@@ArnabChatterjee-ni9qy ya i did its just that you know tantheta=-1/a so then u gotta do the trig stuff by squaring both sides unless im missing an easier way
@@hjker Make a right angle triangle with sides a,1 and sqrt(1+a^2). now since tan is negative, either sin or cos is negative. Take accordingly as the sign of distance and information of location of point
Sorry Ellie. you lost me at "just a bit of manipulation" at 5:38. Would you be kind enough to explain the steps involved please? In particular, I don't get how you got the implied identity -xb/a = +1/a^2 in the fifth equation down, or am I missing something obvious here?
Indeed her argument doesn't work. Instead, once you get to (a-x_b)^2*(1+1/a^2) = d^2, simply divide both sides by (1+1/a^2), then take the square root and you're done.
@@amritlohia8240 I think you're right, but it was the step immediately before that I didn't get - how you get to (a-x_b)^2*(1+1/a^2) = d^2 in the first place.
@@Treviscoe Oh - for that, we have (x_b-a)^2 + (y_b-a^2/2)^2 = d^2 from the distance condition, and now substituting y_b = -x_b/a + 1 + a^2/2 from the equation of the normal gives (x_b-a)^2 + (-x_b/a+1+a^2/2-a^2/2)^2 = d^2, i.e. (x_b-a)^2+(1-x_b/a)^2 = d^2. Now rewrite 1-x_b/a as (a-x_b)/a, so we get (x_b-a)^2+(a-x_b)^2/a^2 = d^2, and since the square of a number is the same as the square of its negative, we can change this to (x_b-a)^2+(x_b-a)^2/a^2 = d^2. Finally, factoring out (x_b-a)^2 gives (x_b-a)^2*(1+1/a^2) = d^2.
Any good books you could recommend for those of us who have forgotten/never really gotten the hand of this? I’ve noticed most books just tend to throw problems at you instead of actually establishing the rules/explaining properly.
1st step: you use the fact that AB and the tangent at A are perpendicular, etc, you find the equation of the tangent at A (using calculus), you do the dot product of the vector between B= (x, y) and A = (a, 1/2 a^2), etc any vector having the same slope as the tangent, etc...set it equal to zero, etc...then find what x and y have to be if the length of this vector is d, etc, you know, setting sqrt( x-coordinate^2 + y-coordinate^2)=d, etc...should be able to find x and y from all that, maybe there is a simpler way, that's what came to mind, lol...
For the second one, just find the intersection of the line issued from A = (..., ...) that passing through B and the parabola "on the other side", lol...find set that intersection point equal to what was found for d, I'm sure the identity will emerge (like, set the x-axis equal to the intersection, set the y-axis equal to the intersetion, etc...a lot of calculations, I would mess something up, but it's fairly straightforward)...
For the 3rd one, solve for d, etc...d=(1+a^2)^3/2/a^2, etc...replace all the a^2 's by t, etc...raise everything to the 2/3, should cancel out the exponent at the top, etc...call that f(t)...
The function, I would suppose, will be differentiable, lol, for a>0 (so that d>0 too)...take the derivative, set it equal to 0, the whole shabang (not sure that's how it's written, lol, shabang)...find the minimum value...it's what they mean afterward I'm wondering about, lol...
Ok, Jesus, lol, yeah, d^2/3 = f(t), so the minimum value of f(t) is the minimum value of d^2/3...d is positive, etc, should be easy to see what the minimum value of d has to therefore be, etc...yeah, don't know why, reading that last bit confused me, lol...
Hi Ellie! I hope this comment finds you well! :) Ive been watching your channel for a little while now and convinced my 7 year old cousin to as well - he is watching you on the big screen every day! :) As much as I love math, I feel like my knowledge as of yet is so little (currently a freshman @HS) and I really want to learn more but whenever I sit down to i am confused on where to start since in school we have set books. Anyhow, I feel like you would be able to provide me with invaluable guidance on how i should start! Thank you and I look forward to your reply! 🚀
Aaaah that’s the cutest thing ever!! 🥺 I’ll make a video on this! It’s a topic I’ve been wanting to cover for ages! Feel free to message me on instagram and we can chat ☺️
Hello i did my master's in financial engineering in korea. In financial engineering master's program, i encountered many students who did pure mathematics as a bachelor. Did you ever think about going into finance or economics phd as those fields use a lot of math and can potentially offer lucrative career opportunities?
At 11:13, I would argue you should show that t = 2 is indeed a minimum either by looking at concavity or showing that the gradient changes from negative to positive. Great video by the way!
