City of London Academy 1 C4: QUESTIONS FROM PAST PAPERS - PARAMETRICS 1. The diagram above shows a sketch of the curve C with parametric equations x = 5t 2 – 4, y = t(9 – t 2 ) The curve C cuts the x-axis at the points A and B. (a) Find the x-coordinate at the point A and the x-coordinate at the point B. (3) The region R, as shown shaded in the diagram above, is enclosed by the loop of the curve. (b) Use integration to find the area of R. (6) (Total 9 marks) 2. The diagram above shows a sketch of the curve with parametric equations
37
Embed
C4: QUESTIONS FROM PAST PAPERS - … of London Academy 1 C4: QUESTIONS FROM PAST PAPERS - PARAMETRICS 1. The diagram above shows a sketch of the curve C with parametric equations x
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
City of London Academy 1
C4: QUESTIONS FROM PAST PAPERS - PARAMETRICS
1.
The diagram above shows a sketch of the curve C with parametric equations
x = 5t2 – 4, y = t(9 – t
2)
The curve C cuts the x-axis at the points A and B.
(a) Find the x-coordinate at the point A and the x-coordinate at the point B. (3)
The region R, as shown shaded in the diagram above, is enclosed by the loop of the curve.
(b) Use integration to find the area of R. (6) (Total 9 marks)
2.
The diagram above shows a sketch of the curve with parametric equations
City of London Academy 2
x = 2 cos 2t, y = 6sint,
(a) Find the gradient of the curve at the point where .
(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. (a) Using the identity cos2θ = 1 – 2sin2θ, find
(2)
The diagram above shows part of the curve C with parametric equations
x = tanθ, y = 2sin2θ, 0 ≤ θ <
The finite shaded region S shown in the diagram is bounded by C, the line x = and the x-axis.
This shaded region is rotated through 2π radians about the x-axis to form a solid of revolution.
(b) Show that the volume of the solid of revolution formed is given by the integral
where k is a constant. (5)
(c) Hence find the exact value for this volume, giving your answer in the form pπ2 + qπ √3,
where p and q are constants. (3)
(Total 10 marks)
20
t
3
t
.dsin 2
2
3
1
6
0
2 dsin
k
City of London Academy 3
4.
The curve C shown above has parametric equations
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)
(Total 12 marks)
5.
The diagram above shows the curve C with parametric equations
.
23 ,8 tyttx
xO
y
P
l
C
4
R
20,2sin4,cos8
ttytx
City of London Academy 4
The point P lies on C and has coordinates (4, 2√3).
(a) Find the value of t at the point P. (2)
The line l is a normal to C at P.
(b) Show that an equation for l is y = –x√3 + 6√3. (6)
The finite region R is enclosed by the curve C, the x-axis and the line x = 4, as shown shaded in the
diagram above.
(c) Show that the area of R is given by the integral .
(4)
(d) Use this integral to find the area of R, giving your answer in the form a + b√3, where a and
b are constants to be determined. (4)
(Total 16 marks)
6.
The curve C has parametric equations
The finite region R between the curve C and the x-axis, bounded by the lines with equations
x = 1n 2 and x = 1n 4, is shown shaded in the diagram above.
(a) Show that the area of R is given by the integral
.
(4)
(b) Hence find an exact value for this area. (6)
(c) Find a cartesian equation of the curve C, in the form y = f(x). (4)
2
3
2 dcossin64
ttt
y
O ln 2 ln 4 x
C
R
1,)1(
1),2(ln
t
tytx
ttt
d)2)(1(
12
0
City of London Academy 5
(d) State the domain of values for x for this curve. (1)
(Total 15 marks)
7. A curve has parametric equations
(a) Find an expression for 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
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)
(Total 12 marks)
8.
The curve shown in the figure above has parametric equations
(a) Find an equation of the tangent to the curve at the point where .
(6)
(b) Show that a cartesian equation of the curve is
(3)
(Total 9 marks)
.2
0,sin,tan2 ttytx
x
y
d
d
.4
t
–1 –0.5 O 0.5 1
0.5
.),(sin,sin226πππ ttytx
6t
11),1(2
1
2
3 2
xxxy
City of London Academy 6
9.
The curve shown in the figure above has parametric equations
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)
10.
The curve shown in the figure above has parametric equations
x = t – 2 sin t, y = 1 – 2cos t, 0 t 2
(a) Show that the curve crosses the x-axis where
(2)
The finite region R is enclosed by the curve and the x-axis, as shown shaded in the figure above.
x
y
a
O
B
12
a
A
.6
0,sin,3cos
ttaytax
x
y
O
R
.3
5and
3
tt
City of London Academy 7
(b) Show that the area of R is given by the integral
(3)
(c) Use this integral to find the exact value of the shaded area. (7)
(Total 12 marks)
11.
