PremiumMatthew K. GCSE Maths tutor, GCSE Physics tutor

Matthew K.

£24 - £26 /hr

Studying: General Engineering (Masters) - Warwick University

5.0
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.

119 reviews| 184 completed tutorials

Contact Matthew

About me

About Me I am a student studying a general engineering masters degree at Warwick University. I am fun and friendly and have a passion for maths and physics with a little bit of rugby on the side. I helped my brother whilst he was doing GCSE maths and physics alongside helping younger boys with homework and discovered how fun it is helping people to realise there is no reason to be afraid of numbers and equations.

About Me I am a student studying a general engineering masters degree at Warwick University. I am fun and friendly and have a passion for maths and physics with a little bit of rugby on the side. I helped my brother whilst he was doing GCSE maths and physics alongside helping younger boys with homework and discovered how fun it is helping people to realise there is no reason to be afraid of numbers and equations.

Show more

About my sessions

My Sessions What we do in any session is determined by what you want to do. I will be there to guide you and show you that, once you understand, exam questions will be a walk in the park. I will try and make sessions as interesting and fun as possible, with diagrams, pictures and maybe even a video here and there, to ensure that once the 55 minutes are up you’ll be on the ball and ready to take on any test sent your way! What to do? If you have any questions or queries don’t hesitate to send me a message through ‘WebMail’ or book a ‘Meet the Tutor’ session (both done on this website) and I’ll be ready and waiting to reply. Please remember to tell me your exam board and what you want help with. I look forward to meeting you!

My Sessions What we do in any session is determined by what you want to do. I will be there to guide you and show you that, once you understand, exam questions will be a walk in the park. I will try and make sessions as interesting and fun as possible, with diagrams, pictures and maybe even a video here and there, to ensure that once the 55 minutes are up you’ll be on the ball and ready to take on any test sent your way! What to do? If you have any questions or queries don’t hesitate to send me a message through ‘WebMail’ or book a ‘Meet the Tutor’ session (both done on this website) and I’ll be ready and waiting to reply. Please remember to tell me your exam board and what you want help with. I look forward to meeting you!

Show more

No DBS Icon

No DBS Check

Ratings & Reviews

5from 119 customer reviews
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.

Juliet (Parent)

September 10 2017

Great session

Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.

Andrea (Parent)

May 19 2017

Billy really pleased with the session - gained a good understanding of maths problem.

Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.

Phip (Parent)

May 11 2017

Harry was buzzing after this session! Thank you.

Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.

Eddie Thomson (Student)

May 9 2017

really good

Show more reviews

Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A
PhysicsA-level (A2)A*

General Availability

Before 12pm12pm - 5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
MathsA Level£26 /hr
MathsGCSE£24 /hr
PhysicsGCSE£24 /hr
Maths13 Plus£24 /hr
Maths11 Plus£24 /hr

Questions Matthew has answered

How do I find the coordinates of maximum and minimum turning points of a cubic equation?

Finding a maximum and minimum involves differentiating the function in order to determine when the gradient is at zero as this will be when the function has its maximum and minimum turning points.

take the function: f(x)=x3+6x2+9x+2

we must first obtain the gradient function through differentiation

d/dx = 3x2+12x+9 using standard differentiation technique (multiplying the power of the unknown by its coefficient to generate its new coefficient and then reduce its power by one to find the new term)

now that we have the gradient function it must be set to zero, since the gradient of the function is zero at a turning point

3x2+12x+9=0

this is a quadratic and we must solve for x to find the x values at which the gradient is zero

by inspection:

(3x+3)(x+3)=0 and therefore when the gradient of the function is zero:

x=-1 and x=-3

to determine which of these is maximum and is which is minimum we can differentiate the gradient function again. This generates a function that tells us the rate of change of the gradient at any given value of x

d/dx=3x2+12x+9

d2/dx2=6x+12 (using the same method as before)

if we substitue in our two x values:

when x=-1 d2/dx2= -6+12 = 6

since this is positive, the gradient is changing in a positive direction and so when x= -1 it is a minimum.

when x=-3 d2/dx2= -18+12 = 6

since this is negative the gradient is changing in a negative direction and so when x= -3 its a maximum.

to find the y values of these x values we simply subsitute them into the original function:

f(x)=x3+6x2+9x+2

f(-1)=-1+6-9+2=-2

f(-3)=-27+54-27+2=2

so our maximum is (-3,2)

and our minimum is (-1,-2)

Finding a maximum and minimum involves differentiating the function in order to determine when the gradient is at zero as this will be when the function has its maximum and minimum turning points.

take the function: f(x)=x3+6x2+9x+2

we must first obtain the gradient function through differentiation

d/dx = 3x2+12x+9 using standard differentiation technique (multiplying the power of the unknown by its coefficient to generate its new coefficient and then reduce its power by one to find the new term)

now that we have the gradient function it must be set to zero, since the gradient of the function is zero at a turning point

3x2+12x+9=0

this is a quadratic and we must solve for x to find the x values at which the gradient is zero

by inspection:

(3x+3)(x+3)=0 and therefore when the gradient of the function is zero:

x=-1 and x=-3

to determine which of these is maximum and is which is minimum we can differentiate the gradient function again. This generates a function that tells us the rate of change of the gradient at any given value of x

d/dx=3x2+12x+9

d2/dx2=6x+12 (using the same method as before)

if we substitue in our two x values:

when x=-1 d2/dx2= -6+12 = 6

since this is positive, the gradient is changing in a positive direction and so when x= -1 it is a minimum.

