My First Stanford Math Tournament Problem
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- čas přidán 24. 06. 2024
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This problem is from the Stanford Math Tournament. The goal is to rotate the parabola y=x^2 so that the curve passes the point (5,0). I first thought to convert the Cartesian equation to polar but it was a bit too hard than it had to be. Check out the video to see how I did it. Also, there's the equation and the graph of the rotated parabola at the end of the video. If you want to see more math for fun videos like this, then be sure to subscribe 👉 bit.ly/3o2fMNo and hit the notification bell. #stanford #blackpenredpen #maths #math
0:00 Stanford Math Tournament: sumo.stanford.edu/smt/
1:03 converting y=x^2 to polar
3:04 how I solved this problem
6:52 check out Brilliant
7:39 bonus part: equation of the rotated parabola and its graph
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Congratulations.
@@reidflemingworldstoughestm1394 for what????????/
I want to ask how to do any tricky questions
Your beard looks weird
I’m not sure anyone else has mentioned this but I came up with a really simple solution:
Rather than making a circle, just create a right triangle with base x and height x^2 (height is y)
This would seem useless, except that we’re actually given the value of the hypotenuse straight up. Its the distance from the origin to the point (5,0), so five.
Now we have a right triangle with sides x, x^2, and 5
From here we just use pythagorean theorem to create an equation for x-
25=x^4+x^2
0=x^4+x^2-25
This is just a simple way to reach the halfway point in the video- from there just solve as shown
Same
Yeah it also seemed to me like he could’ve skipped half the blackboard with the geometric reasoning. The theta, x, x^2 triangle left the problem self explanatory. But he solved it like a madman, proving he does not only know how to solve this exact problem but any problem with an analytic approach and useful math tools
Ya I did exactly that then to find the angle did inverse tan f(x) / x which worked
This is exactly the same as creating a circle, because the equation of a circle is found using a right triangle. x^2 + y^2 = r^2, it doesn't matter if you think of it as a right triangle or a circle.
I am so great to see that. I thought of that sooooo fast and it will work
Over the years you’ve grown my Love for mathematics! So much so, in fact, I’ve picked up a major in applied mathematics and physics while doing my degree in pharmacy. I very much appreciate your videos
Glad to hear! Best wishes to you, Cobalt Chemistry!
@@blackpenredpen thank you :)
W brotha
OMG I CANT BELIEVE I ACTUALLY DID IT
I attempt to do the same way but I converted cos²θ to 1-sin²θ, then I assume I add a value 'a' in θ will make parabola rotate to that shape, so sub in quadratic in sine with θ=0, r=5 to solve a and I get 1.131 rad too.
Yeah I don't know why he said it doesn't help? I thought he'd substitute r=5 and end up with a cubic equation or something
@@tgwnn yeah, check out my comment about it if you want, but you can just do this:
rsinθ = r^2 cos^2(θ), cancel 1 r, use cos^2(θ) = 1 - sin^2(θ), put r=5
You'll get a Quadratic in sinθ, then you can solve for tanθ, using the fact that they are both positive as θ is acute, and thus tanθ = sinθ /√(1+sin^2(θ))
Exact way I did it
@@tgwnn it doesn’t help in the sense that it doesn’t make the problem easier
@@MikehMike01 that's arguable right? I'd say it's a more straightforward way of getting the same equation so it makes it easier (in a way)
The first problem of your channel that I solved correctly is an unique emotion! Thank you, your videos are really helping me to improve my math abilities!
Glad to hear : )
The polar form can also be used to solve this. Just add an offset to the theta of polar form equation, put r=5, theta=0 and solve for the offset and you’ll get same answer
Or you can just plug in 5 for r in the equation and use the identity cos^2(theta)+sin^2(theta)=1 to solve
That's how I learnt to do it
Finally rotations. Now go for quaternions, and explain them for me so I can understand what that numbers are.
Nobody understands quaternions lol
@@tgwnn same lol
There are also books online and other videos, you know😁
@@tgwnn 3blue1brown would like to know your location
@@SimonPetrikovv I watched his videos (re-watched a little now too), and they're definitely very well done, and everyone should watch them, but I'd still say that it doesn't really come close to "understanding" them. It's still some scary frankenstein method that just works. But this is just my impression and my intuition, if yours works differently, more power to you :)
You were my professor for the alg1/2 combo class at pierce I think 2014-2015! you were one of my favorites! I think I got a A in your class lol
Wow it’s been a while! Hope all is well!