The curve C has parametric equations
x = , y = , t < 1.
(a) Find an equation for the tangent to C at the point where t = .
(7)
(b) Show that C satisfies the cartesian equation y = .
(3)
The finite region between the curve C and the x-axis, bounded by the lines with equations x =
and x = 1, is shown shaded in the figure above.
(c) Calculate the exact value of the area of this region, giving your answer in the form
a + b ln c, where a, b and c are constants. (6)
(Total 16 marks)
12. A curve has parametric equations
x = 2 cot t, y = 2 sin2 t, 0 < t .
(a) Find an expression for in terms of the parameter t.
3
5
3
.dcos212
tt
x
y
O
C
123
t1
1
t1
1
21
12 x
x
32
2
x
y
d
d
City of London Academy 8
(4)
(b) Find an equation of the tangent to the curve at the point where t = .
(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)
13.
The diagram above shows a sketch of the curve C with parametric equations
x = 3t sin t, y = 2 sec t, 0 t < .
The point P(a, 4) lies on C.
(a) Find the exact value of a. (3)
The region R is enclosed by C, the axes and the line x = a as shown in the diagram above.
(b) Show that the area of R is given by
6 (tan t + t) dt.
(4)
(c) Find the exact value of the area of R. (4)
(Total 11 marks)
4
O
y
a x
C
P
R
2
3
0
City of London Academy 9
14.
The diagram above shows a cross-section R of a dam. The line AC is the vertical face of the dam,
AB is the horizontal base and the curve BC is the profile. Taking x and y to be the horizontal and
vertical axes, then A, B and C have coordinates (0, 0), (3 2, 0) and (0, 30) respectively. The area
of the cross-section is to be calculated.
Initially the profile BC is approximated by a straight line.
(a) Find an estimate for the area of the cross-section R using this approximation. (1)
The profile BC is actually described by the parametric equations.
x = 16t2 –
2, y = 30 sin 2t, t .
(b) Find the exact area of the cross-section R. (7)
(c) Calculate the percentage error in the estimate of the area of the cross-section R that you
found in part (a). (2)
(Total 10 marks)
15. The curve C is described by the parametric equations
x = 3 cos t, y = cos 2t, 0 t .
(a) Find a cartesian equation of the curve C. (2)
(b) Draw a sketch of the curve C. (2)
(Total 4 marks)
16. The cartesian equation of the circle C is
x2 + y
2 – 8x – 6y + 16 = 0.
(a) Find the coordinates of the centre of C and the radius of C.
R
A B
C
x (metres)
y (metres)
4
2
City of London Academy 10
(4)
(b) Sketch C. (2)
(c) Find parametric equations for C. (3)
(d) Find, in cartesian form, an equation for each tangent to C which passes through the
origin O. (5)
(Total 14 marks)
17.
The curve shown in the diagram above has parametric equations
x = cos t, y = sin 2t, 0 t < 2.
(a) Find an expression for in terms of the parameter t.
(3)
(b) Find the values of the parameter t at the points where = 0.
(3)
(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 < is
y = 2x(1 – x2).
(3)
(e) Write down a cartesian equation for the part of the curve where t < 2. (1)
(Total 12 marks)
1
1–1
–1
x
y
d
d
x
y
d
d
City of London Academy 11
MARK SCHEME
1. (a)
t = 0, 3,–3 Any one correct value B1
At t = 0, x = 5 (0)2 –4 = –4 Method for finding
one value of x M1
At t = 3, x = 5 (3)2 – 4 = 41
At A, x = –4; at B, x = 41 Both A13
(b) Seen or implied B1
M1 A1
M1
A = 2∫y dx=648 (units2) A16
[9]
2. (a) = –4sin 2t, 6 cos t B1, B1
= – M1
At t = , accept equivalents, awrt –0.87 A14
Alternatives to (a) where the parameter is eliminated
B1
03–3–90 2 ttttty
414–3–5,3–At2
xt
tt
x10
d
d
dt10–9dtd
dd 2 ttt
t
xyxy
ttt d10–90 22
32432–3305
10–
3
90 53
0
353
tt
t
x
d
d
t
y
d
d
x
y
d
d
tt
t
sin4
3–
2sin4
cos6
3
2
3–
4
3–
2
3
m
1 2
1
)9–18( xy
)9(–)9–18(2
1
d
d2
1–
xx
y
City of London Academy 12
At B1
M1 A14
y2 = 18 – 9x
B1
At B1
M1 A14
(b) Use of cos2t = 1 – 2sin2t M1
M1
Leading to cao A1
– 2 ≤ x ≤ 2 k = 2 B14
(c) 0 ≤ f(x) ≤ 6 either 0 ≤ f(x) or f(x) ≤ 6 B1
Fully correct. Accept 0 ≤ y ≤ 6, [0, 6] B12 [10]
3. (a) M1 A12
(b)
M1 A1
M1
1–3
2cos,
3
xt
2
3–9–
)27(
1
2
1
d
d
x
y
2
9–d
d2
x
yy
333
sin6,3
yt
2
3–
332
9–
dx
dy
6sin,
22cos
yt
xt
2
62–1
2
yx
))–2(3()9–18( xxy
)(2sin4
1–
2
1d)2cos–1(
2
1dsin 2 C
2secd
dtan
xx
dsec)2sin2(dd
dd 2222 x
yxy
d
cos
)cossin22(2
2
City of London Academy 13
k = 16π A1
x = 0 tanθ = 0 θ = 0, B15
(c) M1
Use of correct limits *M1
A13
[10]
4. (a) At A, x = – 1 + 8 = 7 & y = (–1)2 = 1 A(7,1) A(7, 1) B11
(b) x = t3 – 8t, y = t
2,
Their divided by their M1
Correct A1
At A, m(T) = 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 = or finds c and uses dM1
y = (their gradient)x + “c” .
