You keep saying "inflection point" for what us mathematicians would call a critical point, turning point, or stationary point. We define an inflection point as a point where the graph changes from concave to convex or vice versa.
@@samyakgupta5773f'' changes sign => inflection point, but not the other way. When a function is not twice differentiable, it may still have inflection points. But for double differentiable functions it's equivalent.
12:06 It's actually still easy without the simple cases. Consider a prime p, consider p^2 and p^3. It is clear that p^2 * p^3 = p^5. As p is prime, this is neither a square nor a cube. Also, consider again p^3 and p^2. (p^3)^2 * (p^2)^3 = p^12 which is both (p^4)^3 and (p^6)^2, so it's both a square and a cube. Therefore it can be both a square and a cube, or it can be neither a square nor a cube. It's trivial it can be either or as well.
@@ChOwToo I really don't think it was though.... like every part was a well-defined concept. Like a square number, a cube number, the square number * a cube number. A boolean function that returns 1 iff a number is a square / cube are all well defined concepts and regularly used in number theory...
You certainly solved the first question a lot more elegantly than I did. I had a very convoluted method. First, I factored the cubic as (x - a)(x^2 + bx + c) = 0. Since it is a cubic, and complex solutions come in pairs, this either has only one solution or three solutions, or two solutions with repeated roots. Expanding the cubic and comparing terms to the original one gives a = b, ac = 3000, and 300 = a^2 - c. Now I want to figure out whether the quadratic x^2 + bx + c has two real solutions - if it does, I've got three roots, and if it doesn't, I don't. The discriminant is b^2 - 4c = a^2 - 4c = 300 - 3c. I want to figure out the sign of the discriminant, which means I need the sign of 100 - c. If c > 100, then the discriminant is negative and I have only one root. Now, c > 100 implies ac > 100a (a must be positive if c > 100 since ac = 3000), which implies 3000 > 100a, and thus a < 30. Hence, if there is a root between 0 and 30, it must be unique. Plugging in x = 0, I get -3000, and plugging in x = 30 will give me something positive since 30*30*30 > 30*10 + 30*100 = 30*110. Therefore there is a sign change, and by the intermediate value theorem, I conclude that a solution exists for x < 30, thereby implying that only one root exists.
Hey I just wanted to say thank you so much for all the help with the physics revision tips and questions you posted for the May/June batch. I got my results for AS and ended up getting a very high A! Cheers mate and thank you once again!
What an awesome video!! It was extremely fun to try and solve along haha. On that last question we even could divide that N+1 by 2, because the coefficients have to be symmetric, so we can't use different integers for the first and second halfs of coefficient spots, but it doesn't matter in the end.
13:20 I just turned the function into an infinite geometric series which works out to be tan^2(x) and analyzed the graph of tan^2(x) from there as having asymptotes at +/pi/2 + kpi and zeros at kpi where k is an integer. It's also always positive due to being squared so d is the only correct answer.
34:10 Did he not do the derivative of the sin functions incorrectly? He treated sin^(3)x as if it were x^(3) and used the power rule. And then the derivative of sin is cos? Unless I’m missing something because Tom didn’t interject
41:48 "There are 2 possible sin solutions for every value." What about sin(x) less than or equal to -1?? Then there are 1 or 0 solutions. So to solve this completely, you should check f(-1).
For the quadratic in G it's quicker to multiply by 2 to give 2c^2+3c-2 = 0 and then factorise to give (2c-1)(c+2)=0 and then it's clear that c=1/2 is the only solution.
There are many cases in the exam when a far simpler approach would have generated the correct answer e.g problem F time 40:45 can be solved by drawing a graph approximating the functions and the answer is obvious. I recommend always draw a picture and this often leads to a solution.
40:48 I think I made the equation sin^3 x = - cos^2 x and draw each of those graph since its really easy to sketch if there are only one trig function to some power. Turn out there is only 2 solution from the sketches!
For question F the problem could be written as sin(x) tan^2(x) = -1, and from the graphs of the sin and tan functions one can see that there two solutions over the interval 0 to 360 degrees.
