#### Related to [[Integration]] --- ## Integration and [[Motion dot diagrams|physics]] ##### *Velocity, acceleration and displacement* - Since velocity is a measure of the **rate of change of position** it should follow that the *derivative* of an equation describing position $x$ in terms of time $t$ is equal to the velocity $v$. Likewise, acceleration is a measure of the **rate of change of velocity** so the *derivative* of an equation describing velocity $v$ in terms of time $t$ is equal to the acceleration $a$. - Since integration **is the inverse of derivation we can apply these principles backwards as well.** The indefinite integral of **an acceleration vs time equation is equal to *velocity vs time equation* and the indefinite integral of a velocity vs time equation is equal to *position vs time equation*.** The definite integral of acceleration is equal to the **net change in velocity** and the likewise, the definite integral of acceleration is equal to the *total displacement*. - This relates to the **net change theorem**, for more info see [[Integration]]. $\large\int a(t)\,dt = v(t)\;\;\text{\color{grey} and }\;\int v(t)\,dt =x(t)$ $\newline$ ##### *Work done* - In physics work is a measure of **force over distance** *(see [[Energy]])*. If the force exerted on an object is continuously changing we can find the work done over that **distance through *integration***. *NOTE: This only works if the force exerted on the object can be represented as a function of either distance or time.* - If the mass of an object is **given in pounds you do not have to multiply it by the acceleration due to gravity g**. This is because pounds is already *a unit of force* unlike g and kg which is a unit of **mass**. - The work done to an object is equal to the **integral of a function describing the force per distance - $x$.** In this case, both a and b represent the **positions** which bound the area over which we are trying to find the total force. $\large w=\int_a^b F(x)\,dx$ $\small\color{grey}\text{where }F(x)\text{ is force as a function of distance, denoted }x$ - If the integrand instead **describes force as a function of *time* we must add an additional function that describes distance or position per time $t$.** In this situation the bounds $[a,b]$ denote quantities of time *(the starting $a$ and ending $b$ time values).* $\large w =\int_a^b F(t)\cdot x(t)\,dt$ $\small\color{grey}\text{where }F(t)\text{ is force as a function of time and }x(t)\text{ is position as a function of time, denoted as }t$ --- ## Average value of a function - To find the average of a set of continuously changing data that data must be able to be represented by a function. This is because for a **perfect average we need to sample an infinite amount of data points.** - The variable that this function is in terms of, is usually time, given by $t$. However this function can be in terms of any value so long as we can use that value to dictate the start and end of the bounds on which we will sample the function's average. - By taking the integral of the function and dividing the results by the difference between the bounds *(the "range of the integral")* we can find the average value of that function. $\newline$ $\large\text{Formula for average of a function}$ $\newline$ $\large \lim_{n\rightarrow\infty}\sum_{i=0}^{n}f\biggl(\frac{(a-b)i}{n}\biggl)\;\frac{1}{n}$ $\therefore\,\large\text{Average value of }f(x) =\color{LightCyan}\frac{1}{b-a}\int_a^bf(x)\;dt$ $\newline$ >[!tip] >##### *Average value tip* >- The **integral of the average value of a function is ==equal to the integral of the function itself *(on the same bounds)*.==** In other words the area under the average value of a function is equal to the area the same function. This should be somewhat geometrically intuitive. > >![[Integral applicationsavgvaluething.png|850]] --- ## Arc length - The **arc length of a function** is the length of the line that the function makes when graphed out on an **x, y plot**. - Changing the location of a function **rigidly** does not effect the arc length of that function as long as you also shift the bounds by the same amount. $\text{Arc length of a function: }\;\Huge\boxed{\normalsize\color{LightCyan}l=\int_a^b\sqrt{\Big(f'(x)\Big)^2+1}}$ $\newline$ --- ## Center of mass - To find the center of mass of a set of point masses in a **2D plane** we have to use the general center of mass equation *(See [[Rotational and Circular motion]])* for both the **x component and y component of each point masses' position**. - *In other words we calculate the center of mass along the x axis and the center of mass along the y axis and the resulting two answers should be equal to the position on the 2D plane of the center of mass.* - *NOTE: This also works in 3D.* - A thin sheet is known as a **lamina**. - The **"moment"** of a shape about some axis is essentially the average position of that shape along that axis taking into account how the mass of that object is distributed. ##### *Finding the center of mass of a set of discrete masses (2D or 3D):* - The moment about the y axis is: $My = \sum_{i=1}^n m_i\,x_i$ where $m_i$ is the mass of the object at distance $x_i$ from the y axis. - The moment about the y axis is: $M_x=\sum_{i=1}^n m_i\,x_i$ where $m_i$ is the mass of the object at a distance $y_i$ from the x axis. $\text{Center of mass}\;\;\;\Huge\boxed{\normalsize\color{LightCyan}x_{cg}=\Biggl(\frac{M_y}{\sum m_i},\frac{M_x}{\sum m _i}\Biggl)}$ ##### *Finding the center of mass of 2D objects* - If an object is not made up of individual **discrete point masses** we must use integration to find the center of mass of that object. This is because the mass is distributed **continuously throughout the object**. - To do this we must first find the moment about the y and x axis. Then we must find the total mass of the object and combine that with the x and y moments to get a set of coordinates representing the center of mass of the object as a whole. - In the following equations the variable $\small\rho$ is used to denote the **density of the object**. - In the following equations we are assuming that the density of the object in the **same at all points**. $\newline$ $\text{Moment about the y axis: }\;\Huge\boxed{\normalsize M_y=\int_a^b \rho\Big(f(x)-g(x)\Big)x\;dx}$ $\text{Moment about the x axis: }\;\Huge\boxed{\normalsize M_x=\int_a^b \frac{1}{2}\rho\Big(f(x)^2-g(x)^2\Big)\,dx}$ $\text{Total mass: }\;\Huge\boxed{\normalsize M_{total} = \int_a^b \rho\Big(f(x)-g(x)\Big)\,dx}$ $\newline$ $\text{Center of mass: }\;\color{LightCyan} X_{cg}=\Biggl(\frac{M_y}{M_{total}},\frac{M_x}{M_{total}}\Biggl)$ $\newline$ >[!bug] >##### *Important note:* >- If the 2D shape you are examining is bounded by one function you should still use the above equations. Instead of plugging in a function for $\small g(x)$ or $\small f(x)$ *(which ever is smaller)* you can just plug in the value of the line *(on x or y)* that binds the top or bottom of the object. > - For example, if the object was bounded between a function and $\small x=1$ you would plug in 1 for $\small g(x)$ *(assuming that the function is greater on the bounds than $\small x=1\;$)* --- #subpage