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let's start a discussion with vector
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calculus
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vector calculus is very useful in fluid
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mechanics
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it helps us to identify how the flows
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move from point a to point b
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to do that uh let's
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uh before we do anything else let's
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let's def let's have uh less to feel
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that um
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let's give some definitions here
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the first definition is what happens
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when we differentiate
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a vector
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differentiation
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of a vector
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okay so let's consider a vector
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of the following form let's suppose i
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have a vector
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uh a okay this vector
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a is a function of uh
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the magnitude a and it's in the
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direction
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unit direction e t let's suppose that
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both functions of time
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so that means that if i want to take the
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differentiation of
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a respect to t
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using the chain the the chain rule
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you know this will be the derivative of
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a respect to t
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times to second term
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plus a times
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the derivative of e respect to time
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okay this is this term right here
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represents the change in magnitude
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this represents change in direction
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okay so basically each one of these
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terms
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have their own representation right okay
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let's suppose we have various cases
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just for sake of this exercise
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let's say your vector a is a function of
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time
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and you have another vector b that's a
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function of time
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you have a sorry vector b is a function
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of time we have vector c
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that's a function of time as well
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let's suppose you have another vector u
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that's uh
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that's also this is just a constant
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and it's a it's it's a function of time
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okay so we have those
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this definition so if i wanted to do
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uh the derivative let's say
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of u times a remember this is a scalar
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it's a function of time just a scalar so
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if i use the chain rule
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this is d u d t
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times a plus
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u d a dt
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however since this is a scalar
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but it is a function of time so this
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remains there if it was not a function
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of time that this term would go to zero
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if i was going to take the derivative of
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time
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of a dot product let's say dot product
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of a with
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dot product of b then it applies the
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same rule
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is a dot product of a respect to time
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times dot b
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plus a don't forget the dots there's a
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dot product there
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then the dot product of b respect to
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time
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if i'm going to do the cross product the
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cross product works same
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cross product is if you have a and you
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want to take the cross product
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with a respect to b then that would be
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the uh the derivative of the vector a
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cross product b plus
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b uh sorry a
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cross product the derivative of b
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respect to time
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okay so many of these chain rules apply
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as
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we are used to applying them
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okay uh so these terms becomes very
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useful and very handy as we are
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moving on right so another thing that's
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very useful
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to understand if i have the vector a
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now let's express the full vector a
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that's a function of time
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and let's say this is a function of a1
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e1 everything is a function of time
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e 2 e 2
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[Music]
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sorry a 3 e 3
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and i want to take the derivative of a
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with respect to time that would give me
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the derivative of
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a1 respect to time
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e1 plus
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derivative of uh
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times a this is a vector don't know this
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is a scalar sorry
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a1 times derivative of
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e1 respect to time note that if these
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were
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not functions of time that this term
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will go to zero
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okay so going to the next term that will
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be this term
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will be plus da2
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respect to time e2
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plus a2
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d2 respect to time
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okay
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so if you go to the last term becomes
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then
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plus derivative for 3
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respect to time e3
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plus a 3 derivative of
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e3 respect to time
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so this is what your differentiation
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will look like
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this is a quick review of your calculus
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for inner bacterial notation okay
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so um the couple more
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if you're working a cylinder coordinate
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in cylindrical coordinates that becomes
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very useful and we use this
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a lot because the fluids are many times
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circular and cylindrical approximation
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and
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spherical approximation become very
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handy
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um so the cylindrical coordinates
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tell me the following okay so
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let's suppose i have a vector and
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this vector uh this small variation of
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this vector respect to time
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with respect to position is given by
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the following r d theta
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theta plus d theta
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d theta times z okay dz
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times z okay so now
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let's take a position vector
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and
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yeah okay
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so i have y i have x
