Parametric Equations

1. Find the turning points of the curve with parametric equations

 Hint

Differentiate using chain rule:

Set the derivative equal to zero to find turning points.

 Solutions

Differentiate with respect to t:

Differentiate with respect to t:

Now, we use the chain rule:

We can simplify this expression:

To find the turning points, we set the derivative equal to zero:

Taking the square root of both sides gives us:

So, the turning points occur at and .

Now we find the (x,y) coordinates for each value of t:
When
For
For
So, the first turning point is
When
For
For
So, the second turning point is .

2. A curve is defined by the parametric equations .

(a) By eliminating the parameter, find the cartesian equation of the curve.

 Solutions

The goal of eliminating the parameter is to get a single equation involving only x and y, without t. It’s usually easiest to isolate t from the simpler equation. In this case, equation (2) is perfect for that:

A curve is defined by the parametric equations:

From y=4t, we can solve for t:

Now, substitute this expression for into equation (1):

Let’s simplify the expression to get our final Cartesian equation:

We can also rearrange this to:

(b) Find, in the form , the equation of the tangent to the curve at the point A with parameter .

 Solutions

A curve is defined by the parametric equations:

We want to find the equation of the tangent to this curve at point A, where , in the form .

Step 1: Find the coordinates of point A
First, we determine the and coordinates of point A when .
Substitute into the parametric equations:

So, the point A is .

Step 2: Find the derivative
To find the slope of the tangent line, we calculate . For parametric equations, we use the chain rule:

First, let’s find and :

Now, substitute these into the formula for :

Step 3: Calculate the slope of the tangent at point A
The slope of the tangent line at point A is the value of when :

Step 4: Use the point-slope form of a line
We have the point A and the slope . We use the point-slope form of a linear equation:

Substitute the values:

Step 5: Convert the equation to the form
To remove the fraction and rearrange the terms, multiply both sides by 2:

Now, move all terms to one side to match the form:

3. A line segment is defined by the parametric equations

(a) Find .

 Solutions

Differentiate with respect to

Form the quotient

(b) Find the cartesian equation of the line segment, stating its domain.

 Solutions

Method 1: Eliminate the parameter
Eliminate the parameter using the identity , since , so:

Substitute to obtain the Cartesian equation:

Rearranging gives the line equation in standard form:

Method 2: Use the derivative and point-slope form
From part (a), we have , which is the slope of the line segment. To find the equation of the line, we need a point on the line. We can use to find a point:
When :

So, the point is . Using the point-slope form of a line:

Substituting the point and slope :

Domain
Regarding the domain, since satisfies , so the domain is

4. A circle is defined by the parametric equations .

(a) Sketch the circle.

 Solutions

a circle centred at (1,3) with radius 2

(b) Find at the point with parameter .

 Solutions

Differentiate with respect to \theta:

Hence

valid for

Show that the equation of the tangent at the point with parameter can be written as

 Hint

Any point on the curve can be written as . So after finding the slope , use the point-slope form of a line to find the equation of the tangent line at point P:
Substitute the values of and into the point-slope form, then manipulate the equation to get the desired form.

 Solutions

Parametric equations

Point on the curve
The point is .

Derivatives

Slope

Point-slope form

Multiply both sides by

Substitute (

Expand and rearrange

Use identity

(d) Find the coordinates of the point where .

 Solutions

Plug in Using

we get

So the point is

(e) Find, in the form , the equation of the normal at the point where .

 Hint

Normal line has slope that is the negative reciprocal of the tangent slope. Substitute the into the derivative function found in part (b) to get the derivative (Slop) of tangent line. Then find the slope of the normal line.

 Solutions

The parametric equations are

At we have

hence the point is

The slope of the tangent is

Thus the slope of the normal is the negative reciprocal:

The normal line through is

Rearrange to the form :

5. A ball is thrown from a fixed point O on a windy day. At time t seconds after it is thrown, the vertical displacement metres and the horizontal displacement metres of the ball from can be modelled by the parametric equations:

(a) Use this model to find the position of the ball after 0.8 s

 Solutions

The parametric model is

Evaluate at s:

Therefore the position of the ball after s is

(b) Use this model to find the height of the ball when it has travelled a horizontal distance of 2.25 m

 Hint

Solve for t from the horizontal equation, then find y.

 Solutions

Solve for t from the horizontal equation, then find y.

Rearrange:

Compute the discriminant:

so

This gives (t=4.5) and (t=0.5). Only (t=0.5) lies in the allowed interval .
Now compute the height:

The height is 3.75 m when the ball has travelled 2.25 m horizontally.

Use this model to find the horizontal displacement of the ball at its maximum height.

 Hint

Find the maximum height by differentiating y with respect to t, setting the derivative to zero, and solving for t. Then substitute this t value into the x equation to find the horizontal displacement.

 Solutions

Find when the height is maximum, then evaluate the horizontal position there.

The vertical displacement is

Differentiate to find the maximum:

Set this to zero to get the time of maximum height:

Now evaluate the horizontal displacement at this time:

(d) Explain the significance of the domain of the parametric equations,

 Solutions

t=0 represents the time when the ball is thrown from point O, and t=2 represents the time when the ball lands back on the ground. Therefore, the domain indicates the entire duration of the ball’s flight from launch to landing.
If t<0, it would represent a time before the ball was thrown, which is not relevant to the motion being modeled. If t>2, it would represent a time after the ball has already landed, which is also not relevant to the motion being modeled.

(e) Later in the day, the wind strength increases. It is still blowing from the same direction and the ball is still thrown with the same velocity. Explain how you could refine the model to reflect the change in conditions.

 Solutions

Wind strength increase can be modelled by acceleration in the horizontal direction. We can calculate the the starting velocity based on acceleration, and velocity can be used to find the new position in the horizontal direction.
No change in vertical motion.