Are we all going to ignore how this guy crazily writes the letter X? I’ve never seen someone essentially connect a backwards C and a regular C … instead of the traditional crossing of two lines. Yet when he writes a multiplication sign (also an X) he uses the traditional method of just crossing the lines. Wild.
(Question) Here at 1:11:00, could you think about it as: - If amax = 2k/(N+1), while 2k = a0 + a1 + ... + aN, there are 2 possible cases All terms are equal, then amax = 2k/(N+1), because It's the arithmetic mean of these factors OR There is a term that is greater than every other term and the arithmetic mean must be lesser than this mean So, in both cases, amax is equal to the mean or greater than the mean?
“there are 2 possible cases: All terms are equal… OR There is a term that is greater than *every* other term”. These two cases are not exhaustive! If you changed *every* to *at least one*, then you’d be absolutely right. Very smart answer!
@@thedogatemyhomework8 Nope, you are not, I was treating sin(x) as it's own function, say s, and ignoring the further x dependence to see how it behaves. Hope this makes sense
@@zhelyo_physics to add to this, it doesnt make me fell mor confident on the premise that ure bad at these problems or anything like that, im way closer to the time i took maths classes than u are yet i wouldve fared worse haha
@@bobfake3831 absolutely! I think YT needs more realistic problem solving videos. : ) I made a point to leave lots of the thinking time, little mistakes, getting stuck etc. Glad you have enjoyed!
Question C: y is clearly a geometric series, with initial term sin^2(x), and common ratio sin^2(x), use the formula to get y=tan^2(x). The answer is clear then.
Q2: isn't the general case quite trivial? my first thought was prime factors. given a^2*b^3, if factors of a are a subset of factors of b, always a cube, otherway around always a square? and both, if its the same set? or am I thinking too simplistically? 😳
Haha, adjectives such as “trivial” and “simple” means different things to different people, so I won’t comment on that. Your reasoning is correct and would, for this question, lead to the correct answer. And more than that, you’ve correctly identified the exact special conditions for three out of the four possibilities: (a), (b), and (e). Just for completeness, the remaining possibility (d) occurs if and only if a and b are coprime (have no common factors).
23:20 this problem doesn't have 2 solutions. The solution to (2/3)a^3 - a^3 = 9.... (-1/3)a^3 = 9.... a^3 = -27.... a = -3 is a unique solution. The a = 3 solution doesn't work. The symmetrical parabola with that same area between the curves that goes through y-intercept at "a" would not have the equation of y=x^2+2ax+a. The equation would look different than that to produce the symmetry of the parabola.
Indeed! From the work done in the video, we already know the intercept is negative. Now, it is easy to see that the value of the function when x=-1 is negative (-1), implying that the function has not yet crossed the x-axis at x=-1 (while traversing the graph from left to right). Therefore, the intercept must be greater than -1, which means there must be a feasible solution!
interesting question, while olympiad success looks phenomenal I imagine the vast majority of students have done very well on the exam and the interview. I think willingness to learn, self study of topics and mathematical interests and skill are most important. But that's coming just from my point of view of a teacher not affiliated with the university.
This guy has no idea, 20:00 you come out with 0 as a limit of integration? when you solve the equation you're dividing by x, so x=0 is the other solution you're looking for,
Great video just making my way through! :) BSc Physics student here, a question for all you wonderful mathematicians out there! Is this valid for E? 🤔 @tomrocksmaths @ZPhysics I rearranged to sin y + cos^2 y = sin x + cos^2 x Simplifying to f(y)=g(x) We see f()=g() and as such the equality can be reduced to 2*pi*n*y=2*pi*m*x (As both functions have periodicity of 2 pi) ny=mx ; where n and m are natural numbers So e Feel like there’s some issues with this, like I’m breaking some rules (but I don’t know why aha)
I'm also a physicist (at least by training... now I'm a maths teacher 😂), and I've never heard a physicist use the term "inflection point" to mean a stationary point, nor have I ever read about it in the physics literature. An inflection point is a point at which the curvature changes sign... also in physics 😇 But of course I don't know all the physicists or all the physics literature 😅 I have a suspicion that he got something mixed up...
