C4 Differentiation - Parametric differentiation .

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Transcript of C4 Differentiation - Parametric differentiation .

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    1. A curve C has parametric equations

    x = sin2 t, y = 2 tan t, 0 2 π

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

    =t .

    (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)

    (Total 10 marks)

    3.

    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

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

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

    (Total 12 marks)

    4. A curve has parametric equations

    x = 7cos t – cos7t, y = 7 sin t – sin 7t, 38 ππ

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    11),1( 2 1

    2 3 2

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    7. 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)

    8.

    y

    xO

    C

    P

    Q

    R

    Edexcel Internal Review 5

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    The diagram shows a sketch of part of the curve C with parametric equations

    x = t2 + 1, y = 3(1 + t).

    The normal to C at the point P(5, 9) cuts the x-axis at the point Q, as shown in the diagram.

    (a) Find the x-coordinate of Q. (6)

    (b) Find the area of the finite region R bounded by C, the line PQ and the x-axis. (9)

    (Total 15 marks)

    9. (a) Use the identity for cos (A + B) to prove that cos 2A = 2 cos2 A – 1. (2)

    (b) Use the substitution x = 2√2 sin θ to prove that

    ⌡ ⌠ −

    6

    2

    2 d)8( xx = 31 (π + 3√3 – 6).

    (7)

    A curve is given by the parametric equations

    x = sec θ , y = ln(1 + cos 2θ ), 0 ≤ θ < 2 π

    .

    (c) Find an equation of the tangent to the curve at the point where θ = 3 π

    .

    (5) (Total 14 marks)

    10. The curve C has parametric equations

    x = a sec t , y = b tan t, 0 < t < 2 π ,

    where a and b are positive constants.

    Edexcel Internal Review 6

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    (a) Prove that x y

    d d =

    a b cosec t.

    (4)

    (b) Find the equation in the form y = px + q of the tangent to C at the point where t = 4 π .

    (4) (Total 8 marks)

    11.

    1

    1–1

    –1

    The curve shown in the diagram above has parametric equations

    x = cos t, y = sin 2t, 0 ≤ t < 2π.

    (a) Find an expression for x y

    d d

    in terms of the parameter t.

    (3)

    (b) Find the values of the parameter t at the points where x y

    d d

    = 0.

    (3)

    Edexcel Internal Review 7

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    (c) Hence give the exact values of the coordinates of the points on the curve where the tangents are parallel to the x-axis.

    (2)

    (d) Show that a cartesian equation for the part of the curve where 0 ≤ t < π is

    y = 2x√(1 – x2). (3)

    (e) Write down a cartesian equation for the part of the curve where π ≤ t < 2π. (1)

    (Total 12 marks)

    Edexcel Internal Review 8

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    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

    2. (a) t x

    d d = –4sin 2t,

    t y

    d d 6 cos t B1, B1

    x y

    d d = – 

     

      =

    tt t

    sin4 3–

    2sin4 cos6 M1

    At t = 3 π ,

    2 3–

    4 3–

    2 3 =

    × =m accept equivalents, awrt –0.87 A1 4

    Alternatives to (a) where the parameter is eliminated

    1 2 1

    )9–18( xy =

    )9(–)9–18( 2 1

    d d 2

    1– ×= x

    x y B1

    At 1– 3

    2cos, 3

    === ππ xt B1

    2 3–9–

    )27( 1

    2 1

    d d

    =××= x y M1 A1 4

    2 y2 = 18 – 9x

    9– d d2 =

    x yy B1

    At 33 3

    sin6, 3

    === ππ yt B1

    Edexcel Internal Review 9

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    2 3–

    332 9– =

    × =

    dx dy M1 A1 4

    (b) Use of cos2t = 1 – 2sin2t M1

    6

    sin, 2

    2cos ytxt ==

    2

    6 2–1

    2   

      =

    yx M1

    Leading to ))–2(3()9–18( xxy == cao A1

    – 2 ≤ x ≤ 2 k = 2 B1 4

    (c) 0 ≤ f(x) ≤ 6 either 0 ≤ f(x) or f(x) ≤ 6 B1

    Fully correct. Accept 0 ≤ y ≤ 6, [0, 6] B1 2 [10]

    3. (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 ==

    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 ––1)7( ==⇒+ cc or finds c and uses dM1

    y = (their gradient)x + “c” .

    Hence T: 5 9

    5 2 –xy =

    gives T: 2x – 5y – 9 = 0 AG A1 cso 5

    Edexcel Internal Review 10

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    (c) 2(t3 – 8t) – 5t2 – 9 = 0 Substitution of both x = t3 – 8t and y = t2 into T M1

    2t3 – 5t2 – 16t – 9 = 0

    (t + 1){(2t2 – 7t – 9) = 0}

    (t + 1){(t + 1)(2t – 9) = 0} A realisation that (t + 1)is a factor. dM1

    {t = –1(at A) t = 2 9 at B} 2

    9=t A1

    x = ( ) ( ) 55.1awrt or 55.12536–8– 8441872929 2

    2 9 ===

    Candidate uses their value of t to find either the x or y coordinate ddM1

    ( ) 20.3awrt or 20.25481229 ===y One of either x or y correct. A1 Both x and y correct. A1 6

    Hence ( )4818441 ,B awrt [12]

    4. (a) x = 7cos t – cos 7t, y = 7 sin t – sin 7t,

    tt tt

    x y

    tt t ytt

    t x

    7sin7sin7 7cos7cos7

    d d

    7cos7cos7 d d,7sin7sin7

    d d

    +− −

    =∴

    −=+−=

    Attempt to differentiate x and y with respect to t to give

    t x

    d d in the form ± A sin t ± B sin 7t

    t y

    d d

    in the form ± C cos t ± D cos 7t M1

    Correct t y

    t x

    d d and

    d d A1

    Candidate’s t x t y

    d d d d

    B1ft 3

    Edexcel Internal Review 11

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    Aliter Way 2

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

    t tt tt

    tt tt

    x y

    tt t ytt

    t x

    4tan )3sin4cos2(7 )3sin4sin2(7

    7sin7sin7 7cos7cos7

    d d

    7cos7cos7 d d,7sin7sin7

    d d

    = −

    −− =

    +− −

    =

    −=+−=

    Attempt to differentiate x and y with respect to t to give

    t x

    d d in the form ± A sin t ± B sin 7t

    t x

    d d in the form ± C cos t ± D cos 7t M1

    Correct t y

    t x

    d d and

    d d A1

    Candidate’s t x t y

    d d d d

    B1ft 3

    (b) When t = 6

    7 6

    6 7

    6

    sin7sin7 cos7cos7

    d d)(m,

    6 ππ πππ

    +−

    − ==

    x yT ;

    ( ) 1.73awrt3

    7 37

    2 7

    2 7

    2 37

    2 37

    −=−= −

    = −−

    −− =

    Hence m(N) = 3

    1or 3

    1 −

    − = awrt 0.58

    When t = 6 π ,

    4 2 8

    2 1