Parametric Curves

Position Vectors

A parametric function of a parameter

is defined as

While vectors are considered free arrows, able to be placed anywhere in space without changing their definition, position vectors are assumed to always have their initial point situated at the origin.

Lines in Space

A line in space can be described uniquely using a vector

parallel to the line and a point on the line. If is any other point on the line then

the vector

must be parallel to . Hence we can say that

In vector form we can write this as

or in parametric form

Yellow Question: Find the point of intersection of the two lines

and

STEP 2007, Math I, #7: The line has vector equation

The line has vector equation

Show that the distance

between a point on and a point on can be expressed in the form

Hence determine the minimum distance between these two lines and find the coordinates of the points on the two lines that are the minimum distance apart.

The line has vector equation

Determine the minimum distance between these two lines, explaining geometrically the two different cases that arise according to the

value of

STEP 2013, Math I, #3: For any two points

and , with position vectors and respectively, is defined to be the point with position vector

, where is a fixed number.

(i) If

and are distinct, show that and are distinct unless takes a certain value (which you should state).

(ii) Under what conditions are and distinct?

(iii) Show that, for any points

, and ,

and obtain the corresponding result for .

(iv) The points are defined by and, for , . Given that and are distinct and that , find the ratio in which

divides the line segment .

Skill Set

Parametric Curves

No. 1

STEP 1998, Math III, #8: (i) Show that the line

, where is a unit vector, intersects the sphere at two points if

Write down the corresponding condition for there to be precisely one point of intersection. If this point has position vector

, show that .

(ii) Now consider a second sphere of radius

and a plane perpendicular to a unit vector . The centre of the sphere has position vector and the minimum distance from the origin to the plane is

. What is the condition for the plane to be tangential to this second sphere?

(iii) Show that the first and second sphere intersect at right angles (i.e. the two radii to each point of intersection are perpendicular) if

STEP 2002, Math II, #7: In 3-dimensional space, the lines

and pass through the origin and have directions and , respectively. Find the directions of the two lines

and that pass through the origin and make angles of with both and . Find also the cosine of the acute angle between

and .

The points A and B lie on

and respectively, and are each at distance units from O. The points P and Q lie on and respectively, and are each at distance 1 unit from O. If all the coordinates (with respect to axes

, , and ) of A, B, P and Q are non-negative, prove that:

(i) there are only two values of

for which AQ is perpendicular to BP;

(ii) there are no non-zero values of

for which AQ and BP intersect.

STEP 2010, Math II, #5: The points A and B have position vectors and , respectively, relative to the origin O. Find where is the

angle

AOB.

(i) The line

has equation . Given that is inclined equally to OA and to OB, determine a relationship between , , and . Find also values of

, , and for which is the angle bisector of AOB.

(ii) The line

has equation . Given that is inclined at an angle to OA, where AOB, determine a relationship between , and .

Hence describe the surface with Cartesian equation

STEP 2007, Math II, #8: The points

and have position vectors and , respectively, relative to the origin , and , and are not collinear.

(i) The point

has position vector . Describe the locus of when .

(ii) The point

has position vector , where and are non-zero, and .

The line

cuts the line at . Show that : = : .

(iii) The line

cuts the line at , and the line cuts the line at .

Show that

STEP 2003, Math II, #5: The position vectors of the points A, B, and P with respect to an origin O are

, , and , respectively, where , , and are all non-zero. The points E, F, G, and H are the mid-points of OA, BP, OB, and AP, respectively. Show that the lines EF and GH intersect.

Let D be the point with position vector

, where is non-zero, and let S be the point of intersection of EF and GH. The point T is such that the mid-point of DT is S. Find the position vector of T and hence find

in terms of if T lies in the plane OAB.

Red Question: For an independent parameter

let

for

be a vector function of

. Find a unit vector in the same direction as . Then find and verify that is orthogonal to for all .

Solution: First

So a unit vector in the direction of is

Then

We can now verify that

Velocity

The velocity of a particle is defined to be the rate of change of position with respect to time. So, without much ado, we can say that if the position of a particle is given by the vector function

then the velocity vector is given by

The velocity vector is a free vector that is imagined to travel at the tip of the position vector at all times so that the velocity vector always points in the direction of motion and hence is always tangent to the curve.

We can also define speed to be the magnitude of velocity

Speed is a measure of how fast the particle is moving and is a scalar quantity having no direction.

Example: Calculate the position and velocity vectors and the speed of the parametrized curve at .

and

These vectors are drawn at right. Notice that the velocity vectors shows that at we are moving towards the left at about 1.732 units of distance per unit of time.

Skill Set

Parametric Curves

No. 2

Tangent and Normal Vectors

We know already that the velocity vector is tangent to the curve defined parametrically. So we easily can find a tangent vector for any curve.

How would we find a vector that is everywhere normal to the graph?

Suppose we're just working in the xy-plane. Then the velocity vector is given by

We can easily see that, to get a vector everywhere orthogonal to this one, we just need to switch the components and throw in a negative sign

We could have also used

You can verify that

and

since

are parallel and opposite.

