# Maths Extension 1:Applications of Calculus to the Physical World

 BikiCrumbs: Applications o… Physical World

# The Equation $\frac{dN}{dt} = k(N - P)$

The differential equation $\frac{dN}{dt} = k(N - P)$ has important and interesting applications, most notably in this course is the use of it in Newton's Law of Cooling and exponential growth and decline.

The differential equation has an analytical solution of $\ N = P + Ae^{kt}$ (where P, A and k are constants), which can be derived straight from separating the variables of the differential equation and integrating, as follows:

$\frac{dN}{dt} = k(N - P) \Rightarrow \; \frac{dN}{N - P} = k dt$

Integrating both sides:

$\int_{}^{} \frac{dN}{N - P} = \int_{}{} kdt$

$\ ln (N - P) = kt + C$ (for some constant C)

$\ N - P = e^{kt + C}$

$\ N = P + e^{kt + C}$

$\ N = P + e^{C}e^{kt}$

$\ \mbox{Let }A = e^C$ (since C is a constant, therefore $\ e^{C}$ is also a constant)

$\ N = P + Ae^{kt}$

# Projectile Motion

Projectile motion describes the motion of objects whose path can be described as a parabola. Objects undergoing are influenced by the force of gravity, which acts downwards in the vertical direction. There are no horizontal forces that act on the object. Because of the nature of horizontal and vertical forces, projectile motion is most commonly described parametrically. The equations of motion are easily derived with the initial conditions that horizontal acceleration $\ (\; \ddot x \;)$ is 0 and vertical acceleration $\ (\; \ddot y \;)$ is $\ -g$, the acceleration due to gravity. (Note: the negative sign is chosen, by convention, to indicate anything going in the downwards direction, whilst upwards is taken as positive. Taking downwards as positive and upwards as negative will still work so long as the notation is consistent throughout the calculations)

## Deriving Equations of Motion

A vector can be divided into its horizontal component, and its vertical component. If we let the velocity of the projectile be V, the vertical component of the velocity be y' and the horizontal component of the velocity be x', then by using trigonometry if the acute angle of the velocity vector is measured from the horizontal plane (let this be @), then sin@ = y/V and cos@ = x/V

### Horizontal Component

Therefore, the Horizontal Component is x' = Vcos@

Integrating with respect to time, gives the function for horizontal displacement. x = Vtcos@

### Vertical Component

Therefore, the Verticle Component is y' = Vsin@ There is also gravity(g) due to acceleration, therefore y' = Vsin@ -gt (where t=time)

Integrating with respect to time, gives the fuction for vertical displacement. y = Vtsin@ - g(t^2)/2

## Cartesian Equation of the Path

Make 't' the subject of the horizontal component/equation and substitute it into the vertical component to find 'y' in terms of 'x' or the Cartesian equation

## Range, Maximum Height and Other Properties of Projectiles

### Range

You must understand that the range is when the graph or 'projectile' touches the x-axis for the second time. In other words, find the x-intercept that is not 0.

Most simply put, using the Cartesian equation, let y=0 and find x .

Otherwise find the variable 't' when y=0 and substitute this value into the horizontal component/equation

### Maximum Height

You must understand that the maximum height of a 'projectile' is the greatest vertical point on the graph of the equation defining its motion.

Finding the maximum height of a projectile is much like finding the maximum value for any standard graph; find when its derivative is equal to 0. Or y'=0 in mathematical language.