Non-moderated
Achievement Standard 2.3
Calculus
1.
Find the gradient to the curve f(x) = x2 + 2x +3 at the point when x = 0.
2.
Find the coordinates of the point on the curve g(x) = x2 + 8x + 15 where the
gradient is zero.
3.
Evaluate
4.
Find the equation of the curve h(x) given that h’(x) = 3x2 – 2x + 5 and that h(x)
passes through the point (1 , 7).
5.
The expression for the vertical velocity of a dart is given by
v(t) = 3t2 – 18t + 15 cm/sec.

4
2
(x 2  6x  5)dx
a) When is the acceleration of the dart equal to zero?
b) The dart is 150cm from a reference point when t = 0. Find the height of the dart
3 seconds after timing began.
c) What is the maximum height reached by the dart in the first five seconds?
6.
The cubic graph y = x(x – 1)(x + 2)
= x3 + x2 – 2x
is drawn to the right.
Use calculus to find the area
shaded in the diagram.
7.
An enclosed rectangular garden with an area of 18m2 is surrounded by a path. The
path is 2 metres wide on one pair of opposite sides and 1 metre wide on the other
two sides.
Use calculus to find the dimensions of the garden
which minimise the area of the path and find the
area of the path in this case.
2m
1m
Area = 18m2
2m
1m
Criteria
Achieved
Find and use
straightforward
derivatives and
integrals
1
2
Achievement with Merit
Achievement with Excellence
f’(x) = 2x + 2
f’(0) = 2 gradient = 2
No alternative
3 out of Q1,
2, 3 or 4
g’(x) = 2x + 8
2x + 8 = 0 x = -4
g(-4) = -1 point = (-4 , -1)
No alternative
h(x) = x3 – x2 + 5x + 2
6t – 18 = 0 t = 3 sec
No alternative
y = t3 – 9t2 + 15t + 150
when t = 3 y = 141cm
No alternative
units not req’d
5c)
t2 – 18t + 15 = 0
t = 1 or t = 5
Max height when t = 5 is 157cm
6)
Area =
Working
required
Evidence of a
test for max
needed
5a)
5b)
7
[
0
Sufficiency
Achieved:
or equivalent,
accept rounding
with working
4
Apply
differentiation
techniques to
solve optimisation
problems
Judgement
x3
 3x 2  5x ]04
3
1
= 7
3
3
Apply calculus
techniques to
solve problems
Evidence
No alternative
1
Replacement:
Q5a) and 7
for 1 and 2
Q5b) and Q6
for 3 and 4
Achieved
with Merit:
Achieved plus
Three from
Q5a), 5b),
5c) or Q6
Evidence of two
integrals needed
Replacement:
Q7 for one of
Q5 a), b) or
c)
Full working
required
No test needed.
Achieved
with
Excellence:
A’ = 4 – 36x
A’ = 0 gives x = 3 and y = 6
Final statement
of area of the
path needed.
Merit plus
Q7
Dimensions of the garden = 3 x 6
Area of the path = 32m2
Allow one minor
error
x 4 x 3

x 4 x 3


 x 2  

 x 2

3
3
 4
 2  4
 0
8 5 37
= 

3 12 12
xy = 18
18
A = (x + 2)(
+ 4)
x
= 4x + 36x-1 + 26
-2