Has my life really resolved to watching these kinds of videos in my spare/relaxation time before heading to my Analysis class?....😅😂
The way you made it seem like such a simple process doing things I could never have thought of. Amazing!
This was a great problem! First I tried to find the arc length of the parabola until I realize that since it was a curved line the arc length would be greater than 5 at x=5. But then I used the circle method as well.
Nice problem and a great explanation!!! 😍
Thanks, SyberMath!
@@blackpenredpen No problem, man! I love your teaching style!
At first i wondered why he's holding a pokeball, then i remembered it is a Mic.
My two solutions that I really like:
1) Just like how he showed in the video, get r = sec(θ) tan(θ) , then substitute r=5 (you can think about this as rotation of axes, if it doesn't immediately make sense intuitively) , square both sides, use sec^2(θ) = 1 + tan^2(θ) , and solve for tan(θ) (you'll get a Quadratic in terms of tan(θ))
2. (I already wrote a comment about this before watching the video, where I've explained it in detail)
Imagine instead of solving in this situation, where the parabola is rotated, we rotate the parabola back by θ, and along with the portion of the x-axis that it cuts between the origin and (5,0), now this rotated line segment is a part of a straight line passing through the origin, so it's equation is of the form y=kx, where k is the slope of this line, which is also equal to tanθ, also note that k is a positive real number, as θ is acute. Now, using some standard Co-ordinate geometry and algebra, we find the intersection point of y=x^2 and y=kx, in terms of k, and now we set the distance of this point from the origin equal to 5, as by rotating, the length of the line segment (secant inside the parabola, I guess you could say) does not change. And after simplifying a bit by squaring and all that you get the same Quadratic in k, as you did in tanθ in the first way.
(The Co-ordinates of point are (k,k^2), and the Quadratic in both cases (in terms of tan^2(θ), or k^2 in the second case, and not in terms of just k, or just tanθ) is of the form t^2 + t - 5^2 =0, solving which you will get t=(√(101)-1)/2, taking only the positive value of t)
I get yours comment; That's great that you written such long comment bro 🙏🏻🙏🏻;
I want to discuss my idea in short; I am weak in such typing skills.
@@sr.tarsaimsingh9294 ok, what do you wanna discuss?
HI. (1) no need to square, express in sine and cosine... you have your quadratic already
@@donc9836 ok
I would probably just multiply by a matrix and substitute in x and y to get simultaneous equations in terms of sin(a) and cos(a) then solve for a. seems like the most intuitive method off the top of my head (from the perspective of an A level further maths student)
Ahh I see how you solved it. I solved it by making a circle of radius 5, and finding the intersection between that circle and the original equation. This gave me a point on y=x^2 that when rotated by an angle, would produce (5,0), from there I treated the point on y=x^2 that intersected with the circle as a vector from the origin to that point, I then treated the point (5,0) as a vector from the origin as well. From there I used the dot product identity to find cos(theta) and then find theta.
A great question for any calc student to practice such as myself, thank you
This is one of the first problems on your channel that I actually managed to solve
What a great question! I paused the video, solved it, got off the toilet, and let me say: it was definitely satisfying!
Awsome . This is amazing. Though I am not good at math but this video made me smile . I watch ur video everyday its entertaining.
You could also use the polar form and set "r = 5" and solve for theta
you could also do it by converting to polar,
from 2:10 :
r = sin/(1-sin^2)
and r is just 5 in this case(since the rotated parabola crosses x-axis at 5)
just solve for sin(thena) and we get
sin(theta) = (-1+root(101))/10
theta = 64.8
Yes, we can also solve for tan theta from here, as asked originally
Why is r=5? r would be 5 only if theta was 0 no?
@@Darkev77 why would r be 5, if the angle is 0, if the angle is zero, then r is also 0 because the only point on the parabola which is at an angle 0 from origin and x-axis, is the vertex of the parabola
@@anshumanagrawal346 thanks for the response! But isn’t theta=0 at r=5 (x-intercept)?