Thank you! Yes, you could do that but given the function has one value of t for a stationary point and the question says to find the minimum, it’s more or less implied. The question would say ‘determine if there exists a minimum’ if it wanted you to do that extra work. It’s an Oxford admission test so being fast on these is crucial which is why I left that part out :)
@@EllieSleightholm I figured that was the case. My professors always made us justify trivial things for the sake of completeness, which I appreciate to some extent. Old habits die hard!
Hii ellie Hopefully you are doing great My humble request to you, could you solve UPSC mathematics optional paper please? It is one of toughest exam in india. Hopefully you will enjoy the solving such kinda of difficult question
"Physicists know that on the smallest scales of the Universe pairs of particles pop into existence, suddenly appearing out of the vacuum only to rapidly recombine and disappear again. They are called ‘virtual particles’ and Stephen Hawking wondered what would happen if this process unfolded right on the event horizon of a black hole. If one particle crosses the event horizon then it is forever separated from its companion and can never recombine with it. The particles that are left outside the event horizon are called ‘Hawking radiation’. However, the particles had to ‘borrow’ energy from empty space to appear in the first place. Normally this debt is repaid when they recombine, but as that can’t happen in this case they effectively default on the loan. The repayment has to come from somewhere: the black hole. So over time a black hole slowly loses mass due to the constant need to cover these energy debts to empty space. It means that a black hole slowly evaporates over time." - Colin Stuart for BBC Sky at Night Magazine Jan 2024
I'm making a follow up video going through the entire structure of the exam and diving into a few more questions, so subscribe to stay updated! :)
I will wait😊
I will also wait . however you explain the math so easy and smoothly I understand all the explanation of the math thanks 🥰
❤
Pure matematics is awesome Ellie!! 😊👍 Your resolutions is the best thing in every video, always very explanatory.
Thank you so much 🥺
Thank you so much Ellie!!❤
Well first part could be done easily using polar forms. x-rcostheta, y+rsintheta, where tantheta=slope of normal, r=d and x,y are coordinates of A
yes this works but its kinda annoying sifting through the trig identities to find sin and cos values through tan^2, but yh works
@@hjker Simply make a right angle triangle.
@@ArnabChatterjee-ni9qy ya i did its just that you know tantheta=-1/a so then u gotta do the trig stuff by squaring both sides unless im missing an easier way
@@hjker Make a right angle triangle with sides a,1 and sqrt(1+a^2). now since tan is negative, either sin or cos is negative. Take accordingly as the sign of distance and information of location of point
@@ArnabChatterjee-ni9qy ahhh yeah I didn’t label the sides lol I just used the fact tantheta = -1/a etc. so yeah that would’ve been way easier. Thx!
Sorry Ellie. you lost me at "just a bit of manipulation" at 5:38. Would you be kind enough to explain the steps involved please?
In particular, I don't get how you got the implied identity -xb/a = +1/a^2 in the fifth equation down, or am I missing something obvious here?
Indeed her argument doesn't work. Instead, once you get to (a-x_b)^2*(1+1/a^2) = d^2, simply divide both sides by (1+1/a^2), then take the square root and you're done.
@@amritlohia8240 I think you're right, but it was the step immediately before that I didn't get - how you get to (a-x_b)^2*(1+1/a^2) = d^2 in the first place.
@@Treviscoe Oh - for that, we have (x_b-a)^2 + (y_b-a^2/2)^2 = d^2 from the distance condition, and now substituting y_b = -x_b/a + 1 + a^2/2 from the equation of the normal gives (x_b-a)^2 + (-x_b/a+1+a^2/2-a^2/2)^2 = d^2, i.e. (x_b-a)^2+(1-x_b/a)^2 = d^2. Now rewrite 1-x_b/a as (a-x_b)/a, so we get (x_b-a)^2+(a-x_b)^2/a^2 = d^2, and since the square of a number is the same as the square of its negative, we can change this to (x_b-a)^2+(x_b-a)^2/a^2 = d^2. Finally, factoring out (x_b-a)^2 gives (x_b-a)^2*(1+1/a^2) = d^2.
@@amritlohia8240 Thanks, it's a bit late now and I'm a bit tired but I'll look at that soon.
Can you make a Tutorial video about how to take effective math notes step by step? Btw love your content
To be honest, I think that this question is quite appealingly written, which I find somewhat surprising
Please please make a video on good maths books for selfstudy
Use parametric equation and enjoy
Could you please tell me what is the difference between MMath and MASt and is one more preferable than the other?
Thanks from Azamgarh, up, India
Any good books you could recommend for those of us who have forgotten/never really gotten the hand of this? I’ve noticed most books just tend to throw problems at you instead of actually establishing the rules/explaining properly.
