“Does the set containing all sets contain itself?”
- Bertie
begone, Russell.
Does the set of all sets not containing themselves, contain itself?
“gud😌”
- ——– 12 Grader
thancc.
”*thancc
P.S THIS WEBPAGE WAS AMAZING AND THANKS FOR DOING ALL THIS AND YOURE WONDERFUL.
P.P.S look at all that high effort.”
- admin
oof, thancccccccc.
P.S. thancc.
P.P.S. looked.
“They dont teach relativity in PH107 why because its ““difficult”” then why do they are allowed and interested to teach a whole lot of difficult complicated and sophisticated things in MA105”
- Artemis
Why they chose to not teach relativity in PH107 is not for me to judge. (Even though I am extremely happy that they don’t. :))
Coming to MA105, there was nothing “difficult complicated” that was taught. You will most likely use Calculus at some point of time in your courses here. So it is important to know the basic theory that builds up these concepts. If not, you’d be left with just some hand-wavy and half-baked-rigour-less knowledge of limits and the sort.
Now you may argue that as an engineer (I’m assuming you’re getting an engineering degree), you may not be required to know these fundamentals. In which case, I am saddened by your attitude.
“You showed exemplary courage in the times of war (exam prep?). I thank you very much. May this never stop!
P.S. All the Avengers’ doubts came from just one guy :p”
- Steve Rogers
Hehe, I’m not too surprised to know that it was just one person.
It was fun having this little running gag. All the best for your exam!
Maybe we shall continue this during the days of MA106 or MA108.
“A friendly suggestion here. I feel that the tutorials can be improved a little bit by giving more focus to tutorials 11-14. They were rushed through in the last 2 weeks and given that they consist of such important topics, I feel they should be emphasized on more rather than giving so much time to differentiation and integration in multiple variables. Hope you take this into consideration. Good going with the website, it’s really helpful. Thanks for everything :)”
- Batman
Hey, I can see where you’re coming from. I hope you entered this in your course feedback as I have no control over how these things go but raising this issue there may make a difference.
As for the last part, you’re most welcome :)
“Thanks a lot for this! Really helped me (especially considering my TA doesn’t explain concepts well).
Also, a small typo. In the last question of week 10 slides(Sheet 7, Q10), it’s written maximize instead of minimize.”
- MythV
I’m happy to know that I have been of help! :)
Thanks for pointing that out! Have fixed it now.
“nice”
-
such low effort. thank.
“Could you please upload the solution of question 1, tut 13? Thanks!”
- A confused nibba
Here’s an outline for part (a):
You most likely would be having trouble with parameterising the line integral part. Mainly because of a messy \(z\) coordinate.
Note that the line integral can be written as:
\(\displaystyle\oint_C (x-y)dx + (x+y)dy + (y+z)dz = \oint_C xdy - ydx + \oint_C \nabla\left(yz + \frac{1}{2}(x^2 + z^2)\right)\cdot d\mathbb{r}\).
The second line integral will simply be \(0\) as it’s a smooth grad field along a smooth closed loop. The first can now be simply done as you don’t need to worry about \(z\). The projection of that curve onto the XY-plane can be simply parameterised as \(\left(\frac{a}{2}\cos\theta, \frac{a}{2} + \frac{a}{2}\sin\theta\right)\). (Or you can observe this as twice the area of the projected circle.)
For the surface integral part, note that \(\nabla\times\mathbf{F} = (0, 0, 2)\).
Note that the surface is part of the cone given by \(z = f(x, y) = \sqrt{x^2 + y^2}\). Thus, the normal is given as \((-z_x, -z_y, 1)\) which gives \((\nabla\times\mathbf{F})\cdot\mathbb{n}=2\). Now, find the appropriate domain for \(x\) and \(y\) to get the answer. (Same projection as last which would again give twice the area as the answer.)
“Thank you so much for the slides, also the memes are spicy”
- MelonMusk
danke
“When we write n.dS while evaluating surface integrals, do we take n to be unit vector in case of planes? And what exactly is dS?”
- Clint Barton
\(\mathbf{n}\) is always taken to be the unit vector at the point. It’s just that \(\mathbf{n}\) is constant in the case of planes.
\(dS\) is just notation, it does not have any more meaning than that (for us). In differential notation, we may also write \(dS = \|\Phi_u\times\Phi_v\|d(u, v)\). Again, I’ll emphasise that this is just notation.