@@Darkev77 You have to decide, are you analysing the case with the rotated parabola, or rotated axes?
The way I did it was to put the rotated version of the cartesian plane on the original one (but labeling x' and y' on the other axis), then I saw that the angle between the axis x and x' was exactly θ, so I needed to use the rotation matrix like this:
[x] = [cos(θ) -sin(θ) ] [x']
[y] [sin(θ) cos(θ)] [y']
Since in the x' y' coordinated system the point (5,0) is in the parabola, we can put (5,0) as the input (x',y') then we'd get (x,y) = (5cos(θ),5sin(θ))
Since y = x^2, then we'd have 5sin(θ) = 25cos^2(θ) => sin(θ) = 5cos^2(θ) => 5sin^2(θ)+sin(θ) = 5(cos^2(θ) + sin^2(θ)) = 5 => 5sin^2(θ)+sin(θ)-5 = 0 => sin(θ) = (-1+-sqrt(101))/10
Since θ is in the first quadrant, we'd have to choose + instead of -, so we'd get sin(θ) = (sqrt(101)-1)/10, since sin^-1 is defined in the interval [-π/2,π/2], then we can use sin^-1 since the sine is positive and we'd get θ = sin^-1((sqrt(101)-1)/10) and we're done
If you're asking if it's the same answer as in the video, we can find cos(θ) from sin(θ) = 5cos^2(θ), which would be 5cos^2(θ) = (sqrt(101)-1)/10 => cos(θ) = +-sqrt((sqrt(101)-1)/50), since θ is in the first quadrant, then we'd get cos(θ) = sqrt((sqrt(101)-1)/2)/5
Since sin(θ) = 5cos^2(θ) => tan(θ) = 5cos(θ) => tan(θ) = sqrt((sqrt(101)-1)/2), which would yield the same result ^^
how did u use the rotation matrix tho
@@aghaahmed5946 if you rotate the axis by a angle of theta, having new axis x' and y', the relation between those coordinates are given by the formula I used there
The thing was to make sure it was really a rotation of theta in the counterclockwise direction, which was the way I've choose to represent the two sets of axis to get the point (x',y') = (5,0) into the parabola
But the way it was shown in the picture, you rotated the parabola instead of the axis, I just made things the other way around in other to use the rotation matrix
@@aghaahmed5946 but if the question is how to get the rotation matrix (maybe without using linear algebra, only geometry), I can show you by sending a link to imgur with the explanation (which I have yet to write xD)
@@SimonPetrikovv like i know abt the matrix but i’ve never used it like this
@@aghaahmed5946 neither did I! I saw the problem and thought it might've worked, and it really did! It was a nice trick in order to do this specific question xD
What a beautiful question
Thanks for that amazing video, I've already solved it.
More importantly, I want to thank you for letting me know about this competition.
Sadly, I won't be able to participate because I'm not a student at the USA, but at least I would be able to enjoy the problems.
Thanks again.
I really like problems like this, where there is a visual trick/strategy to come up with the correct answer.
Learned something new today :)
Definitely very cool to come look at problems after I understand the math behind the concepts so I can really know what is going on. This is literally just algebra with some trig
I never pause to do the problem, but today I did, and I got it right!!
Nice!!!!
wow a bprp video I was actually able to solve using the same method as the one shown in the video!
This is a very cool problem. I saw it instantly because of you
Using multivariable techniques, you can parametrize the parabola as r(x)= and so the vector that points to the point (x0,x0^2) that will end up pointing to (5,0) must necessarily have the same length as the vector namely 5. So ||=5 => sqrt(x0^2+x0^4)=5 => x0^4+x0^2-25=0. When you solve for this quartic by keeping in mind that x0 is positive, you get x0=sqrt((-1+sqrt(101))/2). You can then use the formula a*b/(|a||b|)=cos(theta) to solve for the angle
That was really easy to understand
this is the first problem on this channel i was able to solve! (out of the ones i have watched)
Lmao this was what I thought when first learning about quadratic equation "I wonder if I can rotate this graph freely?"
