__Motivation and Experience:__

At school, Maths was by far my favourite subject. My sole motivation for tutoring Maths is to instill the same passion I had for maths at school with those who are looking for some extra help.

I am very familiar with the GCSE syllabus as my younger brother is currently studying for his Maths GCSE, and I'm often his first port of call when he needs help having things explained or taught to him.

Also, I sat my Maths A-Level in year 11, so throughout sixth form I offered informal tutoring to friends studying for their Maths A Level exams, and am extremely familiar with the A Level syllabus, in particular the core modules.

__Tutoring Sessions:__

I will ensure that tutorials I run are engaging by using a number of sources; videos, simple examples, diagrams and past exam questions. I believe that Maths is much more than passing exams and hope to convey this passion for the subject to those I tutor.

**Extra Information:**

At A Level, I also studied Physics and Economics and am willing to offer help on both those two subjects if requested to.

If you have any further questions for me, please do not hesitate to contact me at em13531@my.bristol.ac.uk. I look forward to meeting you.

__Motivation and Experience:__

At school, Maths was by far my favourite subject. My sole motivation for tutoring Maths is to instill the same passion I had for maths at school with those who are looking for some extra help.

I am very familiar with the GCSE syllabus as my younger brother is currently studying for his Maths GCSE, and I'm often his first port of call when he needs help having things explained or taught to him.

Also, I sat my Maths A-Level in year 11, so throughout sixth form I offered informal tutoring to friends studying for their Maths A Level exams, and am extremely familiar with the A Level syllabus, in particular the core modules.

__Tutoring Sessions:__

I will ensure that tutorials I run are engaging by using a number of sources; videos, simple examples, diagrams and past exam questions. I believe that Maths is much more than passing exams and hope to convey this passion for the subject to those I tutor.

**Extra Information:**

At A Level, I also studied Physics and Economics and am willing to offer help on both those two subjects if requested to.

If you have any further questions for me, please do not hesitate to contact me at em13531@my.bristol.ac.uk. I look forward to meeting you.

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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The **normal line** to a curve at a particular point is the line through that point and *perpendicular* to the tangent.

A simple trick to remembering how to find the normal gradient, ** n**, is that the slope of any line perpendicular to a line that has a gradient,

__Example:__

**Find the normal gradient to the curve **y=2x^{3 }+3x+7 **at the point (1,1).**

So firstly, let’s recap how to calculate the gradient of the tangent line:

By differentiating y=2x^{3 }+3x+7 , we find

dy/dx = 6x^{2} +3

Then, by substituting in our point, at x=1 we yield **dy/dx=9**. This is the tangent gradient of the curve (m=9).

Finally we substitute this into our formula for calculating the normal gradient n=-1/m.

Therefore **n=-1/9**.

Now, let’s try another example which demonstrates how we use the normal gradient to find an equation for the normal line.

We will use the formula (y-y_{0}) = n(x-x_{0}), where (x_{0},y_{0}) is a given point.

__Example:__

**Consider a curve **y=x^{5}+3x^{2} +2. **Find the equation of the normal to the curve at the point (-1,2). Leave your answer in the form y=mx+c.**

By differentiating the curve, we have dy/dx = 5x^{4 }+6x.

To find the gradient of the tangent line we substitute in x=-1, which yields

dy/dx = 5(-1)^{4} +6(-1)

= 5-6

=-1 = m

Therefore, we know that the normal gradient is n=-1/m

So **n=1**

Finally, we substitute this into our formula for the normal line (y-y_{0}) = n(x-x_{0}):

In our example, (x_{0}, y_{0}) = (-1,2)

So y-2 = 1(x+1)

And my rearranging, we find **y = x+3.**

The **normal line** to a curve at a particular point is the line through that point and *perpendicular* to the tangent.

A simple trick to remembering how to find the normal gradient, ** n**, is that the slope of any line perpendicular to a line that has a gradient,

__Example:__

**Find the normal gradient to the curve **y=2x^{3 }+3x+7 **at the point (1,1).**

So firstly, let’s recap how to calculate the gradient of the tangent line:

By differentiating y=2x^{3 }+3x+7 , we find

dy/dx = 6x^{2} +3

Then, by substituting in our point, at x=1 we yield **dy/dx=9**. This is the tangent gradient of the curve (m=9).

Finally we substitute this into our formula for calculating the normal gradient n=-1/m.

Therefore **n=-1/9**.

Now, let’s try another example which demonstrates how we use the normal gradient to find an equation for the normal line.

We will use the formula (y-y_{0}) = n(x-x_{0}), where (x_{0},y_{0}) is a given point.

__Example:__

**Consider a curve **y=x^{5}+3x^{2} +2. **Find the equation of the normal to the curve at the point (-1,2). Leave your answer in the form y=mx+c.**

By differentiating the curve, we have dy/dx = 5x^{4 }+6x.

To find the gradient of the tangent line we substitute in x=-1, which yields

dy/dx = 5(-1)^{4} +6(-1)

= 5-6

=-1 = m

Therefore, we know that the normal gradient is n=-1/m

So **n=1**

Finally, we substitute this into our formula for the normal line (y-y_{0}) = n(x-x_{0}):

In our example, (x_{0}, y_{0}) = (-1,2)

So y-2 = 1(x+1)

And my rearranging, we find **y = x+3.**