Currently unavailable: for regular students
Degree: Natural Sciences (Physical) (Bachelors) - Cambridge University
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.
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!
|Maths||A Level||£22 /hr|
|Physics||A Level||£22 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Catherine (Parent) January 14 2016
Catherine (Parent) November 10 2015
Catherine (Parent) November 3 2015
Catherine (Parent) October 20 2015
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.
-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.see more