Are You able to derive "from zero" an inverse transform to Laplace transform formula? He created a transform and found an inverse formula, but how he get it?
Can someone explain the logic with derivatives in the first question? I am missing a step, for example if equation is x^3-300x = 0, we can apply same reasoning, get same conclusion, but equation now has 3 real solutions. The method is valid, but the 0 crossing is irrelevant, relevant part is 3000 crossing (constant) and that is missing.
Excellent question. Applying the same logic to x^3-300x=0 we would get stationary points at x=+10 and x=-10. However now let's plug those numbers back into the original function. When x=10 we get a stationary point at: 10^3-300*10 gives a stationary point at y=-2000. When x=-10 we have a stationary point at -1000-300*(-10)=+2000. Now if we plot those onto a graph our cubic will come from negative infinity when x is large and negative, keep going up, cross the x axis (root 1), then have a stationary point at (-10,2000) . Our next stationary point is negative in terms of y so then the wiggle will go down, cross the x axis again at (10,-2000) this is root 2. Okay, but we know that the cubic will shoot off towards positive infinity as x gets big so we need to cross the x axis again (root 3). For x^3-300x-3000=0 our wiggle is shifted down the y axis so the wiggle happens entirely for negative y but it eventually crosses the x axis once for positive infinity. I recommend visualising this with wolfram alpha or a graph plotting software. Hope this is helpful!
Hi, yes, I treated the function separately, as sin(x) being y(x) or call sin(x) being s and ignoring the further dependence on x. Hence the comment from Tom later on notation 😂
@@zhelyo_physics Physicists always seem to have terrible notation. It took me five years and a course in differential geometry to understand the notation in thermo - but if the notation used was standard like the one used in mutivariable calc, I'd have understood it instantaneously.
Awesome vid!! For the first question, i think the answer of 1 real solution is correct, but the derivation was a bit unclear? The critical points were indeed -10 (local max) and 10 (local min). The second order derivative around the min makes the curve inflect upwards, pointing to a root lying past x=10. With a calculator, the only real root is approx around 21.05. Edit: I now heard Tom, he basically says the same 😂 sorry I need to be more patient with my posts.
Great vid! So just to confirm for question D, is the answer ONLY a=-3? That's what I got as when I equated the 2 parabolas, I got roots of x=0 and x=-a and integrated between those two points. I think the mistake you made was by dividing by x (as x could be 0) instead moving all the term on to one side and factoring which got me 2x = 0 and x + a = 0 But unclear whether the answer was actually (b) or (c) as Tom said (b) was correct, but I disagree (but I'm just an undergrad student lol) 😊
a = 3 also works, if you try it. Note that 0 < -a precisely if a < 0, so in the a < 0 case, you integrate from 0 to -a, while in the a > 0 case, you instead integrate from -a to 0. Therefore the answer is (b).
@@amritlohia8240 ah I think I understand now, thank you. So I did the definite integral with the upper bound as 0 and lower bound -a, but I also needed to do it with those reversed because I don't know that -a < 0
I think it depends on which way round you subtract the parabolas. They way they did it gives a = -3. If you do it the other way you need a = 3 to get a positive area.
Yeah, he didn't get to it but the last question of part 3 basically requires you to find a way to extend the solution you obtain on some finite interval to to another finite interval that's larger. Now, the physicist in me screamed the obvious solution: the interval length is just a matter of units (if there were any) and since units are arbitrary it should be possible to extend the interval to any desired length. And a unit change is just a rescaling, so that's the answer.
I am wondering why these two guys are in shock...As physics and maths are corelated, they are no different, especially in physics there is a lot of mathematics so there must be no difficulty for a physics teacher!
