# Edexcel Core Mathematics 4 Parametric . · PDF fileEdexcel Core Mathematics 4 Parametric...

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Edexcel Core

Mathematics 4 Parametric equations.

Edited by: K V Kumaran

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Co-ordinate Geometry A parametric equation of a curve is one which does not give the relationship between x and y directly but rather uses a third variable, typically t, to do so. The third variable is known as the parameter. A simple example of a pair of parametric equations: x = 5t + 3 y = t2 + 2t Converting to Cartesian You need to be able to find the Cartesian equation of the curve from parametric equations, that is the equation that relates x and y directly. To do this you need to eliminate the parameter. The easiest way to do this is to rearrange on parametric equation to get the parameter as the subject and then substitute this into the other equation. A circle with an origin (a, b) has the parametric equation:

= + = +

You can use the result sin2 + cos2 = 1 to derive these. As before, is the parameter instead of t in the equations. You need to be able to recognise these as parametric equations of circles in the exam.

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Example The curve C is described by the parametric equations x = 5cost, y = cos2t, 0 t a) Find a Cartesian equation for the curve C. b) Draw a sketch of the curve C. a) From C3 again you should remember that: cos2t cos2t sin2t 2cos2t 1 Therefore: y = 2cos2t 1

If x = 5cost then cost = x

5

So finally y = 22x

25- 1

b) This is simply a quadratic that is symmetrical about the y axis and intercepts with the y axis at -1.

-1

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Example The parametric equations of a curve are x = 14 sin t, y = 14t cos t

where 0 < t < 2

. Find

dx

dy in terms of t, and hence show that the gradient of the

curve is zero where t

1ttan

Since the curve is given parametrically we can use the chain rule to find dx

dy

dy dy dt

dx dt dx

So by differentiating the parametric:

x = 14 sin t y = 14t cos t (a Product)

dx

dt= 14 cos t

dy

dt= 14 cos 14t sin t

dy dy dt

dx dt dx

dy 14 cost - 14t sin t

1 ttantdx 14cost

When the gradient is zero

1 ttant 0

1tant

t

This equation could be solved by iterative methods (C3).

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Example 4 A curve is given by the parametric equations

x = 7 sin3 t, y = 6 cos 2t, 0 < t < 4

.

Show that tsin8

7

dy

dx

By chain rule

dx dx dt

dy dt dy

2dx 21sin tcostdt

dy

12sin2t don't forget the 2.dt

By C3 trig identities sin 2t = 2 sin t cos t

2dx dx dt 21sin tcost

dy dt dy 24sintcost

dx 7 sint

dy 8

The final example in this section deals with tangents and normals to curves. Example 5

The curve C is described by the parametric equations

x = tan t y = sin 2t 2

t2

a) Find the gradient of the curve at the point P where t= 3

b) Find the equation of the normal to the curve at P.

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a) Find the gradient of the curve at the point P where t= 3

Using chain rule:

tsecdt

dx 2 t2cos2dt

dy

dx dx dt

dy dt dy

t2costcos2tsec

t2cos2

dx

dy 22

Let t= 3

Grad = 25.03

2cos

3cos2

2

b) Find the equation of the normal to the curve at P. We are asked for the equation of the normal therefore the gradient will be 4 (why?).

Using t= 3

the x and y coordinates are 3 and

2

3respectively.

Using y = mx + c

2

3 = 4 3 + c

c = 2

37

Therefore the equation of the normal is

2

37x4y

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Differentiating ax

This function describes growth and decay, and its derivative gives a measure of the rate of change of this growth/decay. Since = , taking logs of both sides gives ln = ln = ln . Using implicit differentiation to differentiate ln :

1

= ln

= ln = ln

This result needs to be learn, and is not given in the formula sheet.

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C4 Parametric differentiation past paper questions

1. A curve has parametric equations

x = 2 cot t, y = 2 sin2 t, 0 < t 2

.

(a) Find an expression for x

y

d

d in terms of the parameter t. (4)

(b) Find an equation of the tangent to the curve at the point where t = 4

. (4)

(c) Find a cartesian equation of the curve in the form y = f(x). State the domain on which the curve is

defined. (4)

(C4 June 2005, Q6.)

