[Music] so that those equations are true no matter what coordinate system we use we could choose polar coordinates uh rectangular components we could choose cylindrical components any kind of component system you could think of that works because we've just represented those in terms of vectors and the vectors can be represented however we want typically um a lot of applications it's quite convenient to use rectangular components we exist in threedimensional space and most thing most things uh are quite convenient to model or write mathematically in terms of reque rectangular or cartisian components so how we do this um is typically we'll write a vector and it's XYZ components which are rectangular components and they're always relative to a fixed reference frame so here we have XYZ is our fixed reference frame and we have our position Vector we wrote that as R and in this case we can say R is equal to distance in x times our X Direction unit Vector so we'll call that I hat which is a unit Vector in the X Direction plus our y component plus our Direction in the y direction plus our Z component times our unit Vector in The Zed Direction so i j k j k these are unit vectors in the X Y and Z directions and then X which is a function of time Y which can be a function of time Z can also be a function of time r just the uh components of the magn in those directions so these are our X Y Zed components so we described before the magnitude think of magnitude as distance the total distance away from An Origin so the length of a vector is the magnitude so we can write the magnitude as the magnitude of R is equal to the sum of the squares of the components all square rooted so square root of X2 + y^2 plus z s this should be familiar just thean theorem um and then lastly because it's a vector it has a magnitude and a direction we can write the direction as a unit Vector a unit Vector direction that is along that position Vector line so we can say the direction um we'll Define that as Ur which is our unit Vector in the direction of our position vector and that is equal to to our Vector rid the magnitude of R so that's position can essentially do the exact same analysis again with velocity so let's start with our definition of velocity we said that velocity was equal to time derivative Dr DT we've written our uh position Vector as components X Y and Z so let's substitute that in we get the time derivative D DT of our position Vector so our position Vector was x i hat plus y j hat plus K hat so since we have taken a fixed reference frame the components are constant they're constant in magnitude they're unit vectors they're always going to have a magnitude of one and they're constant in Direction because our reference frame isn't moving just the particles is moving with respect to our fixed XY Z so if I J and K aren't moving with respect to time then they can come outside of our derivative and then we were left with a simplified version and we can say velocity Vector is equal to DX DT * I plus Dy DT J DT * K so this would be our definition of our velocity with cartisian coordinates now what we call each of these these uh values here dxdt this is our X component of velocity which we can also just call x dot if we ever see a variable with a DOT above it uh that's referring to the time derivative if you see two dots above it it's the second time derivative so VX is our X component of velocity same thing with our dydt this is equal to our VY which is equal to Y Dot and then d z DT is equal to v z is equal to Z dot so those are the X Y and Z components of our velocity Vector B same thing as with position our position Vector has a magnitude and Direction so we can also write the magnitude of velocity is equal to the square root of the sum of the components squared BX squar b y^ 2 plus b z^ 2 and we can write the direction is equal to D direction is equal to that velocity Vector divided by that magnitude and it all it is always tangent to the path so that's velocity three straight forward it's just the time derivative of our position Vector we have rectangular components VX v y VZ which are XY Z components of velocity and we have a definition for our magnitude um and our Direction label this here all right lastly acceleration and this is where sometimes it could get a bit confusing but acceler if we just follow the same steps can break down and it doesn't become too difficult so we set acceleration is equal that time derivative dbdt which is equal to d^2 r squared and if we go by the same process as before or we can rewrite this as uh the acceleration components ax i a y j plus a z k so ax a y and a z are all the components of acceleration acting in the XY Z Direction and we can write ax is equal to of our velocity in the X Direction which is equal to the double time derivative of our X position which we can also VX DT so if we know our component velocity in the X Direction you take time to that we get our component velocity in the X Direction and we can write the same thing for all the other directions a y VY dotal to Y double dot equal the derivative of Dy DT and AZ is equal to VZ dot or our Z component of our position Z double dot tbz DT so notice when we write these in component form I'm not drawing Vector arrows above these because these are no longer vectors these are scalar quantities and they're just the components of uh the vector quantity in each Direction so they're the coefficients that would go in front of our I's JS and KS they're just scalar values so again we same thing with magnitude total magnitude of our acceleration X2 a y^ 2 plus a z^ 2 all to the power 12 and we can write the direction is equal to the magn vector aided by its magnitude so we've talked about position velocity acceleration now let's write an example so a box sliding down a ramp we have an equation for the path Y is equal to 05 x^2 that's in meters um it gives us a velocity component minus 3 meters perss and ax is equal to minus 1.5 m/s squared at X is equal to 5 m and then it's asking find the Y components of the velocity and acceleration of the box at that same point when X is equal