@@tfg601 are you Indian and have to give that exam? Because of not then it's quite easy to say an exam is not that hard if you don't have to give it. It has math that probably you learn at university elsewhere....
@@chemicalnamesargon yeah I know but seeing the questions they are visibly a lot easier than jee. Jee has a lot more formulas you have to memorize and generally requires 2 years of consistent studies. I agree that our education system is messed up and yours is a 100 times better. The things asked in jee should be taught in college and not in entrance exams.
thanks for the feedback! I feel it was important to show I can’t solve these problems instantly though so I left in thinking time and wrong approaches/mistakes.
@@zhelyo_physics just really surprising someone from Oxford, particularly maths/physics related, would struggle with any of this. If MAT is a problem, you two would get finished by STEP 💪
@@zhelyo_physicsI think the way you took time and explained your approaches made for a very entertaining and educational video. I wouldn't have minded a longer video where you go through the whole paper with Tom because every question you did taught us to think in a different way.
Good to see everyone forgets maths they haven't used in a while
You keep saying "inflection point" for what us mathematicians would call a critical point, turning point, or stationary point. We define an inflection point as a point where the graph changes from concave to convex or vice versa.
@@amritlohia8240 ooops 😀 thanks for mentioning it! It's been a while since I've done this in depth.
ahh yes its the point where the sign of double derivative changes can we say that?
@@samyakgupta5773f'' changes sign => inflection point, but not the other way. When a function is not twice differentiable, it may still have inflection points.
But for double differentiable functions it's equivalent.
@@samyakgupta5773 That's one definition, which is equivalent to the one I gave in cases where the function is twice differentiable.
A point where d²y/dx²=0, dy/dx!≠0 and d³y/dx³≠0
12:06 It's actually still easy without the simple cases.
Consider a prime p, consider p^2 and p^3.
It is clear that p^2 * p^3 = p^5. As p is prime, this is neither a square nor a cube.
Also, consider again p^3 and p^2. (p^3)^2 * (p^2)^3 = p^12 which is both (p^4)^3 and (p^6)^2, so it's both a square and a cube.
Therefore it can be both a square and a cube, or it can be neither a square nor a cube. It's trivial it can be either or as well.
fantastic, thank you for the comment!
Imho this particular question was badly defined.
@@ChOwToo I really don't think it was though.... like every part was a well-defined concept. Like a square number, a cube number, the square number * a cube number. A boolean function that returns 1 iff a number is a square / cube are all well defined concepts and regularly used in number theory...
What a fun video!!
@@blackpenredpen thank you so much!! It would have been better with you in there as well! Fancy a trip?? 😀
@@zhelyo_physicsone day if there’s a chance!
this would be the greatest collaboration :D
@@blackpenredpen…..so IS there a chance? 😂
@@datboy038 lol
You certainly solved the first question a lot more elegantly than I did. I had a very convoluted method. First, I factored the cubic as (x - a)(x^2 + bx + c) = 0. Since it is a cubic, and complex solutions come in pairs, this either has only one solution or three solutions, or two solutions with repeated roots. Expanding the cubic and comparing terms to the original one gives a = b, ac = 3000, and 300 = a^2 - c.
Now I want to figure out whether the quadratic x^2 + bx + c has two real solutions - if it does, I've got three roots, and if it doesn't, I don't. The discriminant is b^2 - 4c = a^2 - 4c = 300 - 3c. I want to figure out the sign of the discriminant, which means I need the sign of 100 - c. If c > 100, then the discriminant is negative and I have only one root. Now, c > 100 implies ac > 100a (a must be positive if c > 100 since ac = 3000), which implies 3000 > 100a, and thus a < 30. Hence, if there is a root between 0 and 30, it must be unique. Plugging in x = 0, I get -3000, and plugging in x = 30 will give me something positive since 30*30*30 > 30*10 + 30*100 = 30*110. Therefore there is a sign change, and by the intermediate value theorem, I conclude that a solution exists for x < 30, thereby implying that only one root exists.
