After successfully studying Maths at GCSE, I decided to study it at A level. Back at primary school, due to my language development delays, I would find it difficult to understand new concepts and apply new knowledge to given problems. This improved for me over the years since. There had always been confidence in underlying improvements in me, including my language and understanding skills.
Below are examples where I talk about what helped me to understand certain topics when studying the A level maths course. For me, it was about using logic and recognising certain patterns.
What I found easy about this was that you simply multiplied the value of the coefficient by the power, followed by taking 1 away from the power.
y = 3x^2
dy/dx= 6x
y=aX^b
dy/dx = (a×b)X^(b-1)
y = e^x
dy/dx = e^x
y= 2e^2x
dy/dx = 4e^2x
y = In (x)
dy/dx = 1/x
y = 2In(x)
dy/dx = 2/x
y = 3In(4x)
dy/dx = 3/4x
Y = ae^(b×c)
dy/dx = (a×b)e^(b×c)
y = a(In) x
dy/dx = a/x
a(ln bx) = a/bx
One thing I found about this topic was that it simply involved adding 1 to the power of a number represented by a letter. You then divide by the new power as shown below. The letter ‘a’ represents what is known as the coefficient. I quickly recognised it as the reverse of the process of differentiation above.
∫ ax^b = ( (ax^(b+1))/ (b+1)) + c
∫ x^3 = ( (x^4)/ 4) + c
b
∫ dx^e = ( (dx^(e+1))/ (e+1)) + c
a
b
∫ (d*(b^2)) / (e+1) - (d*(a^2)) / (e+1)
a
4
∫ 3x = ( (3x^2)/ 2) + c
2
= (3*(4^2)) / 2 - (3*(2^2)) / 2
= (3 * 16)/2 - (3 * 4)/2
= 24 - 6 = 18
(Sin A)/a = (Sin B)/ b = (Sin C)/ c
or
a/(Sin A) = b/(Sin B) = c/(Sin C)
What I found easy to remember about these formulae was for the sine rule it was the fact that the Sine of an angle with a certain letter was divided by a side labelled with the same letter. This happens to be a side opposite the angle. I liked the fact that this equaled any combination of side and angle divided together.
a^2 = b^2 + c^2 - 2bcCos A
b^2 = a^2 + c^2 - 2acCos B
c^2 = a^2 + b^2 - 2abCos C
What I found easy to remember about these formulae. Adding the squares of the other two side lengths, then taking away 2 multiplied by the values of these two sides and the value of the angle with the same letter as the side whose value you’re trying to calculate. Note how a side opposite an angle share a common label.
Log c (a) + Log c (b) = Log c(ab)
Log c (a) - Log c (b) = Log c(a/b)
Log c (a^b = b(Log(a))
The letter 'c' is used to represent what is known as the base.
Log10 (10) + Log10 (100)
= 1 + 2 = 3
10^1 = 10 ; 10^2. The number 10 here is known as the base.
Log5 (625) - Log5 (25) = 4/2 = 2
5^4= 625 and 5^2 = 25
5Log2 (2) = Log2 (2^5) = Log2 (32)
The number of ways of arranging the letters of the word ‘Cat’ is 3! which 3 x 2 x 1 = 6, or 3!/1! The 3! represents the number of letters contained in the word, and each 1! is each letter.
For the word ‘Dinners’. It gets more tricky. The number of ways of arranging the letters of the word Dinners is 7!/2!. This is 5,040/2 = 2,520
7! Represents the 7 letters, and 2! the 2 n’s
For the word ‘Million’, there are 7 letters. But, if we look closely, there are 2 i’s and 2 l’s.
Therefore, the number of ways of arranging ‘Million’ is 7!/(2!*2!) = 7!/2*2 = 5,040/4 = 1,260.
1! = 1
2! = 2 * 1 = 2
3! = 3 * 2 * 1 = 6
4! = 4 * 3 * 2 * 1 = 24
5! = 5 * 4 * 3 * 2 * 1 = 120
The number of ways of arranging A objects is A! Or A x (A-1) x (A-2) x (A-3)... until 1.
Do you have autism and did you study A level maths? What teaching methods helped you?
Are you in a teaching capacity and teaching or have you taught maths to A level? If so, what teaching methods did you employ or are employing to help any autistic students you may have come across?
Why not let me know in the comments section below.
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