parametric differentiation - solutions.pdf

8/17/2019 Parametric Differentiation - Solutions.pdf

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AQA Core 4 Parametric equations

1 of 5 18/03/13 © MEI

Section 2: Parametric differentiation

Solutions to Exercise

1. i)

3 2

d3 2 3

d

d

1 3

d

y

y

2

2

d d

3

d d d 3

y y

ii)

3 2 2

3 2 2

d

2 cos 2 3cos sin 6 sin cos

d

d3 sin 3 3 sin cos 9sin cos

d

yy

2

3

2

d d d

9sin cos

tan

d d d 6 sin cos

y y

iii)

2 3

2 3

1 d 2

2

d

d

1 1

d

y

y

3

2

d d d

1

d d d 2 /

y y

2. 3 2

2 2 1 2

d

2 3 2

d

d

5 5 10

d

y

y

2

2

d d d

10

d d d 3 2

y y

When t = 1,

d d d

10 1

9

d d d 3 2

y y

When t = 1,

3

2

1 2 1 1

1

5 1 6

1

y

Tangent has gradient 9 and passes through -1, 6)

Equation of tangent is 6 9 1))

6 9 9

9 15

y

y

y



AQA C4 Parametric equations 2 Exercise solutions

2 of 5 18/03/13 © MEI

Normal has gradient

1

9

and passes through -1, 6)

Equation of normal is

1

9

1))

9 6) 1)

9 54 1

9 53

y

y

y

y

3. When curve crosses x-axis, y = 0 so

2

1 0

1) 1) 0

1 or 1

2 3 2

2

d

1) 3 1

d

d

1 2

d

y

y

2

d d d

2

d d d 3 1

y y

When = 1,

1

2

d

2

d 3 1

y

When = 1,

2

1 1 1) 2

So the tangent has gradient

1

2

and passes through the point 2, 0)

Equation of tangent is

1

2

2)

2 2

y

y

When = -1,

1

2

d

2

d 3 1

y

When = -1,

2

1 1) 1) 2

So the tangent has gradient

1

2

and passes through the point -2, 0)


1

2

2))

2 2

2 2 0

y

y

y

4. 2

d

2

d

d

2 2

d

y

y

d d d

2 1

d d d 2

y y

Tangent has gradient

1

and passes through the point ,2 .




3 of 5 18/03/13 © MEI


2

2 2

2

1

2 )

2

y

y

y

At point A,

2 2

0 0

At point B,

2

0 y y

The coordinates of A are ,0

and the coordinates of B are

, .

5.

1 2

d

4 4

d

d

4 4

d

y

y

2

2

d d d 4 1

d d d 4

y y

At the point P 8, 2), t = 2

Gradient at P

1

4

.

Gradient of normal at P = 4

Normal has gradient 4 and passes through 8, 2)

Gradient of normal is 2 4 8)

2 4 32

4 30

y

y

y

To find the point where the normal meets the curve again, substitute the

parametric equations for x and y into the equation of the normal.

2

2

1

8

4 30

4

4 4 30

4 16 30

8 15 2 0

8 1) 2 ) 0

or 2

y

Since

2 at point P, then at Q

1

8

.

At Q,

1 1

8 2

1

8

4

4

32

y

The coordinates of Q are 2

, 32 .

6. i) At intersections with x-axis,

0 3sin 0 0 or

When 0, 2 cos0 2




4 of 5 18/03/13 © MEI

When 0, 2 cos 2

At intersections with y-axis,

3

2 2

2 cos 0 or

When

2 2

3sin 3

When

3 3

2 2

3sin 3

The points of intersection with the axes are 2, 0), -2, 0), 0, 3), 0, -3)

ii)

d

2 cos 2 sin

d

d

3 sin 3cos

d

y

y

d d d

3cos

d d d 2 sin

y y

Equation of tangent is cos

3sin 2 cos

2sin

y

Tangent passes through the point 4, 0)

3

3

sin

2

2 2

2 2

1

2

5

3 3

cos

4 2 cos

2 sin

2 sin cos 4 2 cos )

2 sin 4cos 2 cos

sin cos 2 cos

cos

or

When

3

,

1

3 2

1 3

3 2 2

2 cos 2 1

3 sin 3 3 3

y

When

5

3

,

5 1

3 2

5 1 3

3 2 2

2 cos 2 1

3sin 3 3 3

y

The points are 2

, 3

and 2

, 3

.

7. i)

d

2 sin ) 2 2 cos

d

d

2 1 cos ) 2 sin

d

y

y

d d d

2 sin sin

d d d 2 2 cos 1 cos

y y




5 of 5 18/03/13 © MEI

ii) When

1

2

, gradient

2

2

sin 1

1

1 cos 1 0

Gradient of normal = -1

When

1

2

,

2

1) 2

2 1 0) 2

y

Equation of normal is

2 1 2))

2 2

y

y

y

When normal meets x-axis, 0 .

Coordinates of A are

0 .

iii)

Gradient of tangent at P is 1

Tangent at P is 2 1 2))

2 2

4

y

y

y

At point of intersection with

4 4

so the point of intersection is

4 .

parametric differentiation - solutions.pdf

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