# 2015

```NCEA Level 3 Calculus (91578) 2015 — page 1 of 6
Assessment Schedule – 2015
Calculus: Apply differentiation methods in solving problems (91578)
Evidence
Q1
Expected Coverage
(a)
30sec 2 (5x)
(b)
2
dy
= 3 4x − 3x 2 ( 4 − 6x )
dx
dy
At x = 1,
= 3 &times; 1&times; −2 = −6
dx
(c)
)
2
x
+
( 1)2
f ′(x) &gt; 0 ⇒ 8 &gt;
( x + 1)2 &gt;
Merit
r
A correct
expression for
the derivative.
(
f ′( x) = 8 −
Achievement
u
Correct
solution with
correct
derivative.
Correct
derivative.
Correct
solution
with
correct
derivative.
Correct
derivative.
Correct
solution
with
correct
derivative.
2
( x + 1)2
1
4
1
−1
or x + 1 &lt;
2
2
−1
−3
x&gt;
or x &lt;
2
2
Either x + 1 &gt;
(d)
f ′(x) =
x(x − 5) − (x + 4)(2x − 5)
2
x2 ( x − 5)
f ′(x) = 0 ⇒ x(x − 5) − (x + 4)(2x − 5) = 0
(
)
x 2 − 5x − 2x 2 + 3x − 20 = 0
−x − 8x + 20 = 0
2
x 2 + 8x − 20 = 0
(x + 10)(x − 2) = 0
x = −10 or + 2
Excellence
t
NCEA Level 3 Calculus (91578) 2015 — page 2 of 6
(e)
Let V = volume (m3)
S = slant height (m)
h = height (m)
cos 30 =
Valid
statement
of the
relationship
between
rates.
dS dV
or
dr dr
correct.
r
S
Correct
solution
with correct
derivatives.
r
cos 30
dS
1
=
dr cos 30
h
tan 30 =
r
h = r tan 30
1
V = π r 2h
3
1 3
= π r tan 30
3
dV
= π r 2 tan 30
dr
dS dS dr dV
=
&times;
&times;
dt dr dV dt
1
1
=
&times;
&times;2
cos 30 π r 2 tan 30
When r = 10 m,
dS
1
1
=
&times;
&times;2
dt cos 30 π10 2 &times; tan 30
= 0.01273 m / minute
S=
N&Oslash;
N1
N2
A3
A4
M5
M6
E7
E8
No response;
no relevant
evidence.
demonstrating
limited
knowledge of
differentiation
techniques.
1u
2u
3u
1r
2r
1t with minor
error(s).
1t
NCEA Level 3 Calculus (91578) 2015 — page 3 of 6
Q2
Expected Coverage
(a)
1
x − 3x 2
5
(b)
dy
16
= 1+ 2
dx
x
dy
At x = 4,
=2
dx
(
(c)(i)
1. x = 1
2. x = –1, 1, 2
3. –1 &lt; x &lt; 1
(ii)
3
(iii)
Does not exist.
(d)
Excellence
t
Correct
solution with
correct
derivative.
−1
2
Two correct
dx
correct.
dL
x+L
L
=
5
1.5
1.5x + 1.5L = 5L
1.5x = 3.5L
7L
x=
3
dx 7
=
dL 3
dx
=2
dt
dL dL dx
=
&times;
dt dx dt
3
= &times;2
7
6
= = 0.857 m s –1
7
Merit
r
A correct
expression for
the derivative.
−4
5
) ⋅ (1− 6x )
Achievement
u
Four correct
Correct
solution
with correct
derivatives.
(Units not
required.)
NCEA Level 3 Calculus (91578) 2015 — page 4 of 6
(e)
Depth of water = x
h = x + 20
1
1
V = h 3 − 20 3
3
3
1
1
3
= ( x + 20 ) − 20 3
3
3
dV
2
= ( x + 20 )
dx
A = ( x + 20 )
Correct
dV
dx
OR
Correct
dV
dx
AND
dA
dx
dA
dx
Correct
solution.
2
dA
= 2 ( x + 20 )
dx
dV
= 3000
dt
dA dA dx dV
=
&times;
&times;
dt dx dV dt
1
= 2 ( x + 20 ) &times;
&times; 3000
( x + 20 )2
When x = 15
dA
1
= 2 &times; 35 &times; 2 &times; 3000 = 171.4 cm 2 min −1
dt
35
N&Oslash;
N1
N2
A3
A4
M5
M6
E7
E8
No response;
no relevant
evidence.
demonstrating
limited
knowledge of
differentiation
techniques.
1u
2u
3u
1r
2r
1t with minor
error(s).
1t
NCEA Level 3 Calculus (91578) 2015 — page 5 of 6
Q3
(a)
Expected Coverage
f ′(x) =
5
10
&times;2=
2x − 3
2x − 3
10
=4
2x − 3
8x − 12 = 10
Achievement
u
Merit
r
Correct
solution with
correct
derivative.
x = 2.75
(b)
f ′(x) =
=
e 3x − x.3e 3x
(e )
3x 2
1− 3x
e 3x
f ′(x) = 0 ⇒ x =
(c)
Correct
solution with
correct
derivative.
1
3
dx
= −3sint
dt
dy
= 3cos 3t
dt
dy 3cos 3t − cos 3t
=
=
dx −3sint
sint
⎛ 3π ⎞
− cos ⎜ ⎟
⎝ 4⎠
π dy
At t = ,
=
=1
4 dx
⎛ π⎞
sin ⎜ ⎟
⎝ 4⎠
Correct
and
dx
dt
Correct
solution
with
correct
derivatives.
dx
dt
Parts (i)
and (ii)
both
correct.
dy
dt
(d)(i)
dx
= −Ak sin kt + Bk cos kt
dt
d2 x
= −Ak 2 cos kt − Bk 2 sin kt
dt 2
= −k 2 ( A cos kt + Bsin kt )
= −k 2 x
(ii)
x(0) = 0 ⇒ A cos 0 + Bsin 0 = 0
A=0
v(0) = 2k ⇒ 2k = −Ak sin(0) + Bk cos(0)
B= 2
Correct
d2 x
dt 2
Consistent with
dx
incorrect dt
Or
Excellence
t
NCEA Level 3 Calculus (91578) 2015 — page 6 of 6
(e)
Differentiate
corretedly
related but
incorrect
expression for
w.
cos A =
Correct
dw
dA
2
f
2
cos A
w
sin A =
5− f
w= (5 − f )sin A
f=
2 ⎞
⎛
= ⎜5−
⎟ sin A
⎝
cos A ⎠
= 5sin A − 2 tan A
dw
= 5 cos A − 2sec 2 A
dA
dw
= 0 ⇒ 5 cos A − 2sec 2 A = 0
dA
5 cos 3 A − 2 = 0
2
cos 3 A =
5
A = 42.5&deg;
Correct
solution.
w = 1.55 m
N&Oslash;
N1
N2
A3
A4
M5
M6
E7
E8
No response;
no relevant
evidence.
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
```