“Hey! Can you help me solve Q9(b) Sheet 14? And yes, please reply Bruce.”
- Natasha Romanoff
Consider \(\psi = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}\).
Note that the outwards unit normal \(\mathbf{n}\) at any point \(\mathbf{r} = (x, y, z)\) on the ellipsoid is in the same direction as \(\nabla \psi\) and one also has \(p = \mathbf{r}\cdot\mathbf{n}\). Also note that \(\mathbf{r}\cdot\nabla\psi=2\). I will use these facts in the following equations. It is your task to see where I’ve used what and how.
\(\displaystyle\iint_S\frac{1}{p}dS\)
\(= \displaystyle\iint_S\frac{1}{\mathbf{r}\cdot\mathbf{n}}dS\)
\(= \displaystyle\iint_S\frac{\mathbf{n}\cdot\mathbf{n}}{\mathbf{r}\cdot\mathbf{n}}dS\)
\(= \displaystyle\iint_S\left(\frac{\mathbf{n}}{\mathbf{r}\cdot\mathbf{n}}\right)\cdot\mathbf{n}dS\)
\(= \displaystyle\iint_S\left(\frac{\nabla\psi}{\mathbf{r}\cdot\nabla\psi}\right)\cdot\mathbf{n}dS\)
\(= \displaystyle\iint_S\left(\frac{\nabla\psi}{2}\right)\cdot\mathbf{n}dS\)
\(= \displaystyle\iiint_W\frac{\nabla^2\psi}{2}dV\)
\(= \displaystyle\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)\iiint_W 1dV\)
\(= \displaystyle\frac{4}{3}\pi abc\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right).\)
Note that I have used Gauss’ divergence theorem and the fact that the volume of \(W\), the solid ellipsoid, is what it is.
“Can u explain in brief or post something about the alternative formulations of greens theorem.”
- Green
Could you be more specific as to what you want? The alternate formulations are there in slides and I’m not sure as to what more I can tell about that.
“Q9 Sheet 12 is making me turn green. Please help!”
- Bruce Banner
Here’s an outline of what to do. Hopefully you can complete it.
The hemisphere satisfies \(z = \sqrt{1 - x^2 - y^2}\) and \(x^2 + y^2 \le 1\).
Using the fact that \(\mathbf{n}dS = (-z_x, -z_y, 1)d(x, y)\).
In this case, \(z_x = -x/z\) and \(z_y = -y/z\).
Thus, \(\mathbf{F}\cdot\mathbf{n}dS = \frac{1 - 2xy - 2y^2}{z}d(x, y).\)
(Note that both the above equations involving \(dS\) are just written in differential notation. I’m not treating \(dS\) as a real number or any such thing.)
Now the mass blah blah blah stuff asked is just the flux which equals:
\(\displaystyle\iint_S\mathbf{F}\cdot\mathbf{n}dS = \iint_D{1 - 2xy - 2y^2}{\sqrt{1 - x^2 - y^2}}d(x, y)\).
Where \(D = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 \le 1\}\). The above can be solved using polar coordinates.
“You could have quoted The Daily Bugle.
I check this page every day, once in the evening and once before going to bed.
I must admit that I envy and appreciate your math skills. Any tips?”
- Peter Parker
Ah, looks like I don’t remember it well enough to quote that.
I guess just be patient and practice.
“Parametization was the problem indeed.
Speaking of revealing my identity, did you even watch my latest movie (till the very end)?”
- Peter Parker
Oh, good to know that I solved your doubt then.
Yes, I had, I was thinking of making a Mysterio reference but couldn’t think of a nice one.
P.S.: How often do you refresh this page to reply quickly enough?
“My Spidey Sense tells me that Q7 Sheet 10 is a threat. Can you help me fight it?”
- Peter Parker
Good job giving away your identity.
I’m not sure what exactly is the problem with this question, though. I guess you had a problem finding a parameterisation for the curve \(C\) because it’s quite straightforward otherwise.
Note that the cylinder is given by \(\left(x-\frac{a}{2}\right)^{2}+y^{2}=\frac{a^{2}}{4}\).
The equation of the sphere tells us that the points lying on the intersection, that is, the points of \(C\) satisfy: \(z^2 = a^2 - ax\).
The first equation leads to the natural parameterisation of \(x = \frac{a}{2} + \frac{a}{2}\cos\theta\) and \(y = \frac{a}{2}\sin\theta\) where \(\theta \in [0, 2\pi]\).