to 5 meters so let's begin just by writing down some equations that we know we're asked to find velocity and acceleration the Y components so what is the Y component of velocity acceleration so y component of our velocity VY this is equal to DX DT or we can just write it as X sorry dydt which is y so we have an equation for y let's use that let's take the time derivative of our function y is equal to 05x 2 so dydt is equal to to the derivative with respect to time [Music] 0.05 X2 how do we take the derivative of a function that's not really a function of time um would you just take the derivative of the first term and then multiply by the x dot yes exactly very good so X is a function of time so we're going to kind of take the derivative of this with respect to time implicit so the derivative we're going to pretend for now X is just the function of time so we would get 2 * 0.05 X and then to complete this we have to write this using the chain rule times dxdt which is equal to x dot we do this we have an equation for our y component of velocity is 2 * 05x * x dot so we have basically we're going to solve when X is equal to five and we're given our X component of velocity VX which is x dot so we can rewrite this as 2 * 0.05 * 5times our X which is minus 3 so if we do this we should get 1.5 m per second so that's our y component of velocity now now we let's do the exact same thing with acceleration so our y component of acceleration a y is equal to derivative d v y DT so we have an equation for our VY which is right here so we can write time derivative 2 * 0.05 let's just write that as 0.1 x * x dot so we take the time derivative of this we use the product rule because we have two x's in here now an X and an x dot take the derivative of one term plus that function with derivative of the other term so we end up with 0.1 let take the derivative of the first x uh which is just one x dox dot and then the derivative the second term we end up with 0.1 x * X so we know x dot is minus three and we're given ax which is X double dot which is minus 1.5 we can substitute in 0.1 * - 3 - 3 plus 0.1 x now is 5 and X double dot isus 1.5 and we end up with a Y is 0.15 m/s squar so this question when you first read it might not seem too obvious which direction to go um but if we just write down some equations we know that have the terms that are given in here um just follow the steps we can find some substitutions and then it actually ends up not being as difficult as it first appears let's look at example two a particle travels along the path y isal 5x^2 when time is equal to zero this particle's path is going through the origin find the particle's distance and magnitude of its acceleration when time is equal to 1 second if the velocity the X component of velocity is equal to 5T feet per second where T is in seconds to solve this problem it might be easier to uh separate this into the X and Y components so let's start with the X component so we know the velocity BX is equal to 5T so at T is equal to 1 second we can write the position or at least the X component of position using our kinematic equations we can say that the integral the X component velocity times DT is equal to the integral from 0 to T 5T DT we solve this the integral of 5T DT would be 5 t^2 over 2 or 2.5 t^2 2 2.5 t^2 so then at one second we can write X is equal to at one second our X position is 2.5 ft then the other part is looking for is it's acceleration so we need to find the X component of acceleration so let's write that the X component of our acceleration is equal to X which equals to VX the time derivative of 5T which is just equal to five so our X component of our acceleration is equal to 5 fet perss squar and that's constant with respect to time so tals 1 it's still 5 ft per second squar two so now we've done our X component let's look at our y components so we know our position as a function of X so we can write our position Y is equal to 0.5 X2 so when T is equal to 1 second we know know that X is 2.5 ft we substitute X is equal to 2.5 ft 2.5 squared and that would give us 3.125 ft that's our y position uh our velocity is our VY how can we find our velocity we use the same processes before we can take the time der derivative of our function of Y which would be equal to 0.5 * 2 2 * x * x dot and at T is equal to 1 second we know our X component of velocity that's up here that's our x dot say at 1 second x dot is equal to five this is just equal to 0.5 * 2times X position five and our x dot is five so I get our y y component of velocity is 12.5 and then lastly acceleration a y isal to b y dot or Y double dot so we can write this as 12 * 2 is just one so we end up with x dot * x dot plus x * X we know X do at tal 1 second is equal to where is that 2.5 so we have 2.5 * 2.5 our position is oh I confused position our x dot is five not 2.5 5 * 5 + 2.5 times our X component of acceleration X component of our acceleration ax is 5 fet per second squared so that's also five and then if we do this we should get [Music] 37.5 feet per second squared so we're asked to find the position or sorry the magnitude uh find the particle's distance and the magnitude of its acceleration so when we hether with distance we should magnitude of position Vector so this is an x and y space so there's no Z term so we can write distance is equal to the magnitude of our position Vector which is equal to the square root of x^ 2 + y^ 2 we know our X position uh is 2.5 2.5 squared and our y position is 3.125 squared so we get our total distance which is the magnitude of our position Vector is four feet and if we want to find the magnitude of our acceleration lastly the magnitude acceleration is the square root of the sum of the components of acceleration squared so we get the square root of ax which is five I think yeah five plus our a y^ 2 which is 37.5 which would give us 37.8 per second squared so again we just followed the same step calculated our components um based on our path function and found the components of our position and acceleration and then calculated the magnitudes of those based on the components [Music]