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

Jonathan D.

Currently unavailable: for new students

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

5.2
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36 reviews| 69 completed tutorials

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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!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!

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Ratings & Reviews

5.2from 36 customer reviews
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Adi (Student)

January 2 2017

Thank you Jonathan for getting Adi up to spped in Physics and achieving the B that he got!

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Shawnee (Parent)

May 16 2017

Started the year off getting D's/C's in class mocks and past papers, Jonathan has had a huge impact on me raising my grades to the top region of A grade boundaries. Jonathan is a fantastic tutor and i would highly recommend him.

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Asfand (Student)

March 29 2017

Always explains everything very clearly and is very patient. Such a great teacher!

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Cindy (Student)

March 28 2017

Great use of mathematical methods to explain things. Simple and easy to understand and very efficient as well!!

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Qualifications

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

General Availability

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

Subjects offered

SubjectQualificationPrices
MathsA Level£22 /hr
PhysicsA Level£22 /hr
ChemistryGCSE£20 /hr
MathsGCSE£20 /hr
PhysicsGCSE£20 /hr

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 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.

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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2 years ago

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