Isaac Physics Maths Question

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Reply 20

Original post

by mqb2766

I didnt go any further but will have a look tomorrow, too tired tonight. The integration should be like the torus so youre integrating wrt dw, dalpha (0..2pi), dtheta (-inf..0) and there is a reasonable amount of simplification in the integrand because of the theta=0 argument. But will look tomorrow,

Have you had a chance to give it another go?

Reply 21

mqb2766

Original post

by sirius_canis

Have you had a chance to give it another go?

Been feeling rubbish, but should get through it tonight, famous last words

Reply 22

mqb2766

Original post

by sirius_canis

Have you had a chance to give it another go?

Yes, after a bit of playing got the ans. What did you get for the dot product?

Reply 23

mqb2766

Original post

by sirius_canis

Have you had a chance to give it another go?

If youre still unsure about it, just let me know. Tbh, it took a bit of guessing and working out what they wanted to actually understand their description which overloads the meaning of variables too much, especially w.

Reply 24

sirius_canis

Original post

by mqb2766

Yes, after a bit of playing got the ans. What did you get for the dot product?

(w*sin(alpha)^2*cos(alpha)+w*sin(alpha)^2*r_0-sin(theta)*cos(aloha)^2*b*r_0-sin(theta)*cos(alpha)*b*w+cos(theta)*cos(alpha)^3*w+cos(theta)*cos(aloha)^2*r_0)

Reply 25

mqb2766

Original post

by sirius_canis

(w*sin(alpha)^2*cos(alpha)+w*sin(alpha)^2*r_0-sin(theta)*cos(aloha)^2*b*r_0-sin(theta)*cos(alpha)*b*w+cos(theta)*cos(alpha)^3*w+cos(theta)*cos(aloha)^2*r_0)

It should be a fair bit simpler than that.

My take on the question is that the notation in 7.6.8
https://isaacphysics.org/questions/maths_ch7_6_q8
was confusing to me. So the general sectional radius they talk about at the start
w e^(b theta)
is allowing w to vary along the radius between 0 and w_0 for a given value of theta, so the actual radius varies between 0 and w_0 e^(b theta). So the w in the differential area vector dA corresponds to w_0 if you were talking about the external surface area. But more generally we can consider any radius between 0 and w_0 when talking about a small, internal parallelepiped to calc the volume.

So when dotting with the vector dr_w (my notation instead of theirs dw), youre really differentiating the vector r wrt w_0, where r is given by 7.6.6C
https://isaacphysics.org/questions/maths_ch7_6_q6
Thats similar to the expression in a previous post expect there is an extra e^(b theta) multiplier. Remembering to set theta=0 (but not the exponential term) in dr_w, you should get a fairly simple vector.

So to calculate the volume you want to do the triple integral of
dr_w.(dr_theta x dr_alpha)
which is something like
# w e^(3b theta) dalpha dw dtheta
where # is a fairly simple expression (function of alpha only) from the dot and cross product and you integrate alpha between 0 and 2pi (around the circle cross section), w from 0 to w_0 (along a radius of the circle cross section) and theta from -inf to 0 (along the horn).

(edited 10 months ago)

Reply 26

sirius_canis

Original post

by mqb2766

It should be a fair bit simpler than that.
My take on the question is that the notation in 7.6.8
https://isaacphysics.org/questions/maths_ch7_6_q8
was confusing to me. So the general sectional radius they talk about at the start
w e^(b theta)
is allowing w to vary along the radius between 0 and w_0 for a given value of theta, so the actual radius varies between 0 and w_0 e^(b theta). So the w in the differential area vector dA corresponds to w_0 if you were talking about the external surface area. But more generally we can consider any radius between 0 and w_0 when talking about a small, internal parallelepiped to calc the volume.
So when dotting with the vector dr_w (my notation instead of theirs dw), youre really differentiating the vector r wrt w_0, where r is given by 7.6.6C
https://isaacphysics.org/questions/maths_ch7_6_q6
Thats similar to the expression in a previous post expect there is an extra e^(b theta) multiplier. Remembering to set theta=0 (but not the exponential term)) in dr_w, you should get a fairly simple vector.
So to calculate the volume you want to do the triple integral of
dr_w.(dr_theta x dr_alpha)
which is something like
# w e^(3b theta) dalpha dw dtheta
where # is a fairly simple expression (function of alpha only) from the dot and cross product and you integrate alpha between 0 and 2pi (over the circle), w from 0 to w_0 (along a radius of the circle) and dtheta from -inf to 0 (along the horn).

Thank you so much!!! I finally solved this question, it's been such a long journey from 7.6.1 to here.

Reply 27

mqb2766

Original post

by sirius_canis

Thank you so much!!! I finally solved this question, it's been such a long journey from 7.6.1 to here.

Well done for sticking with it. Must admit, the dw and ... on this page (7.6.8 ) is badly explained and from a quick google, a few other kids thought so last year
https://math.stackexchange.com/questions/4903858/volume-of-logarithmic-horns
and didnt get to the bottom of it. You may not get a reply, but it may be worth dropping isaac a comment to explain the problems with notation/explanation etc.

Reply 28

mqb2766

Original post

by sirius_canis

Thank you so much!!! I finally solved this question, it's been such a long journey from 7.6.1 to here.

As a bit of reflection about the volume ans, you could note it could be written as
1/3 * base area * height
so unwind the horn into a circular based cone where the height is similar to the curve length s worked out in an earlier part. Strictly speaking there would be some form of log transformation on the height to make it a normal circular based cone, rather than an exponential profile, but not worked it through. A duffers approach to getting the volume without the dot/cross product triple integral stuff.