Search over 10,000 free study notes

We can find the gradient of a tangent to a curve at a point by finding dy/dx at x=1.Firstly we can rearrange the equation of the curve for y.(1) yx^2=1-x^3(2)y=x^{-2}-xThen we can differentiate the equati...

Consider two scenarios before and after. Before there is a velocity of 8m/s (with right being positive) and mass 1kg. after there is a mass of 1kg and a velocity of -4 m/s as right is positive and it boun...

A stationary point can be found when dy/dx = 0. The first thing we need to do is differentiate y to find dy/dx, and solve it for dy/dx = 0. This gives us...

First we make y the subject of the l1 equation. This is so we can have it in the form y=mx +c where m is the gradient and c is your y-intercept. For this example we would take away 2x from both sides leav...

First we need to consider what a straight line looks like:
y = mx + c (if we get something different we've gone wrong somewhere)one way to get the equation is using a point (general point is (x_1...