Hey I just wanted to say thank you so much for all the help with the physics revision tips and questions you posted for the May/June batch. I got my results for AS and ended up getting a very high A! Cheers mate and thank you once again!
wohoo! amazing to hear! thanks a lot for your comment, much appreciated!!
19:35 my heart dropped seeing you divide by x on both sides hahahaha
Mine too after I realized 😂😂
can u tell me why thats bad? was x 0 or was it cus he didnt factor x out then divide?
@@Imliterallyart a solution was 0 but he corrected it later so it worked out
What an awesome video!! It was extremely fun to try and solve along haha.
On that last question we even could divide that N+1 by 2, because the coefficients have to be symmetric, so we can't use different integers for the first and second halfs of coefficient spots, but it doesn't matter in the end.
22:40 Remember if you are about to make a sign error, always make sure to make two.
13:20 I just turned the function into an infinite geometric series which works out to be tan^2(x) and analyzed the graph of tan^2(x) from there as having asymptotes at +/pi/2 + kpi and zeros at kpi where k is an integer. It's also always positive due to being squared so d is the only correct answer.
Just wanna say thanks, got an A* in physics this year thanks to you.
Amazing to hear! Thank you so much for the comment!
C is simply tan^2 (x) - its thus easy to choose by inspection.
"To cook or not to cook" - issac newton
34:10 Did he not do the derivative of the sin functions incorrectly? He treated sin^(3)x as if it were x^(3) and used the power rule. And then the derivative of sin is cos? Unless I’m missing something because Tom didn’t interject
I was thinking the same
Omg your good at math as well, is there something you can't do. Man istg physics teachers are the smartest being to exist
41:48 "There are 2 possible sin solutions for every value." What about sin(x) less than or equal to -1?? Then there are 1 or 0 solutions. So to solve this completely, you should check f(-1).
I came to the comments to point this out!
I don't know how they missed that the derivative of sin^2 is 2sin cos not 2 sin. The function has 7 inflection points not 2!
And they aren't "inflection points" , they are critical points
@@d7home2129 he didnt actually derive d f(x)/dx he did d f(x) / d (sinx) , but didnt notate it correctly.
@@pianissimo7121 true. Actually the other guy explained it. I didn't notice it
For the quadratic in G it's quicker to multiply by 2 to give 2c^2+3c-2 = 0 and then factorise to give (2c-1)(c+2)=0 and then it's clear that c=1/2 is the only solution.
agreed, nicely spotted!
There are many cases in the exam when a far simpler approach would have generated the correct answer e.g problem F time 40:45 can be solved by drawing a graph approximating the functions and the answer is obvious. I recommend always draw a picture and this often leads to a solution.
40:48 I think I made the equation sin^3 x = - cos^2 x and draw each of those graph since its really easy to sketch if there are only one trig function to some power. Turn out there is only 2 solution from the sketches!
amazing!
For question F the problem could be written as sin(x) tan^2(x) = -1, and from the graphs of the sin and tan functions one can see that there two solutions over the interval 0 to 360 degrees.
@@ericerpelding2348 brilliant!
or you could just graph sin^3 and -cos^2
@@icodestuff6241 good luck doing that non calc
Are we all going to ignore how this guy crazily writes the letter X? I’ve never seen someone essentially connect a backwards C and a regular C … instead of the traditional crossing of two lines. Yet when he writes a multiplication sign (also an X) he uses the traditional method of just crossing the lines. Wild.
(Question)
Here at 1:11:00, could you think about it as:
- If amax = 2k/(N+1), while 2k = a0 + a1 + ... + aN, there are 2 possible cases
All terms are equal, then amax = 2k/(N+1), because It's the arithmetic mean of these factors
OR
There is a term that is greater than every other term and the arithmetic mean must be lesser than this mean
So, in both cases, amax is equal to the mean or greater than the mean?
“there are 2 possible cases: All terms are equal… OR There is a term that is greater than *every* other term”. These two cases are not exhaustive! If you changed *every* to *at least one*, then you’d be absolutely right. Very smart answer!
