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  • IGCSE Edexcel
  • Further Pure Mathematics
  • 1. Further Pure Mathematics Notes

Further Pure Mathematics

  • IGCSE Edexcel
  • Revision Notes

1. Further Pure Mathematics Notes


1.1 Logarithmic Functions and Indices>1.2 The Quadratic Function>1.3 Identities and Inequalities>1.4 Graphs>1.5 Series>1.6 The Binomial Series>1.7 Scalar and Vector Quantities>1.8 Rectangular Cartesian Coordinates>1.9 Calculus>1.10 Trigonometry>

1.9 Calculus

1.9.1 Differentiation and integration of sums of multiples of powers of x (excluding integration of 1/x),sin ax,cos ax,eax

1.9.2 Differentiation of a product, quotient and simple cases of a function of a function

 

1.9.3 Applications to simple linear kinematics and to determination of areas and volumes

Type 1: Area between a curve and x-axis (y > 0)

Type 2: Area between a curve and x-axis (y < 0)

Type 3:  Area between a curve and x-axis (-∞ < y <∞) 

Type 4: Area between two curves

1.9.4 Stationary points and turning points

Coordinates of a Stationary Point

Step 1: Differentiate the equation and equate to 0 (f'(x) = 0)

Step 2: Substitute the value of x into the equation to find y

1.9.5 Maxima and minima

1.9.6 The equations of tangents and normals to the curve y = f(x)

1.9.7 Application of calculus to rates of change and connected rates of change

Finding rate of change of a part of a usually 3D shape (e.g. radius):

  1. Eg: Area of volume of cylinder info
    • 50cm³/s (rate of sand poured)
    • V of cone increases in a way that r of base is always 3 times the h of the cone
    • Find rate of change of radius of cone, when radius is 10cm

Working:

dV/dt = 50 (given)

dr/dt = need to find

dr/dt = dV/dt x dr/dV

r = 3h (given)

h = r/3

V = 1/3πr2h

V = 1/3πr²(r/3)

V = 1/9πr³

dV/dr = 1/3πr²

dr/dt = 50 x 1/(1/3π(10)²) = 0.477 cm/s

dr/dt = 50 x 1/(1/3π(10)²) = 0.477 cm/s

Previous1.8 Rectangular Cartesian Coordinates
Next1.10 Trigonometry

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