Stopped the video to try it by myself. Did it almost the same way just a little different way to solve X^4 + x^2 -25 = 0. I liked this Math problem
omg this is the first bprp question I could actually solve
only difference from your solution is that I answered as the sin^-1 (because I dont like sqrt very much) but of course the decimal value is the same
Thank you teacher
wish they teached these cool tricks in my school
great video as always
I did it with complex numbers, but it's basically the same thing as you did, except for the idea of intersection between the circumference and the parabola. With complex numbers I just wrote that each point of the curve is w=|w|*cis(theta) and multiplied it by cis(-theta) cause the rotation is on the antitrig direction. I ended up with the same quadratic equation and solved it just as you did.
Oh, that was nice!
I just found a video of yours where you found the integral of x^i, and after watching it, I became curious about what the derivative of x^i is, so I went looking, but couldn’t find a video on it. If you ever need a new video, could you do the derivative of x^i?
Feeling proud to solve it
I could somewhat follow the reasoning for this. I’m learning!
I calculated it by finding the intersection point of the circle and parable. Then i calculated the distance between the intersection point and (5, 0) and used it to form an isosceles triangle inside the circle, with the base being the line segment between the intersecton point and (5, 0) and the legs being the radius of the circle. Lastly I calculated the angle using the law of cosines.
I understood about 3/4 of what was explained, but I still enjoyed watching this video
Heeey it is a rotation matrix then! 😂
Indeed
I didn’t use any matrices.
I thought of that too.didnt realize the question wanted clockwise rotation and got a negative answer
I agree based on just looking, but I didn't know Sir Blackpenredpen here did a way simpler method.
@@potatochips9019 Matrix is definitely more a simple way
4:04 Thanks for pointing that out :)
😂
I thought about this in 9th grade for the first time and thought it should be trivial. Later I thought it was impossible because how a function is defined. Now I'm looking forward to learn how to solve it.
I used an alternate method that involves using a general formula (it gives the rotation anticlockwise though), xSin(a) + yCos(a) = (xCos(a)- ySin(a))^2. I subbed in the values given, x=5 and y=0 then, I rearranged to calculate the value of feta. Upon expansion on the right with the substituted values you get 5Sin(a) = 25Cos^2(a), take out 5 and then use trig identity of Cos^2(a) = 1 - Sin^2(a). You then get Sin(a) = 5 - 5Sin^2(a). I then solve it quadratically and get Sin(a) = 1 - squrt(101) all over 10. The answer I got was -64.8 degrees or positive rotation of 295.17.
I left-multiplied the vector (x,x^2) by the 2D rotation matrix, set the resultant vector equal to (5,0) and solved for x and theta (need to take the absolute value of the angle in this case since we are rotating clockwise). I got theta = arcsin((sqrt(101)-1)/10) by using the 2D rotation matrix.
You can use r=sec(θ)tan(θ). If the angle of rotation we want is a, then the new rotated graph is r=sec(θ+a)tan(θ+a). We need this to go through the polar coordinate (5,0) so substitute r=5 and θ=0 to get 5=sec(a)tan(a). Multiply through by cos^2(a) to get 5cos^2(a)=sin(a). Use cos^2(a)=1-sin^2(a). 5sin^2(a)+sin(a)-5=0. And then use quadratic formula to solve for sin(a) and then arcsin and we are done!
So , we get theta = arcsin[(-1plus or minus (101)^1/2]÷10....?
Yeah I got the the same values....
I solved this one by realizing that the secant intersecting the parabola at the origin and (5,0) would have the same length in the original position, so I just calculated the coordinates and used tan^-1 - very similar to what you did! :)
I have no idea what he said but I love it!
OMG, this was so simple. I am definitely out of shape... When I was in school I probably would have thought about the intersection with a circle (if I had any idea) but now my first thought was using a rotation matrix... damn...
Your english is really, really good for being (I assume) from China. And your enthusiasm is contagious.
this is surprisingly easy for a tournament problem, just a simple calculation with quadratic equation and some trigonometry :)
Very nice video as usual 👌
Thanks.
@@blackpenredpen I'm an UG and I love your videos. Your channel is so underrated it hurts.
The way I solved it was imagining a circle of radius 5 in order to find out where it intersected the unrotated parabola because that would be the corresponding point on the rotated parabola. Then I simply set up the equations
y=x²
x²+y²=25
Got the intersection point and took the inverse tangent of the x value
Wouldn't you take the inverse tangent of y/x? Like so: arctan(y/x) ?
