NCEA Level 3 Calculus (91578) 2016 — page 1 of 6
Assessment Schedule – 2016
Calculus: Apply differentiation methods in solving problems (91578)
Evidence Statement
Q1
Expected Coverage
Achievement
(u)
(a)
dy
= 1+ x -2 - 2x -3
dx
Correct solution
(b)
dh 3.2π
πö
æ 4π
=
cos ç t + ÷
è 25
dt
25
2ø
æ 36π π ö
= 0.402 cos ç
+
è 25 2 ÷ø
= 0.395 metres per hour
Correct solution
with correct
derivative
(c)
dx
= -4 sin 2t
dt
dy
= 2 tant sec 2 t
dt
dy 2 tant sec2 t
=
dx
-4 sin 2t
2 tant
=
-4 sin 2t cos 2 t
2
π dy
=
At t = ,
2
dx
4
æ 1 ö
-4 ´ ç
è 2 ÷ø
2
=
= -1
-2
(d)
1
( x - 2 )2
4
dy 1
= ( x - 2)
dx 2
y=
dy 1
= ( 6 - 2) = 2
dx 2
dy -1
 At P
=
dx 2
-1 1
= ( x - 2)
2 2
-1 = x - 2
x =1
At Q (6, 4)
dx
or
dt
Correct solution
with correct
derivatives.
dy -1
=
dx 2
Correct solution
with correct
derivative.
Correct
dy
dt
At P
Merit
(r)
Excellence
(t)
NCEA Level 3 Calculus (91578) 2016 — page 2 of 6
(e)
f (x) = e-( x-k )
Correct f ¢(x)
2
f ¢(x) = -2 ( x - k ) e-( x-k )
Correct f ¢¢(x)
Correct solutions
with correct
f ¢(x) and
2
f ¢¢(x)
f ¢¢(x) = -2e-( x-k ) + 4 ( x - k ) e-( x-k )
2
2
2
2
= e-( x-k ) éë 4 ( x - k ) - 2 ùû
2
f ¢¢(x) = 0 Þ 4 ( x - k ) - 2 = 0
2
4(x - k) = 2
2
1
2
±1
(x - k) =
2
1
x=k±
2
( x - k )2 =
NØ
N1
N2
A3
A4
M5
M6
E7
E8
No response;
no relevant
evidence.
ONE answer
demonstrating
limited
knowledge of
differentiation
techniques.
1u
2u
3u
1r
2r
1t with minor
error(s).
1t
NCEA Level 3 Calculus (91578) 2016 — page 3 of 6
Q2
(a)
(b)
(c)(i)
Expected Coverage
f ¢ ( x ) = ln(3x -1) + x ×
3
3x -1
1
y = ( 2x -1) 2
-1
dy 1
= ( 2x -1) 2 .2
dx 2
1
=
2x - 1
dy 1
At x = 5 ,
=
dx 3
1:
2:
3:
4:
–1, 1
–2, –1, 1, 4
–4, 3, x > 4
Achievement
(u)
Merit
(r)
Correct
derivative
Correct solution
with correct
derivative.
2 correct
answers.
3 correct
answers.
Correct
expression for
dr
dt
Correct solution
dr
with correct
dt
– units not
required.
1< x < 4
(ii)
1
(d)
dV
= 4800 cm 3 s-1
dt
4
V = πr 3
3
dV
= 4πr 2
dr
dr dr dV
=
´
dt dV dt
4800 1200
=
=
4πr 2
πr 2
4
V = 288000π = πr 3
3
4 3
288000 = r
3
3
r = 216 000
r = 60 cm
dr
1200
\ =
= 0.106 cm s–1
dt π ´ 60 2
Excellence
(t)
NCEA Level 3 Calculus (91578) 2016 — page 4 of 6
(e)
Correct
expression for
dV
ds
1
Vol = π r 2 h
3
h= 6+s
s2 + r 2 = 62
Correct solution.
r 2 = 36 - s 2
(
)
1
\V = π 36 - s 2 ( 6 + s )
3
1
= π 216 + 36s - 6s 2 - s 3
3
dV 1
= π 36 -12s - 3s 2
ds 3
dV
Max volume when
=0
ds
 3s 2 +12s - 36 = 0
s 2 + 4s -12 = 0
(
(
)
)
( s + 6)( s - 2) = 0
s = -6 or s = 2
s=2
NØ
N1
N2
A3
A4
M5
M6
E7
E8
No response;
no relevant
evidence.
ONE answer
demonstrating
limited
knowledge of
differentiation
techniques.
1u
2u
3u
1r
2r
1t with minor
error(s).
1t
NCEA Level 3 Calculus (91578) 2016 — page 5 of 6
Q3
(a)
(b)
Expected Coverage
f ¢(x) =
-3
1
( 3x + 2 ) 4 ×3
4
Achievement
(u)
Merit
(r)
Correct
derivative.
y = 6x - e 3x
Correct solution
with correct
derivative.
dy
= 6 - 3e 3x
dx
dy
Want
=0
dx
3e 3x = 6
e 3x = 2
x=
(c)
ln 2
= 0.231
3
Area = A(x) = x ( x - 6 )
2
= x -12x + 36x
A¢ ( x ) = 3x 2 - 24x + 36
3
2
Correct
expression for
Correct solution
for maximum
area with
correct
derivative.
Correct
expression for
dy
.
dx
Correct proof
with correct
derivative.
A¢ ( x )
Max when A¢ ( x ) = 0
(
)
3 x 2 - 8x + 12 = 0
3( x - 6 ) ( x - 2 ) = 0
Max when x = 2
Max Area = 2 ´16 = 32
(d)
ex
sin x
dy sin x × e x - e x .cos x
=
dx
sin 2 x
sin x × e x e x .cos x
=
sin 2 x
sin 2 x
x
x
e
e
cos x
=
×
sin x sin x sin x
= y - y × cot x
y=
= y (1- cot x )
Excellence
(t)
NCEA Level 3 Calculus (91578) 2016 — page 6 of 6
(e)
tan a =
15
d
tan (a + q ) =
Correct
expression for
d ( tan q )
dd
20.4
d
tanq = tan ((a + q ) - a )
tan (a + q ) - tan a
1- tan (a + q ) × tan a
20.4 15
d
d
=
20.4 ´ 15
1+
d2
5.4
= 2 d
d + 306
d2
5.4d
= 2
d + 306
=
d ( tan q )
=0
dd
d 2 + 306 ´ 5.4 - 5.4d ´ 2d
or
Correct solution
– units not
required.
dq
dd
Max when
(
(
)
d 2 + 306
)
2
=0
5.4d + 306 ´ 5.4 -10.8d 2 = 0
5.4d 2 - 306 ´ 5.4 = 0
2
d 2 = 306
d = 17.5 m
NØ
N1
N2
A3
A4
M5
M6
E7
E8
No response;
no relevant
evidence.
ONE answer
demonstrating
limited
knowledge of
differentiation
techniques.
1u
2u
3u
1r
2r
1t with minor
error(s).
1t
Cut Scores
Not Achieved
Achievement
Achievement with Merit
Achievement with Excellence
0–7
8–12
13–18
19–24