|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
The first thing to remember is that the tangent to the curve at a given point has the same gradient as the curve at that point.
Let's consider the general case y=f(x) at a point c=(x1,y1).
Step 1:We have to differentiate to find dy/dx.
Step 2: Calculate dy/dx when x=x1. This is the gradient at the point c.
Step 3: Use the equation of a general straight line with gradient m at a point c 'y-y1=m(x-x1)', where m is the value of the gradient calculated in Step 2.
Here is an example. Find the equation of the tangent to the curve y=x^3-4x at the point (1,-3).
In this example x1=1 and y1=-3
Step 1: dy/dx=3x^2-4
Step 2: When x=1, dy/dx =-1, so m=-1
Step 3: Use the equation y-y1=m(x-x1) to obtain y--3=-1(x-1)
We can then rearrange this to the nicer form of y=-x-2.see more
Firstly, note that cos^2(x)+5sin^2(x)= cos^2(x) +sin^2(x) +4sin^2(x).
By trignoemtric identies, cos^2(x)+sin^2(x)=1 and so we can just integrate 1+4sin^2(x) since this is equal to cos^2(x)+5sin^2(x).
Again, by trignometric identities, 4sin^2(x)=4(1/2-1/2 cos(2x))=2-2cos(2x),
and so 1+4sin^2(x)=3-2cos(2x).
We can now integrate this much more easily...
3 integrates to 3x and -2cos(2x) integrates to -sin(2x).
Hence the integral, remembering the constant of integration, is...
3x -sin(2x) +csee more