -ρ² sinφ cosφ

Calculate the second determinant:

(sinφ cosθ)(ρ sinφ cosθ) - (-ρ sinφ sinθ)(sinφ sinθ)

= ρ sin²φ cos²θ + ρ sin²φ sin²θ

= ρ sin²φ

Therefore,

J = cosφ(-ρ² sinφ cosφ) - ρ sinφ(ρ sin²φ)

J = -ρ² sinφ cos²φ - ρ² sin³φ

J = -ρ² sinφ(cos²φ + sin²φ)

J(ρ,θ,φ) = -ρ² sinφ

So,

|J(ρ,θ,φ)| = ρ² sinφ

Let's substitute the jacobian in the expression of the integral. Then the triple integral becomes:

∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(ρ sinφ cosθ, ρ sinφ sinθ, ρ cosφ)  ρ² sinφ dρ dφ dθ.

Change of variables for double integrals

 We have seen that under the change of variables T(u,v) = (x,y) where x = g(u,v) and y = h(u,v), a small region ΔA in the xy plane is related to the area formed by the product 𝜟u 𝜟v in the uv plane by the approximation: 𝜟A ≃J(u,v)𝜟u𝜟

Remember that the double integral is defined as:

In the following figure we divide the region S in small rectangles Sᵢⱼ and the region R in small rectangles Rᵢⱼ. A small rectangle Rᵢⱼ is the image of a small rectangle Sᵢⱼ under the transformation T.

 

Let's substitute f(xᵢⱼ, yᵢⱼ) and ΔA in the definition of the integral:

∫∫_R f(x,y) dA = ∫∫_S f(g(u,v),h(u,v)) |J(u,v)| du dv

Let's substitute J(u,v) in in the expression on the right side of the equality sign, we have:

Theorem

Example

Solution

Let's do the change of variables in the expression √x² + y² . By substituting x = rcosθ and y = sinθ, the expression becomes equal to r.

The expression dydx becomes J(r,θ) drdθ.  In a previous example J(r,θ) was equal to r. The integral becomes when we substitute everything:

Practice