Hence T:
gives T: 2x – 5y – 9 = 0 AG A1 cso5
(c) 2(t3
– 8t) – 5t2 – 9 = 0 Substitution of both x = t
3 – 8t and
dsin16 2
3
1x
63
1tan
dsin16 6
0
2V
6
04
2sin–
2
116
V
)0–0(–
3sin
4
1–
1216
32–3
4
8
3–
1216 2
2–,
3
4 qp
tt
yt
t
x2
d
d,8–3
d
d 2
8–3
2
d
d2t
t
x
y t
y
d
d
tx
dd
x
y
d
d
5
2
3–
2–
8–3
2–
8–)1(–3
)1(–22
59
514
52 ––1)7( cc
59
52 –xy
City of London Academy 14
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 = at B} A1
x =
Candidate uses their value of t
to find either
the x or y coordinate ddM1
One of either x or y correct. A1
Both x and y correct. A16
Hence awrt
[12]
5. (a) At P(4, ) either 4 = 8cost or
only solution is where 0„ t„
4 = 8cost or M1
or awrt 1.05 (radians) only stated in the range 0„ t„ A1 2
(b) x = 8 cos t, y = 4 sin 2t
At P,
Hence m(N) =
N:
N: AG
or
29
29t
55.1awrt or 55.12536–8–8
4418
729
2
92
2
9
20.3awrt or 20.254812
29 y
481
8441,B
32 t2sin432
3
t
2
t2sin432
3
t
2
tt
yt
t
x2cos8
d
d,sin8
d
d
3
3
2
sin8
cos8
d
d
x
y
0.58awrt
3
1
)8(
8
2
3
21
3
1
1or 3
)4(332 xy
363 xy
363432)4(332 cc
City of London Academy 15
so N:
Attempt to differentiate both x and y wrt t to give ±p sin t and
±q cos 2t respectively M1
Correct A1
Divides in correct way round and attempts to substitute their value of
t (in degrees or radians) into their expression. M1*
You may need to check candidate‟s substitutions for M1*
Note the next two method marks are dependent on M1*
Uses m(N) = dM1*
Uses = (their mN)(x – 4) or finds c using x = 4 and
y = and uses y = (their mN)x + “c”. dM1*
A1 cso AG 6
(c)
attempt at A = M1
correct expression (ignore limits and dt) A1
Seeing sin 2t = 2 sin t cos t anywhere in this part. M1
Correct proof. Appreciation of how the negative sign affects
the limits. A1 AG 4
Note that the answer is given in the question.
363 xy
t
y
t
x
d
d and
d
d
x
y
d
d
)m(their
1
T
32y
32
363 xy
3
2
4
0
d)sin8.(2sin4d
tttxyA
tttA
tttA
tttttttA
dcossin.64
dcossin.64
dsin).cossin2(32dsin.2sin32
2
3
2
3
2
2
3
2
3
2
tt
xy d
d
d
City of London Academy 16
(d) {Using substitution u = sin t = cost}
{change limits:
when t = , u = & when t = , u = 1}
(Note that a = , b = –8)
k sin3 t or ku
3 with u = sin t M1
Correct integration ignoring limits. A1
Substitutes limits of either (t = and t = ) or
(u = 1 and u = ) and subtracts the correct way round. dM1
A1 aef isw 4
Aef in the form a + , with awrt 21.3 and anything that