when x=-3 d2/dx2= -18+12 = 6

since this is negative the gradient is changing in a negative direction and so when x= -3 its a maximum.

to find the y values of these x values we simply subsitute them into the original function:

f(x)=x3+6x2+9x+2

f(-1)=-1+6-9+2=-2

f(-3)=-27+54-27+2=2

so our maximum is (-3,2)

and our minimum is (-1,-2)

Show more

2 years ago

6735 views

How do i solve simultaneous equations?

simulatneous equations are solved by eliminating one of the unknowns such as x or y. there are a number of ways to do this such as substitution or subtraction/addition. For this example we will use subtraction addition as i always found that the easier method at GCSE since it often avoids fractions.

e.g. 2y=3x+12 and 3y=5x+14

we need to choose one of the two unknowns to eliminate and then find their lowest common denominator. I will choose x, where the LCD is 15 (3*5=15). now we multiply the entire equation through in order that the number in front of x (called the coefficient) is 15. in the first equation, we multiply through by 5.

(2*5)y=(3*5)x+(12*5)

10y=15x+60

in the second equation we multiply through by 3 like before

(3*3)y=(5*3)x+(14*3)

9y=15x+42

now both the coefficients of x are 15 so we can subtract one equation from the other

(10-9)y=(15-15)x+(60-42)

y=0x+18

y=18

we now have the y value, since x is now zero. to find the x value all we need to do is go back to one of the first equations and put the value y=18 into the equation to resolve for x.

(2*18)=3x+12

36-12=3x

24=3x

x=8

and thats everything. By making sure we can subtract one equation from another to eliminate one of the unkowns we can subsequently obtain both values. in this case, y=18 and x=8 

simulatneous equations are solved by eliminating one of the unknowns such as x or y. there are a number of ways to do this such as substitution or subtraction/addition. For this example we will use subtraction addition as i always found that the easier method at GCSE since it often avoids fractions.

e.g. 2y=3x+12 and 3y=5x+14

we need to choose one of the two unknowns to eliminate and then find their lowest common denominator. I will choose x, where the LCD is 15 (3*5=15). now we multiply the entire equation through in order that the number in front of x (called the coefficient) is 15. in the first equation, we multiply through by 5.

(2*5)y=(3*5)x+(12*5)

10y=15x+60

in the second equation we multiply through by 3 like before

(3*3)y=(5*3)x+(14*3)

9y=15x+42

now both the coefficients of x are 15 so we can subtract one equation from the other

(10-9)y=(15-15)x+(60-42)

y=0x+18

y=18

we now have the y value, since x is now zero. to find the x value all we need to do is go back to one of the first equations and put the value y=18 into the equation to resolve for x.

(2*18)=3x+12

36-12=3x

24=3x

x=8

and thats everything. By making sure we can subtract one equation from another to eliminate one of the unkowns we can subsequently obtain both values. in this case, y=18 and x=8 

Show more

2 years ago

853 views

How to i calculate total resistance in a circuit?

Total resistance in a circuit depends on whether the resistors are in series or parallel since this determines which equation is to be used.

In series, resistors are essentially lined up next to each other so the total is simple since it is just the sum of the resistors resistance in ohms.

e.g. if a circuit has one 10ohm, one 45ohm and one 235ohm resistor in it, all three of which are in series, the total resistance is (10+45+235)ohm = 290ohms

In parallel each resistor is on its own path for the electricity to follow and as a result the equation is not quite the same, we use the equation:

 1/Rtotal = 1/R1 + 1/R+ 1/R..etc

e.g. a circuit has 3 resistors all in parallel. Their resistances are 25ohm, 5ohm and 100ohm respectivley, what is the total resistance of the circuit?

1/Rt = 1/25 + 1/5 + 1/100

1/R= 4/100 + 20/100 + 1/100

1/R= 25/100

100 = 25Rt

R= 4ohms

often the answer will be a fraction when calculating a parallel due to the fractions involved in the total calculation however i used a simpler calculation for the example to demonstrate the equation's use

Total resistance in a circuit depends on whether the resistors are in series or parallel since this determines which equation is to be used.

In series, resistors are essentially lined up next to each other so the total is simple since it is just the sum of the resistors resistance in ohms.

e.g. if a circuit has one 10ohm, one 45ohm and one 235ohm resistor in it, all three of which are in series, the total resistance is (10+45+235)ohm = 290ohms

In parallel each resistor is on its own path for the electricity to follow and as a result the equation is not quite the same, we use the equation:

 1/Rtotal = 1/R1 + 1/R+ 1/R..etc

e.g. a circuit has 3 resistors all in parallel. Their resistances are 25ohm, 5ohm and 100ohm respectivley, what is the total resistance of the circuit?

1/Rt = 1/25 + 1/5 + 1/100

1/R= 4/100 + 20/100 + 1/100

1/R= 25/100

100 = 25Rt

R= 4ohms

often the answer will be a fraction when calculating a parallel due to the fractions involved in the total calculation however i used a simpler calculation for the example to demonstrate the equation's use

Show more

2 years ago

3298 views

Arrange a free video meeting


To give you a few options, we can ask three similar tutors to get in touch. More info.

Contact Matthew

How do we connect with a tutor?

Where are they based?

How much does tuition cost?

How do tutorials work?

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok