**Me**

Hello, I'm Asher! I have recently completed an MEng (4 years) in Engineering Science at Keble College, University of Oxford. My degree is all about **problem solving, **where I've learn to solve all types of both practical and academic problems. It's also full of **maths** and **physics **and** how to apply them to real life!**

I love to **fully understand** **difficult problems** and then **help others do the same**, so we can work together to get to the bottom of everything you don’t understand.

I'm** **creative, cheerful and energetic and I want you all to get to the stage where **you are confident you can achieve your best.**

**The tutorials**

Each session will be determined by **what you want to do**. We will go back over the previous ground on which everything is built on and **fill in any pot holes in your knowledge**. We will then work on your problem until you are **confident** **with tackling it alone**. Throughout this process, you will **learn how to solve all sorts of related problems**, so you will be confident with tackling anything thrown at you in that subject.

I love making dull things **exciting**, so we will use loads of different ways to understand and remember things. The sessions will be **fun**,** rewarding**, and ensure that you **never fall asleep** in one!

And don’t worry, I won’t let you go away until you have fully understood something. I know that would be just far too frustrating…

**Oxbridge?**

Daunted by the long application process, the mass of colleges, the interviews and everything? Well, I’ve been there already, so I am here to **help you know best how to tackle it all**.

**Personal statement?**

Yes, another thing that just seems to take forever. **Let’s discuss it together** - it makes it much **quicker** and more fun, and I’ve had the **chance to speak to tutors about what they look for in one**.

**Any questions?**

Feel free to leave me a **message** with any questions you have, and arrange a 15 minute free session to have a chat. Look forward to meeting you!

**Me**

Hello, I'm Asher! I have recently completed an MEng (4 years) in Engineering Science at Keble College, University of Oxford. My degree is all about **problem solving, **where I've learn to solve all types of both practical and academic problems. It's also full of **maths** and **physics **and** how to apply them to real life!**

I love to **fully understand** **difficult problems** and then **help others do the same**, so we can work together to get to the bottom of everything you don’t understand.

I'm** **creative, cheerful and energetic and I want you all to get to the stage where **you are confident you can achieve your best.**

**The tutorials**

Each session will be determined by **what you want to do**. We will go back over the previous ground on which everything is built on and **fill in any pot holes in your knowledge**. We will then work on your problem until you are **confident** **with tackling it alone**. Throughout this process, you will **learn how to solve all sorts of related problems**, so you will be confident with tackling anything thrown at you in that subject.

I love making dull things **exciting**, so we will use loads of different ways to understand and remember things. The sessions will be **fun**,** rewarding**, and ensure that you **never fall asleep** in one!

And don’t worry, I won’t let you go away until you have fully understood something. I know that would be just far too frustrating…

**Oxbridge?**

Daunted by the long application process, the mass of colleges, the interviews and everything? Well, I’ve been there already, so I am here to **help you know best how to tackle it all**.

**Personal statement?**

Yes, another thing that just seems to take forever. **Let’s discuss it together** - it makes it much **quicker** and more fun, and I’ve had the **chance to speak to tutors about what they look for in one**.

**Any questions?**

Feel free to leave me a **message** with any questions you have, and arrange a 15 minute free session to have a chat. Look forward to meeting you!

Enhanced DBS Check

01/05/20164.9from 19 customer reviews

Jacob (Parent from Cinderford)

October 14 2016

'Good, mild-tempered.' (William)

Manon (Student)

October 13 2016

Very helpful

Pierre (Student)

October 8 2016

Jacob (Parent from Cinderford)

December 12 2016

(The rate at which the temperature of a body falls is proportional to the difference between the body temperature and the room temperature) - this would be given in the question.

This is a typical ordinary differential equation (ODE) question, which can be solved by separation of variables.

It is best looked at in three stages: Write the general ODE, then find the boundary conditons, then solve and find the equation for temperature in terms of time.

__________________________________________

1. General ODE: let's call temperature T, and time t. Change is temperature over time is written as dT/dt. From the question, dT/dt is proportional to (T-18).

So dT/dt = -k(T-18) where k is the constant of proportionality and the minus sign arrises because the pan is cooling not heating.

__________________________________________

2. Boundary conditions:

The pan of water is heated to 100C which is at t =0, so T(t=0) = 100.

After 5 minutes, the pan cools to 80C. So T(t=5) = 80

__________________________________________

3. Solve the equation and substitute in the boundary conditions.

Firstly, we separate the variables and integrate.

dT/dt = -k(T-18)

dT = -k(T-18) dt

dT/(T-18) = -k dt

Add the integral signs (int):

int dT/(T-18) = int -k dt

and solve:

ln(T-18) = -kt + C where C is the constant of integration.

Now do exp of both sides. exp is a way of wirting e to the power of ( ):

T - 18 = exp(-kt + C)

T - 18 = A exp(-kt) where A = exp(C). This is just used to simplify the problem.

