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Achievement Standard 2.7 Calculus Revision

Calculus HOME | Achievement Objectives | Gradient | Gradient Functions | Differentiation | Gradient at a point | Find point with gradient | Equation of Tangent | Second Derivitive | Coordinates of Max & Min | Increasing, Decreasing Functions | Applications | Kinematics with differentiation | Antidifferentiation | Finding 'C' | Kinematics with anti-differentiation | Rates of Change | Mixed Problems | Revision

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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

 

Old standards Exam papers |

NZQA exemplar

Mathematics and Statistics 91262 (2.7) Apply calculus methods in solving problems
91262 Sample Exam Paper 91262 Sample Assessment Schedule 91262 Annotated Exemplars: Excellence 91262 Annotated Exemplars: Merit 91262 Annotated Exemplars: Achievement 91262 Annotated Exemplars: Not Achieved

 

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

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

 

 

 

Basic Differentiation 1

Basic Differentiation 2

Gradient Function 1

Tangents

Increasing and decreasing functions 1

Increasing and decreasing functions 2

Applications 1

Maxima and Minima 1

Rates

Basic Antidifferentiation

Finding the constant of integration

Kinematics

 

Jump down to... Finding a gradient | Equation of Tangent | Equation of a Normal | Maximum & Minimum | Applications | Finding f(x) given f'(x)| Kinematics examples |

Finding a Gradient

Find the gradient of the curve y = x2 + 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

   

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Equation of Tangent Line

Find the equation of the tangent to the curve y = x2 - 3x + 5 at the point (1,3)

Steps:

y = x2 - 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
y' = -1

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
3 = -1x1 + c
3 = -1 + c
4 = c

6) Tangent equation is y = -1x + 4

 

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Equation of Normal line

Find the equation of the normal to the curve y = x2 - 3x + 5 at the point (1,3)

Steps:

y = x2 - 3x + 5

1) General normal equation y = mx + c

 

2) Differentiate to find gradient of tangent m1

y' = 2x - 3

3) Substitute x = 1 into the derivative

y' = 2x1 - 3 = -1

4) The result is the gradient of the tangent m1

m1 = -1

5) Find the negative inverse of m1to give the gradient of the normal line m2

m2= 1

     

Normal equation so far is y = 1x + c

 

6) Substitute in x=1 & y=3 values in to find c

y = 1x + c
3 = 1x1 + c
3 = 1 + c
2 = c

7) Normal equation is y = 1x + 2

 

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Maximum & Minimum

Find the maximum value of the curve y = -x2 - 6x + 8

Steps:

y = -x2 - 6x + 8

1) Differentiate to find gradient function

y' = -2x - 6

2) The derivative = 0 at a min or max
So solve y' = 0 to find the x value for the maximum

0 = -2x - 6
6 = -2x
-3 = x

3) Substitute the x value back to find y

y = -x2 - 6x + 8
y = -(-3)2 - 6x-3 + 8
y = -9 + 18 + 8
y = 15

Maximum value = 15
Coordinates of the maximum (-3,15)

 

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Applications

Find the minimum surface area of a cuboid (box) which has a square end, and a volume of 800cm3


Steps:

 

1) Find an equation for surface area

A = x2 + 4xL

     

We only want one variable so eliminate L using the volume equation

2) Volume equation - rearrange to be L =

V = Lx2
800 = Lx2
L = 800/x2

3) Substitute L = 800/x2 back into the Area Equation

A = x2 + 4xL
A = x2 + 4x800/x2
A = x2 + 3200/x

4) Differentiate the Area equation

A = x2 + 3200x-1
A' = 2x - 3200x-2

5) Make the derivative = 0 and solve for x
(multiply by x2)

0 = 2x - 3200x-2
0 = 2x3 - 3200
3200 = 2x3
1600 = x3
x = 3√1600 = 11.7cm

6) Calculate the surface area

A = x2 + 3200/x
A = 11.72 + 3200/11.7
A = 410.4cm3

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Find the minimum surface area of a cylinder which has a volume of 330cm3

What are the dimensions of the cylinder?

Steps:

1) Find an equation for surface area

2 circles + rectangle
A = 2пr2 + 2пrL

     

We only want one variable so eliminate L using the volume equation

2) Volume equation - rearrange to be L =

V = пr2L
330 = пr2L
L = 330/пr2

3) Substitute L = 330/2пr2 back into the Area Equation

A = 2пr2 + 2пrL
A = 2пr2 + 2пr330/пr2
A = 2пr2 + 660/r

4) Differentiate the Area equation

A = 2пr2 + 660r-1
A' = 4 пr - 660r-2

5) Make the derivative = 0 and solve for r
(multiply by r 2)

0 = 4 пr - 660r-2
660r-2 = 4r
165r-2 = r
165 = r3
r = 3√165 = 5.5cm (1dp)

6) Calculate the surface area

A = 2п5.52 + 660/5.5
A = 309.4cm3 (1dp)

Dimensions

 

Radius

r = 5.5cm (1dp)

Length

L = 330/пr2
L = 330/п5.52
L = 3.5cm

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Finding f(x) given f('x)

Find the equation for the graph of f(x) given that f'(x) = 6x2 + 8 and the graph of f(x) passes through the point (1,8)

Steps:

f'(x) = 6x2 + 8

1) Anti differentiate to find function f(x)

f(x) = 2x3 + 8x + C

2) Substitute in x = 1 & y = 8 from the coordinate and rearrange to find 'C'

f(x) = 2x3 + 8x + C
8 = 2x13 + 8x1 + C
8 = 2 + 8 + C
8 = 10 + C
-2 = C

3) Write out the equation for f(x)

f(x) = 2x3 + 8x + C
f(x) = 2x3 + 8x - 2

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Kinematics

Find the speed of a car at t = 4 seconds, given the distance (metres) is defined by d = t2 - 6t + 15

Steps:

d = t2 - 6t +15

1) Differentiate to find 'speed' function

v = 2t - 6

2) Substitute t = 4 into the 'speed' function

v = 2t - 6
v = 2x4 - 6
v = 8 - 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 = t2 - 6t + 15

Steps:

d = t2 - 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

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Find the acceleration of a car at t = 2 seconds, given the distance (m) is defined by d = 3t3 - 9t2 + 18

Steps:

d = 3t3 - 9t2 + 18

1) Differentiate to find 'speed' function

v = 9t2 - 18t

2) Differentiate again to find the 'acceleration' function

a = 18t - 18

3) Substitute t = 2 into the 'acceleration' function

a = 18t - 18
a = 18x2 - 18
a = 18 m/s/s

Find the distance function of a car, given the speed function is defined by v = 6t2 + 4t + 5 and the car is at a distance of 28m when t = 3 sec

Steps:

v = 6t2 + 4t + 5

1) Anti differentiate to find 'distance' function

d = 2t3 + 2t2 + 5t + c

2) Substitute in d = 28 and t = 3 to find 'c'

d = 2t3 + 2t2 + 5t + c
28 = 2x33 + 2x32 + 5x3 + c
28 = 54 + 18 + 15 + c
28 = 88 + c
c = -60

3) The distance function

d = 2t3 + 2t2 + 5t - 60

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