Wednesday, 26 November 2014

Wednesday the 26th of November

Advanced Higher
Differential Equations
L.I. To solve differential equations that are separable.
S.C. I can seperate the variables and get the equation ready to be integrated
       I can integrate both sides and solve the equation.
 


Tuesday the 25th of November

Advanced Higher
Further integration
L.I. To look at integrals that return to the original integral once you have integrated by parts
S.C. I can integrate by parts
       I can recognise the original integral appearing in the second integration by parts
       I can deal with this by substitution and finding the value of the integral.

Monday, 24 November 2014

Monday the 24th of November

2(1)
Circles
L.I. To use the correct formula for calculating the circumference or the area as asked for in the question
S.C. I can pick the correct formula and use it correctly when asked to find the circumference or a fraction of a circles circumference
       I can pick the correct formula and use it correctly when asked to find the area or a fraction of a circles area
       I can use circles as part of a composite shape and calculate its area and circumference as needed.

Advanced Higher
Further integration
L.I. To continue to look at integration by parts
S.C. I can integrate by parts twice in order to integrate an expression
       I can use integration by parts to find a definite integral.

Friday, 21 November 2014

Friday the 21st of November

2(1)
Circles
L.I. To calculate the radius given the area of a circle
S.C. I understand how the formula for area of a circle can be re-arranged to find the radius
       I can plug the numbers in correctly to calculate the radius of the circle given the area
       I always make sure to put the correct units in my answer
       I can use the radius to calculate other things in the circle (diameter, circumference)

National 5
Simultaneous Equations
L.I. To solve simultaneous equations in context
S.C. I can create the two formula from the example given
       I can multiply up the equations as needed to get one variable ready to be eliminated
       I can solve the equations to find the two variables.

Thursday, 20 November 2014

Thursday the 20th of November

2(1)
Area of a circle
L.I. To use the formula given to calculate the area of a circle
S.C. I can substitute into the formula for area of a circle, using the radius
       I can calculate the radius if I am given the diameter
       I can remember the formula for calculating the area and differentiate it from the formula for calculating the circumference.
National 5
Simultaneous Equations
L.I. To use simultaneous equations to solve problems in context
S.C. I can set up the two equations from the information given
       I can solve the equations simultaneously
       I can put the answers back into the appropriate context.

Wednesday, 19 November 2014

Wednesday the 19th of November

Advanced Higher
Integration by parts
L.I. To use integration by parts to find the integral of expressions that can be written as two functions muliplied together
S.C. I can identify the part of the function to be called u and the part of the function to be called dv/dx
       I can find du/dx and v from these functions
       I can plug the appropriate functions into the integration by parts formula
       I can state my final answer to the integration after using integration by parts



2(1)
Circles
L.I. To calculate the diameter or radius given the circumference.
S.C. I can use the circumference to calculate the diameter of a circle by dividing the circumference by pi.
       I can use this to find the radius.
       I can round to a given accuracy.

Tuesday, 18 November 2014

Tuesday the 18th of November

National 5
Simultaneous Equations
L.I. To solve a set of equations with two variables
S.C. I can multiply up one or both of the equations to get the variables coefficients to be the same
       I can decide whether to add or subtract the two equations to eliminate one of the variables
       I can find the value of the variable I don't eliminate
       I can find the other variables value by substituting into one of my original equations
 
 
 

 
Advanced Higher
Further integration
L.I. Using algebraic long division and partial fractions to simplify an expression to then integrate it
S.C. I can use algebraic long division to simplify an expression
       I can use partial fractions to simplify an expression
       I can integrate this new basic expression



Monday, 17 November 2014

Monday the 17th of November

2(1)
Circle
L.I. To introduce the formula for calculating the circumference of a circle from the investigation they carried out last time
S.C. I know the relationship between the circumference and the diameter
       I can calculate the circumference of a circle given the diameter
       I can calculate the circumference of a circle given the radius
Advanced Higher
Further integration
L.I. To use partial fractions to make integration of rational functions easier
S.C. I can put a rational function into partial fractions
       I can integrate the terms from the partial fraction

Algebraic Long Division
L.I. To use algebraic long division before integrating to make it easier
S.C. I can write an improper algebraic rational function as a sum of a polynomial and a proper rational function.
       I can integrate the expression found
 

Friday, 14 November 2014

Friday the 14th of November

2(1)
Circles
L.I. To use the formula C = pi d to calculate the circumference of a circle
S.C. I know what pi is to 2 decimal places
       I know what the diameter is and it's connection to the radius
       I know the connection between the circumference and diameter of a circle
       I can calculate the circumference of a circle given it's diameter or radius

National 5
Simultaneous equations
L.I. To find the point of intersection graphically and using elimination
S.C. I can sketch the graph of a linear equation
       I can find, by looking, the point of intersection of two lines
       I can set up two equations to be solved simultaneously
       I can eliminate one of the variables in the two equations
       I can solve for the first variable
       I can substitute into one of the original equations to find the other variable


Wednesday, 12 November 2014

Wednesday the 12th of November

Advanced Higher
Further integration
L.I. to learn how to integrate to get inverse trigonometric functions as answers
S.C. I can integrate to get the inverse sine function as an answer
       I can integrate to get the inverse cosine function as an answer






Friday, 7 November 2014

Friday the 7th of November

2(1)
Algebra
L.I. To solve equations that have unknowns and constants on both sides
S.C. I can get all the variables to one side and simplify
       I can get all the constants to one side and simplify
       I can solve the equation

National 5
Straight line
L.I. To state the equation of a straight line between 2 given points
S.C. I can find the gradient of the line joining the 2 points
       I can substitute into the formula y - b = m(x - a) where m =gradient and (a,b) is a point on the line.
       I can simplify this equation and then state it's gradient and it's y-intercept.


Thursday, 6 November 2014

Thursday the 6th of November

2(1)
Algebra
L.I. To solve equations
S.C. I can solve basic equations of the form ax + b = c

National 5
Equation of a straight line
L.I. To understand the relationship between gradient, y-intercept and the equation of a straight line
S.C. I can find the y-intercept (c)
       I can calculate the gradient of a straight line (m)
       I can find the equation of a straight line in the form y = mx + c

Monday, 3 November 2014

Monday the 3rd of November

2(1)
Algebra revision
L.I. To tidy up expressions using the rules of algebra
S.C. I can add parts of expressions that are the same letter(s)
       I can subtract parts of expressions that are the same letter(s)
       I can write an expression in it's most simple form.

Advanced Higher
Differentiation
L.I. To look at related rates in real life examples
S.C. I can identify a related rate that is explicitly defined
       I can indentify a related rate that is implicitly defined
       I can answer questions based on related rates by using different differentiation techniques to solve them