T = 18 + A exp(-kt)

__________________________________________

Seconly, we substitute in the boundary conditions. Use T(t=0) = 100 to start with.

This gives:

100 = 18 + Aexp(0)

so A = 82

T = 18 + 82 exp(-kt)

Now it's time to use the 2nd boundary condition, T(t=5) = 80.

This gives:

80 = 18 + 82 exp(-5k)

Subtract 18 from both sides, and divide by 82:

(80 - 18)/82 = exp(-5k)

Take ln of both sides and then divide by -5

k = -(1/5) ln(82/62)

k = 0.0559

So the final solution is:

T = 18 + 82 exp(-0.0559t)

We now have the general solution, which can be used to find any further information about the quesiton we need.

__________________________________________

The question asks for the temperature at t=10. Simply substitute t-10 into the final solution above.

T = 18 + 82 exp(-0.0559 x 10)

T = 64.0C (three sig fig)

(The rate at which the temperature of a body falls is proportional to the difference between the body temperature and the room temperature) - this would be given in the question.

This is a typical ordinary differential equation (ODE) question, which can be solved by separation of variables.

It is best looked at in three stages: Write the general ODE, then find the boundary conditons, then solve and find the equation for temperature in terms of time.

__________________________________________

1. General ODE: let's call temperature T, and time t. Change is temperature over time is written as dT/dt. From the question, dT/dt is proportional to (T-18).

So dT/dt = -k(T-18) where k is the constant of proportionality and the minus sign arrises because the pan is cooling not heating.

__________________________________________

2. Boundary conditions:

The pan of water is heated to 100C which is at t =0, so T(t=0) = 100.

After 5 minutes, the pan cools to 80C. So T(t=5) = 80

__________________________________________

3. Solve the equation and substitute in the boundary conditions.

Firstly, we separate the variables and integrate.

dT/dt = -k(T-18)

dT = -k(T-18) dt

dT/(T-18) = -k dt

Add the integral signs (int):

int dT/(T-18) = int -k dt

and solve:

ln(T-18) = -kt + C where C is the constant of integration.

Now do exp of both sides. exp is a way of wirting e to the power of ( ):

T - 18 = exp(-kt + C)

T - 18 = A exp(-kt) where A = exp(C). This is just used to simplify the problem.

T = 18 + A exp(-kt)

__________________________________________

Seconly, we substitute in the boundary conditions. Use T(t=0) = 100 to start with.

This gives:

100 = 18 + Aexp(0)

so A = 82

T = 18 + 82 exp(-kt)

Now it's time to use the 2nd boundary condition, T(t=5) = 80.

This gives:

80 = 18 + 82 exp(-5k)

Subtract 18 from both sides, and divide by 82:

(80 - 18)/82 = exp(-5k)

Take ln of both sides and then divide by -5

k = -(1/5) ln(82/62)

k = 0.0559

So the final solution is:

T = 18 + 82 exp(-0.0559t)

We now have the general solution, which can be used to find any further information about the quesiton we need.

__________________________________________

The question asks for the temperature at t=10. Simply substitute t-10 into the final solution above.

T = 18 + 82 exp(-0.0559 x 10)

T = 64.0C (three sig fig)

**Solve for x: 2x ^{2} + 7x + 6 = 0**

The first thing to do is look at the left hand side and factorise it. To do this we use the reverse of FOIL. Then when the expression is factorised, we can solve for x.

1. First find the skeleton with just x's:

(2x )(x )

2. Next, find 2 numbers that mutiply to make 6, and these go at the end of the brackets. This could be 6 and 1 or 2 and 3, or -6 and -1 or -2 and -3.

Since there is a 7x term, the 2 numbers must add to 7 when one of them is multiplied by 2 - due to the 2x term (2x ).

so (2x 3)(x 2) or (2x 2)( 3)

3. Trial and error shows us that (2x + 3)(x + 2) is the right answer.

4. Now set each bracket respectively to 0 and solve for x. The solutions are x = -3/2 and x = -2.

And there we have the answers!

**Solve for x: 2x ^{2} + 7x + 6 = 0**

The first thing to do is look at the left hand side and factorise it. To do this we use the reverse of FOIL. Then when the expression is factorised, we can solve for x.

1. First find the skeleton with just x's:

(2x )(x )

2. Next, find 2 numbers that mutiply to make 6, and these go at the end of the brackets. This could be 6 and 1 or 2 and 3, or -6 and -1 or -2 and -3.

Since there is a 7x term, the 2 numbers must add to 7 when one of them is multiplied by 2 - due to the 2x term (2x ).

so (2x 3)(x 2) or (2x 2)( 3)

3. Trial and error shows us that (2x + 3)(x + 2) is the right answer.

4. Now set each bracket respectively to 0 and solve for x. The solutions are x = -3/2 and x = -2.

And there we have the answers!