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# Achievement Standard 2.7 Calculus Revision

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

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

 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 |

 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

# 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

# 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

# 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)

# 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

 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

# 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