Calculus is a branch of mathematics that focuses on describing change. It was originally discovered by Isaac Newton and Gottfried Liebniz.
Most real life objects in the Universe around us have a variable rate of change. So do, for example curves on a graph, as they have a variable gradient.
We previously learnt how to find the gradient of a curve at a point by drawing a tangent. Differentiation allows us to find the gradient without having to draw a tangent.
All differentiation is a function where we take the input as the equation of a line and output the gradient function.
$$y = ax^n \Rightarrow \dfrac{dy}{dx} = anx^{n-1} $$
Let’s find the gradient function for the following equation of a line:
$$ y = 2x^2 + 3 $$ $$ \dfrac{dy}{dx} = 2 \times 2x^{2 - 1} $$ $$ \dfrac{dy}{dx} = 4x $$
The gradient function equals the gradient so we can use it to either calculate the value for x by setting the equation equal to the gradient OR we can find the gradient at a point by inputting the value of x.
The maximum point, minimum point, stationary point and turning point all mean the same thing; they are a point on a curve where the gradient equals 0. We can find the coordinates for this by setting the gradient function equal to 0. For example, let’s find the stationary point for the above equation:
$$ \dfrac{dy}{dx} = 0 $$ $$ 0 = 4x $$ $$ x = 0 $$
We can now find the y coordinate by subsituting this into the original equation:
$$ y = 2(0)^2 + 3 $$ $$ y = 3 $$
Therefore are stationary point is:
$$ (0, 3) $$
We can tell whether a quadratic curve is a maximum or minimum point depending on the gradient or by looking at the graph itself.
We can also apply differentiation to physics.
Velocity is the rate of change of displacement compared to time (velocity equals change in displacement divided by change in time):
$$ v = \dfrac{\Delta s}{\Delta t} $$
Acceleration is the rate of change of velocity compared to time (acceleration equals change in velocity divided by change in time):
$$ a = \dfrac{\Delta v}{\Delta t} $$
We can find an equation for each by knowing 1 of these using differentiation:
$$ \text{displacement} \Rightarrow \text{velocity} \Rightarrow \text{acceleration} $$
For example we are given this equation:
$$ s = 24t^2 - t^3 $$
where s is displacement and t is the time.
and we are asked to find expressions for the velocity and the acceleration:
$$ \dfrac{ds}{dt} = v = 48t - 3t^2 $$
$$ \dfrac{dv}{dt} = a = 48 - 6t $$
We can find the acceleration or velocity at a particular time by substituting it into the expressions we have made.
More being added soon…