1st step: you use the fact that AB and the tangent at A are perpendicular, etc, you find the equation of the tangent at A (using calculus), you do the dot product of the vector between B= (x, y) and A = (a, 1/2 a^2), etc any vector having the same slope as the tangent, etc...set it equal to zero, etc...then find what x and y have to be if the length of this vector is d, etc, you know, setting sqrt( x-coordinate^2 + y-coordinate^2)=d, etc...should be able to find x and y from all that, maybe there is a simpler way, that's what came to mind, lol...
For the second one, just find the intersection of the line issued from A = (..., ...) that passing through B and the parabola "on the other side", lol...find set that intersection point equal to what was found for d, I'm sure the identity will emerge (like, set the x-axis equal to the intersection, set the y-axis equal to the intersetion, etc...a lot of calculations, I would mess something up, but it's fairly straightforward)...
For the 3rd one, solve for d, etc...d=(1+a^2)^3/2/a^2, etc...replace all the a^2 's by t, etc...raise everything to the 2/3, should cancel out the exponent at the top, etc...call that f(t)...
The function, I would suppose, will be differentiable, lol, for a>0 (so that d>0 too)...take the derivative, set it equal to 0, the whole shabang (not sure that's how it's written, lol, shabang)...find the minimum value...it's what they mean afterward I'm wondering about, lol...
Ok, Jesus, lol, yeah, d^2/3 = f(t), so the minimum value of f(t) is the minimum value of d^2/3...d is positive, etc, should be easy to see what the minimum value of d has to therefore be, etc...yeah, don't know why, reading that last bit confused me, lol...
Thank you, I enjoyed that! (If I may offer a tiny tiny tiny comment? - "amazing" seems to be evolving into a verbal tick for you.......)
Thanks, enjoyed making it! Looool it’s because I find maths amazing 🤣 I’ll switch up the words next time 😭
Hi Ellie!
I hope this comment finds you well! :)
Ive been watching your channel for a little while now and convinced my 7 year old cousin to as well - he is watching you on the big screen every day! :)
As much as I love math, I feel like my knowledge as of yet is so little (currently a freshman @HS) and I really want to learn more but whenever I sit down to i am confused on where to start since in school we have set books. Anyhow, I feel like you would be able to provide me with invaluable guidance on how i should start!
Thank you and I look forward to your reply!
🚀
Aaaah that’s the cutest thing ever!! 🥺 I’ll make a video on this! It’s a topic I’ve been wanting to cover for ages! Feel free to message me on instagram and we can chat ☺️
@@EllieSleightholm thanks so much Ellie!! I sent you a request on LinkedIn with the username 'Mariam S.'
i don't have Instagram :(
Hello i did my master's in financial engineering in korea. In financial engineering master's program, i encountered many students who did pure mathematics as a bachelor. Did you ever think about going into finance or economics phd as those fields use a lot of math and can potentially offer lucrative career opportunities?
At 11:13, I would argue you should show that t = 2 is indeed a minimum either by looking at concavity or showing that the gradient changes from negative to positive. Great video by the way!
Thank you! Yes, you could do that but given the function has one value of t for a stationary point and the question says to find the minimum, it’s more or less implied. The question would say ‘determine if there exists a minimum’ if it wanted you to do that extra work. It’s an Oxford admission test so being fast on these is crucial which is why I left that part out :)
@@EllieSleightholm I figured that was the case. My professors always made us justify trivial things for the sake of completeness, which I appreciate to some extent. Old habits die hard!
@afrolichesmain777 same here, definitely prefer that approach ☺️ the joys of being a mathematician!
Dreamed she was flying a helicopter. Maybe that comes after the PR phase.
Hii ellie
Hopefully you are doing great
My humble request to you, could you solve UPSC mathematics optional paper please?
It is one of toughest exam in india.
Hopefully you will enjoy the solving such kinda of difficult question
IQ 300 admission TEST
😮
Hack all pone numbers gauss 😅
A GENUINE REQUEST : make a video on hawking radiation hypothesis explain it
Otherwise lose your subscriber
"Physicists know that on the smallest scales of the Universe pairs of particles pop into existence, suddenly appearing out of the vacuum only to rapidly recombine and disappear again. They are called ‘virtual particles’ and Stephen Hawking wondered what would happen if this process unfolded right on the event horizon of a black hole. If one particle crosses the event horizon then it is forever separated from its companion and can never recombine with it. The particles that are left outside the event horizon are called ‘Hawking radiation’.
However, the particles had to ‘borrow’ energy from empty space to appear in the first place. Normally this debt is repaid when they recombine, but as that can’t happen in this case they effectively default on the loan. The repayment has to come from somewhere: the black hole. So over time a black hole slowly loses mass due to the constant need to cover these energy debts to empty space. It means that a black hole slowly evaporates over time." - Colin Stuart for BBC Sky at Night Magazine Jan 2024
@@davidnewell3232 thanks got that ^_^
You are so beautiful!!!!!!!❤❤❤❤