Also, substituting this value of \(x\) in the second equation and noting that \(z \ge 0\) gives us that \(z = a\sin\frac{\theta}{2}\).
Thus, we now have a parameterisation for the curve and can (hopefully) easily solve it now.
You just need to compute the integral \(\oint_C \mathbf{F}\cdot d\mathbf{r}\). Check the solutions for the similar tutorial problems if you don’t get it.
“There should be summary of all the content covered in the tutorial in the start as you have done in week1 sheet. It will be very helpful”
- Harry Potter
I’m not quite sure as to what you’re referring to as there was no summary at the beginning of the Week-1 slides. But I guess I can add the questions covered at the beginning for easier reference.
“Tony forgot to ask Q9(ii) Sheet11. Please provide a solution for that too.”
- Thor Odinson
I guess MA 105 questions are an Avengers’ level threat, huh?
However, question 9 is indeed a nice question in my opinion. So, here you have the solution for all parts.
“Iron man has a problem with the solution too! Shouldn’t phi_x=xf(r) instead of xf(x) in the first slide??”
- Iron man
Iron man is indeed correct. The typo has been fixed.
“Please provide a solution for Sheet 10 Q11”
- Iron man
Surely Iron man shouldn’t be having a problem with this.
Anyway, here’s the solution.
“Can you please give the solution to Q7 Sheet 8?”
- Freshie
I can and I shall.
By symmetry, the given volume is 8 times the volume in the positive octant.
In that octant the volume lies above the region \(Q := \{(x, y, 0) \in \mathbb{R}^3 :x \ge 0, y \ge 0, x^2 + y^2 \le a^2\}\) and underneath the cylinder given by \(x^2 + z^2 = a^2\).
Therefore,
\(V = 8\displaystyle\int_0^a\left(\int_0^{\sqrt{a^2-x^2}}\left(\int_0^{\sqrt{a^2-x^2}}1dz\right)dy\right)dx = \dfrac{16a^3}{3}\).
“You are really helpful .I attended tsc class you took.And I really understand the concept .Thanking You”
- Enthu Wala frieshie
That’s what I hoped to do. Glad to know I was successful :)
“Thanks for taking the efforts to actually write out the solutions, they are really helpful. Hope I can do the same for a future batch.”
- Random Guy
You’re welcome!
P.S.: Nothing is stopping you from making your own solutions for your own batch. I’m sure that there are plenty of subjects for which solutions aren’t being made.
“Can a continuous function with bounded and open domain be. integrable “
- A person who is going to revolutionize whole world
Yes, it can. But it’s not necessary. This is because a continuous function need not be bounded on a bounded and open domain. (Think of an example.)
Hence, the function won’t be integrable.
“The solutions are very helpful! Thanks!”
- someoneNotFromD1T5
Ayy, I’m glad you find them helpful!
Also, thanks for not asking me for physics solutions, lol.
“What about question 7 of tut sheet 6? what do they mean by bounded in a disc?”
- CuriousFreshie
Hello, what the question means is that given any \(\epsilon > 0\) and any \(M \in \mathbb{R}\), there will be a point \((x_0, y_0)\) in the disc of radius \(\epsilon\) centered at origin such that the value of \(\mid f_x(x_0, y_0)\mid\) is greater than \(M\). (This is what is means for the derivative to not be bounded in any disc around \((0, 0)\).)
The ‘mail me at this address’ on your website (the MA105 page) has aryamanmaithani@gmial.com instead of aryamanmaithani@gmail.com.
- Aryaman Maithani
Thanks a lot for pointing this out. Nice name!
Hey!! great help. thanx a lot. a request. is it possible for you to upload solutions for phy tuts as well. thanx
-
I’m glad you find it helpful. You’re most welcome. No, I won’t be uploading solutions for physics tuts. I don’t have the time for an extra subject.
You are amazing bro.
- [hiding name to protect privacy]
Thank you!
can you upload solutions of tutorial sheets??
-
I’m not quite sure how you missed the solutions which are pretty much the highlight of the page.
This site is very helpful. It would be nice if you could provide solution for extra question.
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Hey, I’m glad you find the site helpful. However, I won’t be uploading the solutions for extra questions. That would defeat the purpose of the questions. (Purpose being - you must think about them.)
Good job
- God
Thanks, God.