# C4 Differentiation - Tangents and normals C4 Differentiation - Tangents and normals...

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C4 Differentiation - Tangents and normals PhysicsAndMathsTutor.com

1. A curve C has parametric equations

x = sin2 t, y = 2 tan t, 0 2 π

C4 Differentiation - Tangents and normals PhysicsAndMathsTutor.com

3. The curve C has the equation ye–2x = 2x + y2.

(a) Find x y

d d in terms of x and y.

(5)

The point P on C has coordinates (0, 1).

(b) Find the equation of the normal to C at P, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(4) (Total 9 marks)

4.

The curve C shown above has parametric equations

23 ,8 tyttx =−=

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)

Edexcel Internal Review 2

C4 Differentiation - Tangents and normals PhysicsAndMathsTutor.com

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

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

(Total 12 marks)

5. A curve C is described by the equation

3x2 – 2y2 + 2x – 3y + 5 = 0.

Find an equation of the normal to C at the point (0, 1), giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(Total 7 marks)

6.

–1 –0.5 O 0.5 1

0.5

The curve shown in the figure above has parametric equations

.),(sin,sin 226 πππ ttytx

C4 Differentiation - Tangents and normals PhysicsAndMathsTutor.com

7.

x

y

a

O B

1 2

a

A

The curve shown in the figure above has parametric equations

. 6

0,sin,3cos π≤≤== ttaytax

The curve meets the axes at points A and B as shown.

The straight line shown is part of the tangent to the curve at the point A.

Find, in terms of a,

(a) an equation of the tangent at A, (6)

(b) an exact value for the area of the finite region between the curve, the tangent at A and the x-axis, shown shaded in the figure above.

(9) (Total 15 marks)

8. A curve C is described by the equation

3x2 + 4y2 – 2x + 6xy – 5 = 0.

Find an equation of the tangent to C at the point (1, –2), giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(Total 7 marks)

Edexcel Internal Review 4

C4 Differentiation - Tangents and normals PhysicsAndMathsTutor.com

9. A curve has parametric equations

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

.

(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 π

.

(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) (Total 12 marks)

10. The curve C with equation y = k + ln 2x, where k is a constant, crosses the x-axis at the point

A

0,

e2 1 .

(a) Show that k = 1. (2)

(b) Show that an equation of the tangent to C at A is y = 2ex – 1. (4)

(c) Complete the table below, giving your answers to 3 significant figures.

x 1 1.5 2 2.5 3

1 + ln 2x 2.10 2.61 2.79 (2)

Edexcel Internal Review 5

C4 Differentiation - Tangents and normals PhysicsAndMathsTutor.com

(d) Use the trapezium rule, with four equal intervals, to estimate the value of

⌡ ⌠ +

3

1

d)2ln1( xx .

(4) (Total 12 marks)

11. f(x) = x + 5

e x , x ∈ .

(a) Find f ′(x). (2)

The curve C, with equation y = f(x), crosses the y-axis at the point A.

(b) Find an equation for the tangent to C at A. (3)

(c) Complete the table, giving the values of

+

5 e xx to 2 decimal places.

x 0 0.5 1 1.5 2

+

5 e xx

0.45 0.91

(2)

(d) Use the trapezium rule, with all the values from your table, to find an approximation for the value of

xx x

d 5

e 2

0 ⌡

⌠

+

. (4)

(Total 11 marks)

Edexcel Internal Review 6

C4 Differentiation - Tangents and normals PhysicsAndMathsTutor.com

12. The curve C has equation 5x2 + 2xy – 3y2 + 3 = 0. The point P on the curve C has coordinates (1, 2).

(a) Find the gradient of the curve at P. (5)

(b) Find the equation of the normal to the curve C at P, in the form y = ax + b, where a and b are constants.

(3) (Total 8 marks)

13.

y

x

M

O

P

N

R

The curve C with equation y = 2ex + 5 meets the y-axis at the point M, as shown in the diagram above.

(a) Find the equation of the normal to C at M in the form ax + by = c, where a, b and c are integers.

