Search over 10,000 free study notes

Let the lines be: r 1 = (a, b, c) + t(d, e, f) r 2 = (g, h, i) + s(j, k, l) Since t and s above are variables and the rest of the letters are constants, the only way to change the point which the vector equa...

Firstly let's take a right angle triangle. Label its sides a,b, c for the hypotenuse. Then arrange 4 of these triangles into a large square, with a smaller square contained within it. The area of the smaller...

Integrating this expression is a simple trick. We use integration by parts. For this we need a function we can integrate and a function we can differentiate. We know how to differentiate ln(x) which is 1/x. ...

We can start with the identity sin 2 (x)+cos 2 (x)=1 If we divide through the equation by cos 2 (x), we get: sin 2 (x)/cos 2 (x) + cos 2 (x)/cos 2 (x) = 1/cos 2 (x) If we look at the left hand side of the eq...

To get the required tangent equation we need its gradient and the coordinates of a point it passes through. We can then substitute it into the formula y - y 1 = m(x - x 1 ). 1. Finding the gradient: To find ...