At 34:06, Shouldn't the derivative be 3sin^2(x)cos(x) -2sin(x)cos(x) because of the inner derivative of sine? Am I insane or just don't know maths?
@@thedogatemyhomework8 Nope, you are not, I was treating sin(x) as it's own function, say s, and ignoring the further x dependence to see how it behaves. Hope this makes sense
as a physics student this actually made me feel way more confident lmao
great to hear! thank you for the comment!
@@zhelyo_physics to add to this, it doesnt make me fell mor confident on the premise that ure bad at these problems or anything like that, im way closer to the time i took maths classes than u are yet i wouldve fared worse haha
@@bobfake3831 absolutely! I think YT needs more realistic problem solving videos. : ) I made a point to leave lots of the thinking time, little mistakes, getting stuck etc. Glad you have enjoyed!
awesome video :) really enjoyed it
@@4iden.r great to hear!
Question C: y is clearly a geometric series, with initial term sin^2(x), and common ratio sin^2(x), use the formula to get y=tan^2(x). The answer is clear then.
Love this so much!!!
product of square and cube can easily be solved by saying x^2*x^3=x^5 and this can only work when x=0 and 1 so only option c will be correct
"Possibly not an infinite number of sines added together..." There's a Mr. J. Fourier who would like a word with you.
Q2: isn't the general case quite trivial? my first thought was prime factors. given a^2*b^3, if factors of a are a subset of factors of b, always a cube, otherway around always a square? and both, if its the same set? or am I thinking too simplistically? 😳
Haha, adjectives such as “trivial” and “simple” means different things to different people, so I won’t comment on that. Your reasoning is correct and would, for this question, lead to the correct answer. And more than that, you’ve correctly identified the exact special conditions for three out of the four possibilities: (a), (b), and (e). Just for completeness, the remaining possibility (d) occurs if and only if a and b are coprime (have no common factors).
trivial in maths is when the value solves for 0 in additive equations or 1 in multiplicative equations.
23:20 this problem doesn't have 2 solutions. The solution to (2/3)a^3 - a^3 = 9.... (-1/3)a^3 = 9.... a^3 = -27.... a = -3 is a unique solution.
The a = 3 solution doesn't work. The symmetrical parabola with that same area between the curves that goes through y-intercept at "a" would not have the equation of y=x^2+2ax+a. The equation would look different than that to produce the symmetry of the parabola.
The problem does have two solutions, their reasoning simply wasn't that rigorous
th-cam.com/users/shortsLfXIT6zfrzE?si=amC-WT9drzck1FSJ
The graphs of the functions
@@m4riel no, it is not the same equation going through that y-intercept...
Great video- F - would you not also need to show the intercept is -1
Yes, and that's trivial to show using the intermediate value theorem.
Indeed! From the work done in the video, we already know the intercept is negative. Now, it is easy to see that the value of the function when x=-1 is negative (-1), implying that the function has not yet crossed the x-axis at x=-1 (while traversing the graph from left to right). Therefore, the intercept must be greater than -1, which means there must be a feasible solution!
So this exam is for people who just finished high school? I literally already forgot how trigonometry is solved
Is success in the exam and the interview enough to get in or do past academic success matter (olympiads etc.) ?
interesting question, while olympiad success looks phenomenal I imagine the vast majority of students have done very well on the exam and the interview. I think willingness to learn, self study of topics and mathematical interests and skill are most important. But that's coming just from my point of view of a teacher not affiliated with the university.
This guy has no idea, 20:00 you come out with 0 as a limit of integration? when you solve the equation you're dividing by x, so x=0 is the other solution you're looking for,
I think that u should do jee advance maths section
It is hard
thanks for the idea, I do love their physics problems
Great video just making my way through! :) BSc Physics student here, a question for all you wonderful mathematicians out there!