(4)

This normal to C at M crosses the x-axis at the point N(n, 0).

(b) Show that n = 14. (1)

Edexcel Internal Review 7

C4 Differentiation - Tangents and normals PhysicsAndMathsTutor.com

The point P(ln 4, 13) lies on C. The finite region R is bounded by C, the axes and the line PN, as shown in the diagram above.

(c) Find the area of R, giving your answer in the form p + q ln 2, where p and q are integers to be found.

(7) (Total 12 marks)

14. A curve has equation

x3 − 2xy − 4x + y3 − 51 = 0.

Find an equation of the normal to the curve at the point (4, 3), giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(Total 8 marks)

Edexcel Internal Review 8

C4 Differentiation - Tangents and normals PhysicsAndMathsTutor.com

1. (a) 2 d d2sin cos , 2sec d d x yt t t t t = = B1 B1

2

3

d sec 1 d sin cos sin cos

y t x t t t t

= =

or equivalent M1 M1 4

(b) At 3

t π= , 3 4

x = , 2 3y √= B1

2secd 163 d 3sin cos

3 3

y x

π

π π √ = = M1 A1

16 32 3

3 4 y x√

√

− = −

M1

30 8

y x= ⇒ = M1 A1 6

[10]

2. (a) –2 sin 2x – 3 sin 3y x y

d d =0 M1 A1

y x

x y

3sin3 2sin2–

d d

= Accept y

x 3sin3–

2sin2 ,

y x

3sin3 2sin2– A1 3

(b) At , 6 π

=x 13cos 6

2cos =+

yπ M1

2 13cos =y A1

93

3 ππ =⇒= yy awrt 0.349 A1 3

(c) At

9 ,

6 ππ ,

( ) ( ) 3

2 sin3 sin2

3sin3 2sin2

– d d

3

3

9

6 −=−== π

π

π

π

x y M1

=

6 –

3 2–

9 – ππ xy M1

Leading to 6x + 9y–2π = 0 A1 3 [9]

Edexcel Internal Review 9

C4 Differentiation - Tangents and normals PhysicsAndMathsTutor.com

3. (a) e–2x x yyy

x y x

d d22e2–

d d 2– += A1 correct RHS *M1 A1

xxx y x

y x

2–2–2– e2– d de)e(

d d

= B1

(e–2x – 2y) xy x y 2–e22

d d

+= *M1

y

y x y

x

x

2–e e22

d d

2–

–2+ = A1 5

(b) At P, 4– 2–ºe ºe22

d d

= +

= x y M1

Using mm′ = –1

4 1'=m M1

)0–( 4 11– xy = M1

x – 4y + 4 = 0 or any integer multiple A1 4

Alternative for (a) differentiating implicitly with respect to y.

e–2x – 2ye–2x y y x

y x 2

d d2

d d

+= A1 correct RHS *M1 A1

y xyy

y xxx

d de2–e)e(

d d 2–2–2– = B1

(2 + 2ye–2x) = y x

d d e–2x – 2y *M1

x x

y y

y x

2–

2–

e22 2–e

d d

+ =

y

y x y

x

x

2–e e22

d d

2–

2–+ = A1 5

[9]

4. (a) At A, x = – 1 + 8 = 7 & y = (–1)2 = 1 ⇒ A(7,1) A(7, 1) B1 1

(b) x = t3 – 8t, y = t2,

t t yt

t x 2

d d,8–3

d d 2 ==

Edexcel Internal Review 10

C4 Differentiation - Tangents and normals PhysicsAndMathsTutor.com

8–3

2 d d

2t t

x y =∴ Their t

y d d divided by their t

x d d M1

Correct x y

d d A1

At A, m(T) = 5 2

3– 2–

8–3 2–

8–)1(–3 )1(–2

2 === Substitutes

for t to give any of the four underlined oe:

T: y – (their 1) = mr(x – (their 7)) Finding an equation of a tangent with their point and their tangent gradient or 1 = 5

9 5

14 5 2 –

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