2. Figure 2

The curve shown in Figure 2 has parametric equations

x = sin t, y = sin

6

t ,

2

< t <

2

.

(a) Find an equation of the tangent to the curve at the point where t = 6

.

(6)

(b) Show that a cartesian equation of the curve is

y = 2

3x +

2

1(1 x2), 1 < x < 1.

(3)

(C4 June 2006, Q4.)

x 1 0.5 1 0.5

0.5

y

O

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3. A curve has parametric equations

x = 7 cos t cos 7t, y = 7 sin t sin 7t, 8

< t <

3

.

(a) Find an expression for x

y

d

d in terms of t. You need not simplify your answer. (3)

(b) Find an equation of the normal to the curve at the point where t = 6

.

Give your answer in its simplest exact form. (6)

(C4 Jan 2007, Q3.) 4. A curve has parametric equations

x = tan2 t, y = sin t, 0 < t < 2

.

(a) Find an expression for x

y

d

d in terms of t. You need not simplify your answer. (3)

(b) Find an equation of the tangent to the curve at the point where t = 4

.

Give your answer in the form y = ax + b , where a and b are constants to be determined. (5)

(c) Find a cartesian equation of the curve in the form y2 = f(x). (4)

(C4 June 2007, Q6.)

5.

Figure 3

The curve C shown in Figure 3 has parametric equations

x = t 3 8t, y = t 2

where t is a parameter. Given that the point A has parameter t = 1,

(a) find the coordinates of A. (1)

The line l is the tangent to C at A.

(b) Show that an equation for l is 2x 5y 9 = 0. (5)

The line l also intersects the curve at the point B.

(c) Find the coordinates of B. (6)

(C4 Jan 2009, Q7.)

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6.

Figure 2

Figure 2 shows a sketch of the curve with parametric equations

x = 2 cos 2t, y = 6 sin t, 0 t 2

.

(a) Find the gradient of the curve at the point where t = 3

.

(4)

(b) Find a Cartesian equation of the curve in the form

y = f(x), k x k,

Stating the value of the constant k.

(4)

(c) Write down the range of f(x).

(2)

(C4 June 2009, Q5.) 7. A curve C has parametric equations

x = sin2 t, y = 2 tan t , 0 t < 2

.

(a) Find x

y

d

d in terms of t.

(4)

The tangent to C at the point where t = 3

cuts the x-axis at the point P.

(b) Find the x-coordinate of P.

(6)

(C4 June 2010, Q4.)

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8. The curve C has parametric equations

x = ln t, y = t2 2, t > 0.

Find

(a) An equation of the normal to C at the point where t = 3,

(6)

(b) A Cartesian equation of C.

(3)

(C4 Jan 2011, Q6.)

9.

Figure 3

Figure 3 shows part of the curve C with parametric equations

x = tan , y = sin , 0 < 2

.

The point P lies on C and has coordinates

3

2

1,3 .

(a) Find the value of at the point P. (2)

The line l is a normal to C at P. The normal cuts the x-axis at the point Q.

(b) Show that Q has coordinates (k3, 0), giving the value of the constant k.

(6)

(C4 June 2011, Q7.)

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10.

Figure 2

Figure 2 shows a sketch of the curve C with parametric equations

x = 4 sin

6

t , y = 3 cos 2t, 0 t < 2.

(a) Find an expression for x

y

d

d in terms of t. (3)

(b) Find the coordinates of all the points on C where x

y

d

d = 0. (5)

(C4 Jan 2012, Q5.)

11.

Figure 2

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Figure 2 shows a sketch of the curve C with parametric equations

x = 3 sin 2t, y = 4 cos2 t, 0 t .

(a) Show that x

y

d

d = k3 tan 2t, where k is a constant to be determined.

(5)

(b) Find an equation of the tangent to C at the point where t = 3

.

Give your answer in the form y = ax + b, where a and b are constants.

(4)

(c) Find a Cartesian equation of C.

(3)

(C4 June 2012, Q6.)

12.

Figure 2

Figure 2 shows a sketch of part of the curve C with parametric equations

x = 1 2

1t, y = 2t 1.

The curve crosses the y-axis at

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