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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...

finding the roots, means to find the coordinates at which the curve intersects the X axis. At first look you may notice that each function in the equation is divisible by 2, you can do this to make solvin...