Download Implicit Differentiation and Parametric Equations

Implicit Differentiation and Parametric
Equations Answers
1.
Differentiates w.r.t. x to give
3x2,  2x
dy
dy
+ 2y, 4 + 3y2
=0
dx
dx
M1, B1, A1
At (4, 3)
48  (8y + 6)  4 + 27y = 0
M1
38
= 2
19
A1
 y = 
Gradient of normal is
y  3 =
i.e.
1
2
M1
1
(x  4)
2
M1
2y  6 = x  4
x – 2y + 2 = 0
A1
8
[8]
2.
dy
 dy

 2y  6y
2x +  2 x
=0
dx
 dx

dy
= 0  x + y = 0 (or equivalent)
dx
Eliminating either variable and solving for at least one value of x or y.
y2 – 2y2 – 3y2 + 16 = 0 or the same equation in x
y = 2 or x = 2
(2 – 2), (–2, 2)
dy x  y
Note:

dx 3 y  x
M1 (A1) A1
M1
M1
A1
A1
7
Alternative:
3y2 – 2xy – (x2 + 16) = 0
y=
2 x  (16 x 2  192 )
6
dy 1 1
8x
  .
dx 3 3 (16 x 2  192 )
dy
=0
dx
8x
(16 x 2  192 )
64x2 = 16x2 + 192
x = 2
(2, –2), (–2, 2)
M1 A1  A1
= 1
M1
M1 A1
A1
7
[7]
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3.
(a)
(b)
dy
dy
), –6y
=0
dx
dx
dy
dy
At (1 , 2) 10 + (4 + 2 ) – 12
=0
dx
dx
7
dy 14

or 1 52
 = 1.4 or
5
dx 10
10x, +(2y + 2x
The gradient of the normal is 
M1, (B1), A1
M1
A1
5
7
M1
5
(x – 1)
7
(allow tangent)
Its equation is y – 2 = 
y= 
5
M1
5
5
5
19
x + 2 or y =  x +
7
7
7
7
A1cao
3
[8]
4.
(a)
dx
=  sin t,
dt
(b)
2 cos 2t = 0
dy
= 2 cos 2t
dt
2 cos 2t
dy
=
 sin t
dx
M1 A1 A1
3
2t =
 3 5 7
,
,
,
2
2
2 2
M1
So t =
 3 5 7
,
,
,
4
4
4 4
A1 A1
3

 1
  1
 
,  1) 
,1)   
2
2  


M1 A1
2
(c)
 1
  1


,1)  
,  1) 

 2   2
(d)
y = 2 sin t cos t
= 2 1  cos t cos t = 2x 1  x
2
(e)

M1
2
y = 2x 1  x 2
M1 A1
3
B1
1
[12]
5.
(a)
3( x  1)
A
B
, and correct method for finding A or B


( x  2)( x  1) x  2 x  1
A = 1, B = 2
(b)
f(x) = 
M1
A1, A1
1
2

2
( x  2)
( x  1) 2
Argument for negative, including statement that square terms
are positive for all values of x. (f.t. on wrong values of A and B)
3
M1 A1
A1 ft
3
[6]
6.
(a)
(b)
dy
dx
= –2cosec2 t,
= 4 sin t cos t
dt
dt
dy  2 sin t cos t
(= –2sin3 t cos t)

2
dx
cosec t
At t =

, x = 2, y = 1
4
both M1 A1
M1 A1
4
B1
both x and y
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Substitutes t =
dy

into an attempt at
to obtain gradient
4
dx
1
(x – 2)
2
Accept x + 2y = 4 or any correct equivalent
Equation of tangent is y – 1 = –
(c)
 1
 
 2
M1
M1 A1
Uses 1 + cot2 t = cosec2 t, or equivalent, to eliminate t
4
M1
2
2
 x
1+   
y
 2
correctly eliminates t
A1
8
4  x2
The domain ir x…0
y=
cao A1
B1
4
Alternative for (c):
1
1
x
x  y 2
 y 2
sin t =   ; cos t = sin t   
2
22
2
2
y x
y
sin2 t + cos2 t = 1  
 =1
2
4 2
8
Leading to y =
4  x2
M1 A1
A1
[12]
7.
(a)
dx
1
dy
1
and