Is this valid for E? 🤔 @tomrocksmaths @ZPhysics
I rearranged to sin y + cos^2 y = sin x + cos^2 x
Simplifying to f(y)=g(x)
We see f()=g() and as such the equality can be reduced to 2*pi*n*y=2*pi*m*x
(As both functions have periodicity of 2 pi)
ny=mx ; where n and m are natural numbers
So e
Feel like there’s some issues with this, like I’m breaking some rules (but I don’t know why aha)
isnt 1 both a square nubmer and a cube number, which would imply c) ?
I'm also a physicist (at least by training... now I'm a maths teacher 😂), and I've never heard a physicist use the term "inflection point" to mean a stationary point, nor have I ever read about it in the physics literature. An inflection point is a point at which the curvature changes sign... also in physics 😇 But of course I don't know all the physicists or all the physics literature 😅 I have a suspicion that he got something mixed up...
I definitely did 😀 thanks for the comment!
Are You able to derive "from zero" an inverse transform to Laplace transform formula?
He created a transform and found an inverse formula, but how he get it?
Can someone explain the logic with derivatives in the first question?
I am missing a step, for example if equation is x^3-300x = 0, we can apply same reasoning, get same conclusion, but equation now has 3 real solutions.
The method is valid, but the 0 crossing is irrelevant, relevant part is 3000 crossing (constant) and that is missing.
Excellent question. Applying the same logic to x^3-300x=0 we would get stationary points at x=+10 and x=-10. However now let's plug those numbers back into the original function. When x=10 we get a stationary point at: 10^3-300*10 gives a stationary point at y=-2000. When x=-10 we have a stationary point at -1000-300*(-10)=+2000. Now if we plot those onto a graph our cubic will come from negative infinity when x is large and negative, keep going up, cross the x axis (root 1), then have a stationary point at (-10,2000) . Our next stationary point is negative in terms of y so then the wiggle will go down, cross the x axis again at (10,-2000) this is root 2. Okay, but we know that the cubic will shoot off towards positive infinity as x gets big so we need to cross the x axis again (root 3).
For x^3-300x-3000=0 our wiggle is shifted down the y axis so the wiggle happens entirely for negative y but it eventually crosses the x axis once for positive infinity. I recommend visualising this with wolfram alpha or a graph plotting software. Hope this is helpful!
Am I missing something or should you have applied chain rule in F?
Hi, yes, I treated the function separately, as sin(x) being y(x) or call sin(x) being s and ignoring the further dependence on x. Hence the comment from Tom later on notation 😂
@@zhelyo_physics Physicists always seem to have terrible notation. It took me five years and a course in differential geometry to understand the notation in thermo - but if the notation used was standard like the one used in mutivariable calc, I'd have understood it instantaneously.
Awesome vid!!
For the first question, i think the answer of 1 real solution is correct, but the derivation was a bit unclear? The critical points were indeed -10 (local max) and 10 (local min). The second order derivative around the min makes the curve inflect upwards, pointing to a root lying past x=10. With a calculator, the only real root is approx around 21.05.
Edit: I now heard Tom, he basically says the same 😂 sorry I need to be more patient with my posts.
Haha great to see you are so engaged though! Thank you for the comment!
Physics is math. A bit silly for Tom to say 'as a physicist you don't need to be able to do maths'. Physics is maths
For the first problem couldn’t you use b^2-4ac
Cubic not a quadratic
@@vghhhj1657 oh I did not see thank you
Great vid! So just to confirm for question D, is the answer ONLY a=-3?
That's what I got as when I equated the 2 parabolas, I got roots of x=0 and x=-a and integrated between those two points. I think the mistake you made was by dividing by x (as x could be 0) instead moving all the term on to one side and factoring which got me 2x = 0 and x + a = 0
But unclear whether the answer was actually (b) or (c) as Tom said (b) was correct, but I disagree (but I'm just an undergrad student lol) 😊
a = 3 also works, if you try it. Note that 0 < -a precisely if a < 0, so in the a < 0 case, you integrate from 0 to -a, while in the a > 0 case, you instead integrate from -a to 0. Therefore the answer is (b).
