This recording shows a proof of the gradient that rho times the velocity vector in cylindrical
coordinates in this problem you are given the definition of the gradient of the cylindrical
coordinates and two hints. The first, the differential of the unit vector in the radial
direction with respect to data is equal to the unit vector in the theta direction, and
the second is the derivative of the unit vector in the theta direction with respect to theta
is equal to the negative of the unit vector in the radial direction. Also you are given
the final answer. So you are given the gradient that rho times the velocity vector, and you
are just asked to proof this. So show that you could derive this equation. So the first
thing you should do is substitute what we know for the gradient dot rho times the velocity
vector. So we know the gradient, and that is just equal to what is given above. That
is dot rho times the velocity vector, and the velocity vector is just the velocity in
each direction times the unit vector in that direction. So the unit vector in the r direction
times velocity in the radial direction, plus unit vector in the theta direction times velocity
in the theta direction., plus unit vector in the z direction, time velocity in the z
direction. So lets go through and foil this equation. So the first term times all of this.
So this is er hat dot del, del(r), rho and for the second term etheta hat dit 1/r dot
del/del(theta) rho times the same velocity vector. For the third term k hat dot del/del(z),
rho, again times the velocity vector. Now lets go though and evaluate the partial differentials.
For the first one we need to evaluate the partial with respect to r. So this would be
er hat dot rho, times the second term, times derivative of the first. So d(er hat)/d(r)
plus er hat dot rho, times the first term time the derivative of the second. So er hat
d/dr of V. If we look etheta hat, vtheta the rest of these terms are not a function of
r. So their derivatives will be equal to zero. So now lets go on the the next one and evaluate
the derivative with respect to theta. So this first term will be etheta hat dot rho, 1 over
r, second term so Vr time the derivative of the first d(er hat)/dtheta plus etheta hat
dot rho, 1 over r. First term so er hat times derivative of the second. So d/dtheta of Vr.
For this next term etheta hat rho, 1 over r, second term Vtheta, derivative of the first,
d(etheta hat)/dtheta, plus etheta dot rho, 1 over r, first term etheta hat, d/dtheta
of the second term. So Vtheta lets scroll down. I will just continue on to the next
page. Now we will go and evaluate the derivative with respect to k. Similar to above. K and
Vz are not function of theta so their derivatives will be 0. Similarly down here Vr, theta are
not functions of z. So their derivatives will be 0, and we are just left with the two partials
from here. So this will be k hat dot rho, second term Vz, derivative of the first d(k
hat)/dz plus k hat dot rho, second term k hat, derivative of the first dVz/dz. So now
lets go through and evaluate these dot products. Here er hat dot rho Vr this dot product is
equal to 0. We go though the next term er hat dot er hat is equal to 1. So we will be
left with d/dr of rho, Vr. Next if we make our substitution we look at our hint. We are
told that this derivative here is equal to etheta hat. So we evaluate this dot product
etheta hat dot etheta hat will be 1, and we will just be left with rho Vr over r. So we
will look at the next term. We will see that Vtheta hat dot er hat will be equal to 0.
So this term will go to 0. So we look at the next part. We are told that this derivative
here is equal to minus er hat, minus just er hat, just from the second hit from up here.
So we evaluate this dot product. etheta hat dot er hat. That will be equal to 0. So this
term goes to 0, and if we evaluate this term. etheat hat dot etheta hat is equal to 1. So
we will have 1 over r d/dtheta, rho Vtheta. If we look at the next line, this is a plus
sign. On the next line k hat dot rho Vz. This term, this goes to 0. To evaluate this dot
product k hat dot k hat is equal to 1, and we will be left with d/dz, of rho Vz. So now
lets go and see how close we are to the equation that we are given for the gradient dot rho
times the velocity vector. So d/dz rho Vz, is in the equation up here. So that is good.
1 over r D/dtheta rho times Vtheta is in the equation up here. So this is good. Then we
have two terms here, and one term left up here. So we want to ask yourself. are these
two terms equal to the one term above. So lets go and write that. So lets say is d/dr
rho Vr plus rho Vr over r equal to. So this is just asking are these two things equal,
1 over r d/dr, r rho Vr. So that is just the final term that is left in this equation.
Given to us above. So lets just factor our this rho. So this is equal to rho over r times
the derivative of d/dr of r Vr. So if we just evaluate this partial derivative. We will
say that rho over r times Vr dr/dr plus r times dVr/dr. So if we just. The derivative
of r with respect to r. This term is 1. So this leaves us with. We just factor this back
in. This will be, this term goes to rho, Vr over r plus foiling this one. r's will cancel,
and we will be left with del/del(r) of rho Vr. Factoring this rho back in, and then we
will see that this term here is equal to this. So we have shown this proof to be true. That
this gradient, rho, this gradient dot rho times this velocity vector is equal to this
equation that is given here.