2
dt
dt (1  t ) 2
(1  t )

dy  (1  t ) 2
and at t = ½, gradient is –9

dx
(1  t ) 2
M1 requires their dy/dt / their dx/dt and
substitution of t.
2
and y = 2
3
2
Equation is y – 2 = -9 (x – )
3
B1, B1
M1 A1cao
At the point of contact x =
(b)
B1
M1 A1
1
1
– 1 or t = 1 –
(or both)
x
x
Then substitute into other expression y = f(x) or x = g(y) and rearrange
1
1
(or put  1  1  and rearrange)
x
y
Either obtain t in terms of x and y i,e, t =
To obtain y =
x
(*)
2x  1
1
(1  t )
x
Or Substitute into

2
2x  1
1
1 t
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M1
M1
a1
3
M1
3
1
1

2  (1  t ) 1  t
= y (*)
1
x
Area = 
dx
2x  1
2
=
(c)
A1
M1
3
B1
3
=

u  1 du 1
1

1  du
2u 2 4
u
putting into a form to integrate

M1
1
1
1

=  u  ln u 
4
4
1
M1 A1
3
1  1 1 1
=    ln 
4  12 4 3 
1 1
=  ln 3 or any correct equivalent.
6 4
M1
A1
6
[16]
1
Or Area =
x
 2x  1 dx
B1
2
3
=

1
1
 2 dx
2 2x  1
putting into a form to integrate
M1
1
1
1

=  x  ln( 2 x  1)
4
2
2
M1 A1
3
=
1 1 1 1 1 1 1
  ln   ln
2 3 4 3 6 4 3
Or Area =
=
1
1
 1  t (1  t )
A
B
2
dM1 A1
dt
C
 (1  t )  (1  t )  (1  t )
2
6
B1
dt
M1
putting into a form to integrate
1
1
1

=  ln(1  t )  ln(1  t )  (1  t ) 1 
4
2
4

= Using limits 0 and ½ and subtracting (either way round)
1 1
=  ln3 or any correct equivalent.
6 4
1
Or Area =
x
 2x  1 dx then use parts
M1 a1ft
dM1
A1
6
B1
2
3
1
=
1
1
x ln( 2 x  1) 
(2 x  1) dx
2
2
2

M1
3
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1
1 
1
x ln( 2 x  1)   (2 x  1) ln( 2 x  1)  x
2
2 
4
1 1 1 1 1 1
=   ln  ln  
2  3 3 12 3 3 
1 1 1
=  ln
6 4 3
=
8.
(a)
dy y 3
 
dx x 2t
DM1
A1
6
M1
2t
3
Gradient of normal is 
M1
At P t = 2
B1
4
Gradient of normal @ P is 
3
A1
Equation of normal @ P is y –9 = 
(b)
M1A1
4
(x – 5)
3
M1
Q is where y = 0  x =
27
47
(o.e.)
5
4
4
A1
Curved area = ydx = y
dx
dt
dt
M1
= 3(1 + b).2t dt
= [3t2 + 2t3]
Curve cuts x-axis when t = –1
Curved area = [3t2 + 2t3] 21 = (12 + 16) – (3 – 2) (= 27)
6
A1
M1A1
B1
M1
P
1
((a) – 5) × 9 (= 30.375)
2
Total area of R = curved area + 
= 57.375 or AWRT 57.4
Area of
Q
triangle =
M1
M1
A1
9
[15]
9.
(a)
dx
 a sec t tan t
dt
B1
dy
 b sec2 t
dt
B1
dy
b sec 2 t
b

 cosec t
dx a sec t tan t a
(b)
At t 

x  a 2 , y  b;
4
 y  b  b
2
a
M1 A1
dy b 2

dx
a
4
B1B1
x  a 2 
M1 A1
4
b 2x
y
b
a
[8]
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