However, in 3-dimensional space, it's not so easy to find a vector that is orthogonal everywhere to another vector. Also, there are many more ways to be orthogonal and perhaps there is one way that makes more sense or is more useful than other ways.

Recall a result from the previous section that if is a unit vector, that is, a vector that is always one unit in length no matter the value of , then

is always orthogonal to .

The vector is a vector that is always tangent to the curve, so let's define the unit tangent vector as

must be always orthogonal to the tangent vector and hence always normal to the curve. We define the principal unit normal vector as

is parallel to the velocity vector but is always one unit in length, therefore the new vector

We now have a unit vector that is always tangent to the curve and a unit vector that is always normal to the curve. Since these definitions are based on derivatives of the parametrized curve, they are valid in two and three dimensions.

Example: Let be a parametrized curve for any real.

Then so

Therefore the unit tangent vector is

In the first quadrant, where

In the second quadrant, where

To find the unit normal we differentiate this.

This vector is normal to the curve. To get the unit normal, we divide this vector by its magnitude to get the unit normal in the first quadrant

In the second quadrant, the unit normal is

and

we have

and

For example, consider the points

and

we have

And at

These vectors are shown in the diagram at right. Notice that because of our careful formulation of them, the unit normal points in the direction of change of the unit tangent. What this means visually is that the normal points "into" the curve, in the direction that the tangent is rotating.

Keep in mind that , for ease of calculation when working on the xy-plane and not in 3-space, it is often a lot easier to simply look at the graph and to find the principal unit normal by switching the components of the tangent vector and multiplying one of the components by -1.

For our example, above we had

would have to equal either

so we knew that

Skill Set

Parametric Curves

No. 3

STEP 1998, Math II, #6: Two curves are given parametrically by

(1)

. I could have just looked at the graph in the xy-plane and deduced which was correct from the way the graph was curved.

In 3-space I usually can't get away with this and need to use the derivative definition of

in order for

and

(2)

Find the equation of the normal [line] to the curve (1) at the point with parameter

Sketch, using the same axes, the curves for

and

Find the gradients of the tangents to the curves at the points where

. Show that this normal is a tangent to the curve (2) .

STEP 1999, Math III, #6: A closed curve is given by the equation

where

which describes the curve anticlockwise as

is an odd integer and is a positive constant. Find a parametrization

ranges from to .

Sketch the curve in the case

, justifying the main features of your sketch.

The area

enclosed by such a curve is given by the formula

Use this result to find the area enclosed by (*) for

A parabola C is given parametrically by

has exactly one real solution if

. [Hint: In order to carry out the integration you will need a few trig identities.]

STEP 2001, Math III, #5: Show that the equation

Find an equation which must be satisfied by

Find, in Cartesian form, the equation of the locus of the points from which exactly two normals can be drawn to C. Sketch the locus.

STEP 2005, Math II, #7: The position vectors, relative to an origin O, at time

, exactly one normal to C will pass through

. Hence show

that, if

at points on C at which the normal passes through the point

of the particles P and Q are

and

(i) Give a geometrical description of the motion of P and Q.

(ii) Let

respectively, where

. Show that

be the angle POQ at time that satisfies

STEP 2003, Math III, #4: A curve is defined parametrically by

(iii) Show that the total time for which

The tangent at the point with parameter

, meets the curve again at the point with parameter

, where

. Show that

, where

and

Given a point on the curve, with parameter , a sequence of points , , , ... on the curve is constructed such that the

tangent at

meets the curve again at

. Find a second value of

but

show that

Curvature

Curvature is an intrinsic quality of a curve. What this means is that it is not dependent upon the orientation of the curve, like concavity. Curvature is meant to measure how tightly a curve is coiled, or how smooth it is. We would like our definition of curvature to be consistent with what we mean intuitively by curvature. For example, any definition of curvature should tell us that a line has 0 curvature. A circle should have constant curvature. You will see below that the formulation we use does just that.

Imagine a parametrized curve in the xy-plane

but

, for which

, with

with a tangent vector that makes an angle

with the positive x-axis.

. From the picture at left you can see that if the differentials are small enough

Imagine a small differential piece of arc

We define curvature

Our definition of curvature measures the rate of change of the angle

with respect to our movement on the curve.

We now need to get an expression for this that we can calculate with, i.e. one that involves

and any other derivatives of these we might need.

Firstly,

and

, ,

But

Therefore we get a final formulation of curvature as

Curvature of a parametrized curve

Example: Show that a circle has constant curvature.

We can parametrize a circle of radius

centered at a point

and

So its curvature function is

Because of this result, at a point

on a parametrized curve, we define the radius of curvature to be

, at

, shown below for

This radius of curvature defines a tangent circle to the curve, otherwise known as the osculating circle.

Example: Find the equation of the osculating circle to the cycloid

From the definition of the cycloid we get

So the curvature function for this curve is

we have the position vector

Notice that the curvature becomes undefined whenever the curve touches the x-axis which makes sense since there is a sharp turn there.

and

Since the point we're interested in is at a maximum, the osculating circle lies directly below and tangent to the cycloid. Hence a parametrization of the osculating circle is