@@amritlohia8240 ah I think I understand now, thank you. So I did the definite integral with the upper bound as 0 and lower bound -a, but I also needed to do it with those reversed because I don't know that -a < 0
@@michaelcolbourn6719 Yes.
@@amritlohia8240 great, thanks for explaining :)
I think it depends on which way round you subtract the parabolas. They way they did it gives a = -3. If you do it the other way you need a = 3 to get a positive area.
36:00 the derivative of sinx isnt cosx in degrees, only in radians
tom rocks
@@JPEG.really agreed!
You ngas to smart for me
why dont you give him some actual math, that's not even things that he should struggle with that much. Give him some real analysis questions.
great idea! I have not done real analysis since 2010! I remember I very much enjoyed the course then. Would be a great fun video to revisit.
It would have easier if the x values had dimensions, I.e. units.
Yeah, he didn't get to it but the last question of part 3 basically requires you to find a way to extend the solution you obtain on some finite interval to to another finite interval that's larger. Now, the physicist in me screamed the obvious solution: the interval length is just a matter of units (if there were any) and since units are arbitrary it should be possible to extend the interval to any desired length. And a unit change is just a rescaling, so that's the answer.
ohh that sounds interesting, I'll have a look at that question!
Don't worry. He fell victim to circular definition in one of Numberphile's videos 😅
42:09 I would have followed my rule which states "make it easy". I actually went for the equation cos^2=sin/(sin-1)
I quickly ruled out that 0
I think zphysics is gonna smash this paper
Hehe thanks, I did find it very difficult though. Let's see! 😀
thank you sir I got an A in physics because of you
Amazing! Thank you for the comment!!
Zhelyo's maths is amazing!
However, there may be better solutions to a few questions.
@@sl2357 agreed!
Try IIT JEE ADVANCED Physics Paper. It's india's 2nd toughest entrance exam. Imagine a 12th grade student solving this!!!!
I have, I very much enjoy those problems. I have a JEE Playlist here: th-cam.com/play/PLSygKZqfTjPAs0IBTO4cXbTdwtQ1fpiX8.html
I am wondering why these two guys are in shock...As physics and maths are corelated, they are no different, especially in physics there is a lot of mathematics so there must be no difficulty for a physics teacher!
When you study mathematics specifically its VERY different
you must never proof a theorem in mathematics!
watching this gave me anxiety
👏👏👏👍
thank you!
@@zhelyo_physics😊😊👋
Ear longing, and a lip ring ...;goodbye
Вхахахха на превью негр написано 😂
As a jee student this is too easy😂😂
Bro jee is not hard
@@tfg601 are you Indian and have to give that exam? Because of not then it's quite easy to say an exam is not that hard if you don't have to give it. It has math that probably you learn at university elsewhere....
@@arnavthescientist1149 perhaps consider the same logic being applied to your original comment.
@@chemicalnamesargon yeah I know but seeing the questions they are visibly a lot easier than jee. Jee has a lot more formulas you have to memorize and generally requires 2 years of consistent studies. I agree that our education system is messed up and yours is a 100 times better. The things asked in jee should be taught in college and not in entrance exams.
Ok
This was absolutely painful to watch. So sloooooow…
thanks for the feedback! I feel it was important to show I can’t solve these problems instantly though so I left in thinking time and wrong approaches/mistakes.
@@zhelyo_physics just really surprising someone from Oxford, particularly maths/physics related, would struggle with any of this.
If MAT is a problem, you two would get finished by STEP 💪
@@zhelyo_physicsI think the way you took time and explained your approaches made for a very entertaining and educational video. I wouldn't have minded a longer video where you go through the whole paper with Tom because every question you did taught us to think in a different way.
@@apope2087 you do realise TOM did a step paper and absolutely smashed it? on video? He is just guiding zphysics in this video
@@varenbeats fine. Still would’ve expected a physics oxbridge type to eat this for breakfast “making mistakes for educational value” or not.
As an Indian, I don't find it torturing
Nobody cares he's from Oxford lmao, that doesn't make him smarter than any other university.