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Vector Calculus

V Calc.png

I am struggling with this question. So if the j and k components were reversed, it would be a Cylinder, but in this form I'm unsure.

I considered the different values of t (as it goes from 0 to pi) and how j is irrespective to t but aren't really sure what else to do from here or how to go about sketching the surface.

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Original post by bong69rip420
V Calc.png

I am struggling with this question. So if the j and k components were reversed, it would be a Cylinder, but in this form I'm unsure.

I considered the different values of t (as it goes from 0 to pi) and how j is irrespective to t but aren't really sure what else to do from here or how to go about sketching the surface.

Whether or not you swap the j and k components, this isn't a cylinder. (It is a surface of revolution).

Swapping the components really shouldn't throw you. Suppose we have a cylinder of radius R and height H, with the axis of the cylinder in the direction of k.

r(θ,t)=Rcosθi+y=Rsinθj+tk{\bf r}(\theta, t) = R \cos \theta {\bf i} + y = R \sin \theta {\bf j} + t {\bf k}, where 0θ<2π,0tH0 \leq \theta < 2\pi, 0 \leq t \leq H

Swapping j and k will just change this to a cylinder of radius R and height H, with the axis of the cylinder in the direction of j (which is hopefully no real surprise).
Original post by bong69rip420
V Calc.png

I am struggling with this question. So if the j and k components were reversed, it would be a Cylinder, but in this form I'm unsure.

I considered the different values of t (as it goes from 0 to pi) and how j is irrespective to t but aren't really sure what else to do from here or how to go about sketching the surface.

Original post by RDKGames


Thank you for both of your replies, I really appreciate it first and foremost. This really isn't a module I'm particularly strong at - I'm more of a Stats kind of guy, but somehow ended up here so please bear with me :frown:

So the a[1+(s/L)^2] would become a at s=0, so the curve would contract as s increases until s=0 and then would expand as s increases from 0 to h, back to when s=-h.
Original post by bong69rip420
Thank you for both of your replies, I really appreciate it first and foremost. This really isn't a module I'm particularly strong at - I'm more of a Stats kind of guy, but somehow ended up here so please bear with me :frown:

So the a[1+(s/L)^2] would become a at s=0, so the curve would contract as s increases until s=0 and then would expand as s increases from 0 to h, back to when s=-h.

Yes. If you define b = 1/L^2, then a[1+(s/L)^2] is just a(1+bs^2), which is just a parabola.

I know it can be confusing when 2 people give different advice about a question, but to my mind the key thing is to recognize that what you have here is a surface of revolution. In general, when you can define the surface in the form r(s,t)=f(s)sinti+f(s)costj+sk{\bf r}(s, t) = f(s) \sin t {\bf i} + f(s) \cos t {\bf j} + s {\bf k} (or some variation obtained by permuting i, j, k), you have a surface of revolution, and you can visualise the surface by imagining plotting f(s) against s to get a profile, and then rotating that profile 360 degrees around the k axis to make a "vase" shape. (It's the k-axis in the example I'm giving because that is the coordinate corresponding directly to s).
Reply 5
Here is me still struggling with precalculus concepts :frown: :biggrin:.
Original post by DFranklin
Yes. If you define b = 1/L^2, then a[1+(s/L)^2] is just a(1+bs^2), which is just a parabola.

I know it can be confusing when 2 people give different advice about a question, but to my mind the key thing is to recognize that what you have here is a surface of revolution. In general, when you can define the surface in the form r(s,t)=f(s)sinti+f(s)costj+sk{\bf r}(s, t) = f(s) \sin t {\bf i} + f(s) \cos t {\bf j} + s {\bf k} (or some variation obtained by permuting i, j, k), you have a surface of revolution, and you can visualise the surface by imagining plotting f(s) against s to get a profile, and then rotating that profile 360 degrees around the k axis to make a "vase" shape. (It's the k-axis in the example I'm giving because that is the coordinate corresponding directly to s).


I am still slightly unsure as to how to do this.

Plotting f(s) against s gives a parabola crossing at f(0)=a with f(-h) = f(h). How do I rotate around, in my case, the j, axis?
Original post by RDKGames

Can you hint me any further, given my response above? Apologies if I am being dense, I am just very unsure and my notes are nonexistent
I have learnt a lot in the last 5 minutes, will come back with more soon!
Original post by bong69rip420
I am still slightly unsure as to how to do this.

Plotting f(s) against s gives a parabola crossing at f(0)=a with f(-h) = f(h). How do I rotate around, in my case, the j, axis?

https://en.m.wikipedia.org/wiki/Surface_of_revolution

85248909-98B2-428F-A361-850D1DB29CF1.jpeg5ABEEBEC-5239-4178-A784-8595B1D4CE31.jpeg

Could you guide me any further with 1b? Unsure how to proceed
is this a level?
Original post by bong69rip420
85248909-98B2-428F-A361-850D1DB29CF1.jpeg5ABEEBEC-5239-4178-A784-8595B1D4CE31.jpeg

Could you guide me any further with 1b? Unsure how to proceed

Sorry for my awful notation btw :colondollar:
Original post by lilbabypenguin
is this a level?

No, it's a 2nd year Maths/Physics module.
Original post by bong69rip420
No, it's a 2nd year Maths/Physics module.

Can you please explain which formula for the area you're trying to use?

If you can get any screenshots the right way up that would help too.
Original post by DFranklin
Can you please explain which formula for the area you're trying to use?

If you can get any screenshots the right way up that would help too.

Sorry, I just took them on my phone and that's how they came out.

The formula I'm using is Untitled2.png
Original post by bong69rip420
Sorry, I just took them on my phone and that's how they came out.

So, you're supposed to be taking the modulus of ds x dt. I don't know if you've just done that (very) wrong, or if you've done something else entirely, but that's where everything falls apart. (I.e. everything before "A = an integral" looks vaguely plausible, but that line is totally wrong).

Hint/observation: it's at the point of taking the modulus that you'll get the square root you need to match the given integral.
Original post by DFranklin
So, you're supposed to be taking the modulus of ds x dt. I don't know if you've just done that (very) wrong, or if you've done something else entirely, but that's where everything falls apart. (I.e. everything before "A = an integral" looks vaguely plausible, but that line is totally wrong).

Hint/observation: it's at the point of taking the modulus that you'll get the square root you need to match the given integral.

I didn't take the modulus of it, lol. Jesus.

Thanks!
Original post by bong69rip420
No, it's a 2nd year Maths/Physics module.


thank god, got a bit scared then
Original post by bong69rip420
I didn't take the modulus of it, lol. Jesus.

So, out of curiosity rather than ragging on you (we all have brain farts!); am I right in thinking you effectively decided that |ai +bj + ck| = a + b + c?

(I know TSR can be a pain about attachment orientation, but I feel I lose at least 20 IQ points trying to follow something at 90 degrees, so I might be misinterpreting).

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