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and now i'm going to have
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i'm looking from top and i have a
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rotation
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that looks something like this okay it's
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a perfect circle
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and now uh i got the position here
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at theta okay
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but then i have another position here
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and another moment that is
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d theta okay so
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we know that both are r but this is the
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same radius the radius doesn't change
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and this direction um let's change the
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colors here
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so this direction coming in this coming
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in this direction
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is e r okay and we can call this
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a remember just because the direction is
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going to change the magnitude is still
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one
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let's put this prime then what you have
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is
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you you have a vector that here that's
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perpendicular to this
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and 90 degrees have another vector
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that's perpendicular to this and 90
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degrees
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and this is um the direction
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e theta this is the direction
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e theta but it's different because a
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different location
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okay and we are doing all this rotation
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about this point let's call it the point
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a point oh i'm sorry so
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if i want to know the change between
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this
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the change that's happening here the
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theta okay
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this what is this guy over here so
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what happens is you can show
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that delta e theta
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okay is going to equal to your final e
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theta
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this vector minus this vector minus
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prime minus e theta
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okay but that is nothing else
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then if you use your uh
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the rules of the arc right that says
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this distance
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is equal to r times theta so
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and it will be uh minus
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d theta e r
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so your d theta if i'm actually going to
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take a
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a very s small differential
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is equal to minus theta
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er okay so
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what what this is uh this relationship
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is very useful
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um and also and then the same manner
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you can show that dr er
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is also equal to d theta
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e theta okay you can show that i'm not
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going to go into the whole derivation
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here
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uh you can also show that
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a differentiation of the e c the
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c uh will be
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[Music]
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theta but since i can say e z
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and take the cross product of z is the
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same vector
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so that will give me equal to zero
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okay uh so those are some
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uh concepts that just to tell you how it
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works
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now uh let's consider the following case
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let's consider i have a velocity
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okay now this velocity is a function of
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x y
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z and time and i want to find the
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acceleration
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okay so let's suppose i want to find a
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small
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variation of this a small variation of
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this
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is uh equal to using the chain rule will
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be a variation of b
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respect to x d of x
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plus the partial variation of b
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respect to y d y
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plus partial respect to
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v respect to z dz
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plus partial resp uh uh
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derivative respect to v respect to t
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dt okay
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so uh pay close attention here now
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so what happens here is um
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but we are doing this this whole thing i
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can divide this by
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t on both sides
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so let's let's put this in a different
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color
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so this is divided by dt this is divided
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by dt
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divided by dt divided by
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dt divided by dt and note that dt
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over dt this is going to be one so
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your entire equation then for your
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acceleration
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is actually equal to the variation
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that the the partial derivative of v
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respect to t okay plus
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partial derivative of v
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um v respect to x
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what is that that's just nothing else
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than your component in u
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remember that the whole thing here i can
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also write it at
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in terms of u i plus
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v j plus w
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k okay well these are the various
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components
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x y and z and they all can be a function
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of x y and z
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so i can say this is
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uh this so
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this right here is nothing else than
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u this is nothing else than b
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this is nothing i'm sorry okay they're
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related to this
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this is nothing else than w then
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really what you're really having is u
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velocity respect to x
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plus v velocity respect to
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y plus w velocity
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respect to z okay
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so this is what we call local divert
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this is called the local derivative
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this is called the convective derivative
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okay and a very convenient manner to
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write this
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is in terms of the gradient operator so
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let's define what the gradient operator
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does
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so there is the gradient
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operator
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okay the gradient operator says
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your delta is equal to i
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times partial of x plus
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j this is unit vector
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partial of y plus
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k partial of z
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okay that's th this is what your vector
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is going to look like so
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how does this work if i take this vector
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and i um multiply by a scalar fee
