PremiumJonathan D. A Level Maths tutor, GCSE Physics tutor, GCSE Chemistry t...

Jonathan D.

Currently unavailable: for regular students

Degree: Natural Sciences (Physical) (Bachelors) - Cambridge University

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About me

About Me:

I am a second year Natural Sciences student at Cambridge, where I now study primarily physics! I am massively passionate about all things Maths and Science, and I would love to be able to help you achieve success in either subject.

The Session

I am a firm believer that, when it comes to Maths and Science, having a really good understanding of the basic principles is more than half the battle! Personally I have always found past paper questions a great way of preparing for examinations and consolidating knowledge, so I will do my best to incorporate them into the sessions as and when. However, I am very flexible when it comes to the structure of the sessions: THE MOST IMPORTANT THING is that it works for you!!!

If you have any questions, please feel free to ask! I would be delighted to answer them!

Subjects offered

SubjectLevelMy prices
Maths A Level £22 /hr
Physics A Level £22 /hr
Chemistry GCSE £20 /hr
Maths GCSE £20 /hr
Physics GCSE £20 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelA*
PhysicsA-LevelA*
ChemistryA-LevelA*
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

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Ratings and reviews

5from 13 customer reviews

Catherine (Parent) January 14 2016

As ever, Jonathan is very helpful and gives very clear explanations.

Catherine (Parent) November 10 2015

Very helpful, as ever.

Catherine (Parent) November 3 2015

Very helpful, as ever. It helped Samuel's confidence in working with experiment-style questions.

Catherine (Parent) October 20 2015

Thank you Jonathan for another very helpful session.
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Questions Jonathan has answered

How do you find the equation of a tangent to a curve at a particular point?

Imagine being given the equation y=x3-2x+3, and being asked to find the tangent to the curve at the point where x=1. The tangent to the curve will be a straight line, and therefore will take the form y=mx+c. To find m (the gradient of the tangent), it is necessary first of all to differe...

Imagine being given the equation y=x3-2x+3, and being asked to find the tangent to the curve at the point where x=1.

The tangent to the curve will be a straight line, and therefore will take the form y=mx+c.

To find m (the gradient of the tangent), it is necessary first of all to differentiate the equation of the original curve. Doing this gives y’=3x2-2, where y’ is the gradient of the curve at a particular point. We are looking for the gradient at the point where x=1. Therefore, to find m, we must substitute x=1 into our expression. Doing so, we find that m=1.

We now know the equation of the tangent is y=x+c. To find c (the y-intercept), we must first of all know the coordinates of a point that the tangent is going to pass through. In our case, we know that the tangent must pass through the point on the line where x=1. To find the y-coordinate of this point, we can sub x=1 into our original equation of the curve. Doing so, we find that the point we must use is (1,2).

Now that we know a point on the line, we can sub those x and y values into the expression y=x+c. This gives us the equation 2=1+c, and some quick rearrangement shows us that c=1.

Therefore the equation of the tangent is y=x+1.

In summary:

-Equation of tangent is of the form y=mx+c

-To find m, differentiate the equation of the curve to find its gradient at the required point

-Find the coordinates of a point the tangent is going to pass through, and sub into the equation of the tangent to find c.

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1 year ago

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