network profile
| bio | website | tudelft.nl |
|---|---|---|
| location | Delft, Netherlands | |
| age | 25 | |
| visits | member for | 3 months |
| seen | May 5 at 11:23 | |
| stats | profile views | 4 |
I am a PhD student in the field of fluid dynamics and soft matter.
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Apr 11 |
awarded | Teacher |
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Mar 26 |
awarded | Supporter |
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Mar 23 |
revised |
Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ edited body |
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Mar 22 |
comment |
Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ I had tried it, but it gave a very different result. However, looking at the documentation I found that I should use ArcTan[b^2 Cos[phi]],a^2 Sin[phi]] to get the correct result. Thanks! If you could write it as an answer I will definitely accept it. |
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Mar 22 |
revised |
Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ added 258 characters in body |
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Mar 22 |
comment |
Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ @MichaelE2 no I am sure the phase-shift is the correct thing to do. The reason is that ArcTan[(a/b)^2 Tan[[Phi]]] represents the angle the interface normal of an ellipse makes with the x-axis as determined from the angle the local x,y coordinates make with the x-axis (this angle is $\phi$). So I know that the whole thing should run from 0 to 360 deg and that the minima and maxima should coincide with those found when a=b. ---- So maybe I didn't explain enough of the background of the problem in my original question, but this is what I was looking for |
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Mar 21 |
comment |
Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ Thanks for the explanation of how to get the answer to the integral, but I found out that it was in fact not the answer to the integral that was incorrect, but rather what I fed it: the Cos[ArcTan[ Tan[[Phi]]]] which has a modulus at $\pi/2$ |
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Mar 21 |
revised |
Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ added 4 characters in body |
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Mar 21 |
answered | Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ |
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Mar 21 |
revised |
Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ added 183 characters in body |
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Mar 21 |
revised |
Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ some additional information |
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Mar 21 |
asked | Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ |
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Mar 2 |
awarded | Autobiographer |
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Feb 27 |
awarded | Scholar |
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Feb 27 |
accepted | Stop Mathematica from giving imaginary solutions |
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Feb 27 |
comment |
Stop Mathematica from giving imaginary solutions @DanielLichtblau thanks for the explanation, I was not aware of that phenomenon. The Cubics->False option doesn't solve it however |
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Feb 26 |
awarded | Editor |
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Feb 26 |
comment |
Stop Mathematica from giving imaginary solutions I have done what you mentioned, except for the //First//First part, because that will give me the wrong 1 of the 3 answers. |
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Feb 26 |
revised |
Stop Mathematica from giving imaginary solutions A step further in the question |
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Feb 26 |
awarded | Student |