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right what i am saying is that what you
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end up
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is v times dx
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in the eye v
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times of uh sorry let's write it in the
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same form right you're just applying
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to this you're taking applying this
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operators
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that's what you're doing
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okay so
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that's what this really means now what
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if this was a
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vector right now when i have a vector i
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cannot multiply a vector but i can take
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a dot product of a vector
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so let's suppose i take i have a
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velocity field
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and i call this velocity field
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u i plus
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v j plus w
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k and then what if i take
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let's do it two different ways what
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happens if i take this
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dot delta right if i
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take this dot delta note one note
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exactly what's going to happen this side
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over here
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is um you're taking the dot product
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of u v and w
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and then on this side you're taking the
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dot product
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of partial respect to x
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partial respective y
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partial respect to z so
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what's really going to happen here is
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dot product
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so you you're taking this dot product
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the result of this
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is nothing else than the multiplication
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of
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this first times x so this is what you
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end up with
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partial of x plus
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v times partial of y
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plus w times partial of z
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and you can immediately see what's
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happening here right
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you can see that this is not on this
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side it's actually only before
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so this is saying that any operator you
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want to apply to this
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then we'll actually go to that okay for
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instance let me explain this
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let's suppose i wanted to say i'm going
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to write it right here
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this dot this
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times v so you know because applied v
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you take this whole thing and you apply
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feed to it and applying feed to it is
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saying you're applying
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feed to all of this so this is u partial
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of
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phi respect to x plus
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b partial of phi respect to y
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plus w partial of
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phi respect to z
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okay so that's what this means now if i
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do it the
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other way around and if i said now i'm
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going to take this
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i'm going to take another dot product
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let's see what happens then
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so what happens if i say i want to take
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this
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and i want to take this dot product but
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then once i take the dot product i want
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to multiply this times my
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velocity vector v now see what happens
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so if i did that then basically you're
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taking this
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this expression right here
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that's the first expression so it says
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u partial of x
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plus v partial of
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plus w partial of
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z all of this
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and i'm going to multiply b so when i do
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this
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then you basically end up with u
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partial of x times v
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plus v
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[Music]
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partial of y respect to v
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plus w partial of v
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respect to z and
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uh note now that this term right
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here that i just derived for you
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is actually identically equal to this
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term that we have here
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right so in general
00:20:52
what we could do is we could say your
00:20:56
acceleration vector will be nothing else
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than the
00:21:01
derivative of v respect to t
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plus your velocity vector
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take uh
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dot with your gradient vector
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multiplied by v and that's what this is
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going to equal to
00:21:20
the beauty of expressing it like this
00:21:23
now my delta
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operator can be anything my delta
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operator i could express it in terms of
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cartesian coordinates
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in terms of um
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sorry uh i can express it in terms of
00:21:36
cartesian coordinates
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i can express that in terms of um
00:21:42
excellent degree coordinates uh i could
00:21:45
have
00:21:46
expressed that in spherical coordinate
00:21:48
for instance
00:21:49
your del operator in cylindrical
00:21:52
coordinates
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is actually looks like this
00:21:59
cylindrical coordinate this operator
00:22:02
will be e r
00:22:06
partial of r plus
00:22:10
e theta don't forget this one over
00:22:14
r that's often skipped out and forgotten
00:22:19
plus e z
00:22:22
partial of z okay
00:22:25
so you can see that this is more general
00:22:28
and can apply to
00:22:29
any type of problem which is very very
00:22:32
convenient with stuff that we do
00:22:35
okay so uh
00:22:38
given this fact we got some another
00:22:42
property that's very useful
00:22:44
then um what happens if i say
00:22:47
v cross product
00:22:50
of a a vector a let's say okay
00:22:54
so i'm taking the cross product now the
00:22:56
other way around if i
00:22:57
did that then what you happen is i have
00:23:01
various solutions right if i'm working
00:23:03
in now let's say
00:23:05
a uh cartesian coordinates
00:23:14
then you will end up with a
00:23:17
x x
00:23:22
a y respect to y
00:23:26
a z respect to z
00:23:30
okay uh
00:23:33
in in cylindrical coordinates
00:23:45
in cylindrical coordinates you will end
00:23:48
up with 1 over
00:23:49
r partial of r
00:23:52
a r divided by r
00:23:56
plus 1 over r partial of theta
00:24:01
respect to theta plus partial of
00:24:04
a z respect to z
00:24:08
okay now if i want to do the
00:24:11
cross product of this a del operator
00:24:15
what we call this we
00:24:18
we actually call we have a name for this
00:24:21
we call this the divergence
00:24:24
of a okay now if you want to call
00:24:27
to find the curl okay this is very
00:24:30
important especially
00:24:32
when we start doing rotational fluids
00:24:35
you take the cross product of this
00:24:38
then it becomes very interesting as well
00:24:41
when is cartesian
00:24:46
your cartesian coordinates look like
00:24:48
this
00:24:54
i j
00:24:58
k partial of
00:25:01
x partial of
00:25:04
y partial of z
00:25:09
a x a y
00:25:12
a z these are the components of this
00:25:14
vector
00:25:15
okay now if i did this in
00:25:19
cylindrical coordinates
00:25:26
in cylindrical coordinates uh my cross
00:25:29
product will look like
00:25:31
look as follows take one over r
00:25:34
outside this integral outside the
00:25:38
uh the determinant
00:25:42
r times theta don't forget is r
00:25:45
e e z
00:25:48
partial of r partial of theta
00:25:54
partial of z then you have
00:25:57
a r r time e theta
00:26:01
part r times e z a z
00:26:04
sorry okay so you got several cases
00:26:07
depending on
00:26:08
the the coordinate system that we
00:26:11
actually working with
00:26:15
okay so
00:26:23
it should be obvious that
00:26:26
if you take the dot product of this will
00:26:29
not be equal to
00:26:31
eight a dot product of this
00:26:35
e also if i take this cross product of
00:26:38
a will not be the same as taking a
00:26:42
the cross product of delta okay those
00:26:45
are two different properties
00:26:47
and you get two different answers let's
00:26:49
apply some of the concepts to some
00:26:51
problems
00:26:52
and see how this actually pans out
