Not sure what the R 3 means, but the transformation is just the usual polar-Cartesian we all know and love:
x = r cosφ
y = r sinφ
z = z
Can someone help me with this please.
A vector field in cylindrical polar co-ordinates is given by
V = R Rˆ + R 3 s i n ( φ ) c o s ( φ ) φˆ + 3 z kˆ
where Rˆ, φˆ, kˆ are the appropriate unit vectors. Translate this vector field into the
Cartesian x, y, z co-ordinate system.
What is the first step??
Could you please guide me through the steps.
Thanks.
5 answers
yeahhh i get that but how do you translate theta in terms of cos or sine?
i know i need to sub 'unit R^' with cosφ i + sinφ j and z with 'unit K^' with z k.
How about 'unit φ^' ?
What do i need to sub phi with?
I look else where it says sub it with -sinφ i + cosφ j.
But how? and why? can someone explain it to me?
Thankssss
How about 'unit φ^' ?
What do i need to sub phi with?
I look else where it says sub it with -sinφ i + cosφ j.
But how? and why? can someone explain it to me?
Thankssss
The unit vectors are just the vectors perpendicular to the cylindrical surface.
r^ points radially outward
φ^ points tangent to the planar circle
z^ points upward
To convert to Cartesian vectors,
r^ = cosφ i + sinφ j
φ^ = -sinφ i + cosφ j
z^ = z^
You have V = rrˆ + r^3 sinφ cosφ φˆ + 3z kˆ
Now, the position of any point on the curve is
V.(r^,φ^,z^)
= r(cosφ i + sinφ j) + (r^3 sinφ cosφ)(-sinφ i + cosφ j) + 3z k
= (rcosφ-r^3 sin^2φ cosφ)i + (rsinφ+r^3sinφcos^2φ)j + 3z k
= (x-xy^2)i + (y+x^2y)k + 3zk
r^ points radially outward
φ^ points tangent to the planar circle
z^ points upward
To convert to Cartesian vectors,
r^ = cosφ i + sinφ j
φ^ = -sinφ i + cosφ j
z^ = z^
You have V = rrˆ + r^3 sinφ cosφ φˆ + 3z kˆ
Now, the position of any point on the curve is
V.(r^,φ^,z^)
= r(cosφ i + sinφ j) + (r^3 sinφ cosφ)(-sinφ i + cosφ j) + 3z k
= (rcosφ-r^3 sin^2φ cosφ)i + (rsinφ+r^3sinφcos^2φ)j + 3z k
= (x-xy^2)i + (y+x^2y)k + 3zk
Thanks you, Steve. It does make a bit of sense to me know.
But just to clarify, since φ is the points tangent to the planer circle,
thus, φ is the derivative of the points x, y, z?
But just to clarify, since φ is the points tangent to the planer circle,
thus, φ is the derivative of the points x, y, z?