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Revision for Achievement Standard 2.7 Calculus Revision tests and worked answers (Ignore the definite integrals  yet to be updated to new standards) 
Link to Exam Papers (old standards) p211 

NZQA exemplar
2013 Exam paper and answer schedule NZQA Examiners report (pdf) Excellence example 2012 Exam paper and answer schedule NZQA Examiners report (pdf) Excellence example


Find the gradient of the curve y = x^{2} + 3x + 5 when x = 4 

Steps: 

1) Differentiate 
y' = 2x + 3 
2) Substitute x = 4 into the derivative 
y' = 2x4 + 3 
3) The result is the gradient 
y' = 11 
Find the equation of the tangent to the curve y = x^{2}  3x + 5 at the point (1,3) 

Steps: 
y = x^{2}  3x + 5 
1) General tangent equation y = mx + c 

2) Differentiate to find gradient m 
y' = 2x  3 
3) Substitute x = 1 into the derivative 
y' = 2x1  3 
4) The result is the gradient m 
m = 1 
Tangent equation so far is y = 1x + c 

5) Substitute in x=1 & y=3 values in to find c 
y = 1x + c 
6) Tangent equation is y = 1x + 4 
Find the equation of the normal to the curve y = x^{2}  3x + 5 at the point (1,3) 

Steps: 
y = x^{2}  3x + 5 
1) General normal equation y = mx + c 

2) Differentiate to find gradient of tangent m_{1} 
y' = 2x  3 
3) Substitute x = 1 into the derivative 
y' = 2x1  3 = 1 
4) The result is the gradient of the tangent m_{1} 
m_{1} = 1 
5) Find the negative inverse of m_{1}to give the gradient of the normal line m_{2} 
m_{2}= 1 
Normal equation so far is y = 1x + c 

6) Substitute in x=1 & y=3 values in to find c 
y = 1x + c 
7) Normal equation is y = 1x + 2 
Find the maximum value of the curve y = x^{2}  6x + 8 

Steps: 
y = x^{2}  6x + 8 
1) Differentiate to find gradient function 
y' = 2x  6 
2) The derivative = 0 at a min or max 
0 = 2x  6 
3) Substitute the x value back to find y 
y = x^{2}  6x + 8 
Maximum value = 15 
Find the minimum surface area of a cuboid (box) which has a square end, and a volume of 800cm^{3} 

Steps: 

1) Find an equation for surface area 
A = x^{2} + 4xL 
We only want one variable so eliminate L using the volume equation 

2) Volume equation  rearrange to be L = 
V = Lx^{2} 
3) Substitute L = 800/x^{2} back into the Area Equation 
A = x^{2} + 4xL 
4) Differentiate the Area equation 
A = x^{2} + 3200x^{1} 
5) Make the derivative = 0 and solve for x 
0 = 2x  3200x^{2} 
6) Calculate the surface area 
A = x^{2} + 3200/x 
Find the minimum surface area of a cylinder which has a volume of 330cm^{3} What are the dimensions of the cylinder? 

Steps: 

1) Find an equation for surface area 
2 circles + rectangle 
We only want one variable so eliminate L using the volume equation 

2) Volume equation  rearrange to be L = 
V = пr^{2}L 
3) Substitute L = 330/2пr^{2} back into the Area Equation 
A = 2пr^{2} + 2пrL 
4) Differentiate the Area equation 
A = 2пr^{2} + 660r^{1} 
5) Make the derivative = 0 and solve for r 
0 = 4 пr  660r^{2} 
6) Calculate the surface area 
A = 2п5.5^{2} + 660/5.5 
Dimensions 

Radius 
r = 5.5cm (1dp) 
Length 
L = 330/пr^{2} 
Find the equation for the graph of f(x) given that f'(x) = 6x^{2} + 8 and the graph of f(x) passes through the point (1,8) 

Steps: 
f'(x) = 6x^{2} + 8 
1) Anti differentiate to find function f(x) 
f(x) = 2x^{3} + 8x + C 
2) Substitute in x = 1 & y = 8 from the coordinate and rearrange to find 'C' 
f(x) = 2x^{3} + 8x + C 
3) Write out the equation for f(x) 
f(x) = 2x^{3} + 8x + C 
Find the speed of a car at t = 4 seconds, given the distance (metres) is defined by d = t^{2}  6t + 15 

Steps: 
d = t^{2}  6t +15 
1) Differentiate to find 'speed' function 
v = 2t  6 
2) Substitute t = 4 into the 'speed' function 
v = 2t  6 
3) Speed at t = 4 sec 
v = 2 m/s 
Find the acceleration of a car at t = 2 seconds, given the distance (metres) is defined by d = t^{2}  6t + 15 

Steps: 
d = t^{2}  6t +15 
1) Differentiate to find 'speed' function 
v = 2t  6 
2) Differentiate again to find the 'acceleration' function 
a = 2 
3) Substitute t = 4 into the 'acceleration' function 
Note: nowhere to substitute in t = 4 (acceleration here is always 2) 
4) Acceleration at t = 4 sec 
a = 2 m/s/s 
Find the acceleration of a car at t = 2 seconds, given the distance (m) is defined by d = 3t^{3}  9t^{2} + 18 

Steps: 
d = 3t^{3}  9t^{2} + 18 
1) Differentiate to find 'speed' function 
v = 9t^{2}  18t 
2) Differentiate again to find the 'acceleration' function 
a = 18t  18 
3) Substitute t = 2 into the 'acceleration' function 
a = 18t  18 
Find the distance function of a car, given the speed function is defined by v = 6t^{2} + 4t + 5 and the car is at a distance of 28m when t = 3 sec 

Steps: 
v = 6t^{2} + 4t + 5 
1) Anti differentiate to find 'distance' function 
d = 2t^{3} + 2t^{2} + 5t + c 
2) Substitute in d = 28 and t = 3 to find 'c' 
d = 2t^{3} + 2t^{2} + 5t + c 
3) The distance function 
d = 2t^{3} + 2t^{2} + 5t  60 