This how I did it. I find this approach rather elegant and quick.
@@foxtrot8325 Since y=x^2, you get phi=arctan(y/x)=arctan(x)
@@foxtrot8325 Originally that’s what I was going to do, then I realized because the parabola is defined by the equation y=x², which means arctan(y/x)=arctan(x²/x)=arctan(x)
I did this too because I haven't really done polar coordinates. I added extra steps and treated both points as vectors from the origin, where I took the dot product cos(theta)= (a dot b )/ mag(a)*mag(b), and took the inverse of that to find theta.
I didn't pause and try. I'm soooooo ashamed!
I did ! Check out my solution and tell me what you think. Which is better ? Mine or bprp's ?
No worries 😆
And now with x^4 and N1= 0 N2=4
I did, and it was pretty complicated because I had to discover what rotating a graph actually means
I ended up solving a trigonometry equation: 5*(sin teta)^2 + sin teta - 5 = 0
So I kinda understand why you didn't try
Everybody poops
i used a bit lengthy and calculative method. I assumed a parabola tilted at an angle theta and used its property PS=PM to define the locus and then use the point (5,0) to get a trignometric equation.
Your Energy is damn good!!!
Will you solve JEE Advanced Maths questions?(Toughest exam in the world;)
The way I went about solving this is to realize that the point being rotated to (5, 0) is a distance of 5 from the origin. Using 5 = sqrt(x^2+y^2) you can solve for x, and then solve for y.
This is how I did it:
In the second graph, notice the point (0,0) (vertex) and (5,0) are 5 units apart. I just had to find in the first graph which point was 5 units apart in straight line from the vertex of the parabola (0,0).
By Pythagoras: x^2+y^2=5^2=25, and substituting x^2=y (the function), I got to y+y^2=25 : y^2+y-25=0, whose positive solution is [-1+sqrt(101)]/2
We know from the function that y=x^2=[-1+sqrt(101)]/2, so x=sqrt([-1+sqrt(101)]/2). Now the line that goes from the origin to the point (sqrt([-1+sqrt(101)]/2), [-1+sqrt(101)]/2) is just y=([-1+sqrt(101)]/2)*x
We can define the slope of a line (its derivative) as the tangent of the angle it forms with respect to the origin, such that: tan(α)=sqrt([-1+sqrt(101)]/2)
Applying inverse tan on both sides, we get that α=arctan(sqrt([-1+sqrt(101)]/2))
Thank you sir ❤️❤️
I did this using matrices, and it ended up being a lot longer than your method! But one advantage is i was able to find out, if you use a parametrization (t, t^2) and plug it into a standard rotation matrix, along with your theta working, there is a second solution of the form pi-theta, and the t value where the intersect occurs will be -5cos(theta).
you have a great channel . i love your contant
Great video
I had a similar solution except I used sin(θ) = y/5 = x^2/5 at the end rather than tan(θ). Saved an extra square root.
Enjoyed the question.
Saw the still, got the equation for tan theta exactly as shown, wondered what vile sorcery would enable you to calculate theta from it, and... oh
I'm really grateful for your work, I use your videos to pass college lol. Can I ask why the group playlists on your home page is gone? The playlists of ODE was very helpful because it was arranged
I am very sorry. I am re-organizing all my channels these days. If you need help, please let me know.
@@blackpenredpen Wish me luck sir! hahaha, its finals week.
I got it! Sweet. I didn't actually write out all the work to get the answer, I just had an idea, so I paused the video to think about how I would generally go about solving it, and convinced myself that it would work. And it turned out to be the exact same steps you took. Lol
Yep. That's how I did it! I thought it's easier to rotate that point up to the parabola rather than the other way around. Same angle either way.
Exactly, its more simple to just rotate the axes and then just get x²+x⁴=25 where tan (ø)= x
Please make videos other questions questions of this paper too
I like that. My first intuition was to draw a circle of radius 5. Then I have 2 equations and I can solve.
Imo the question becomes pretty easy once you understand and visualize the turning of the graph..
Instead of rotating the y=x² graph...its better to rotate the axis instead...since both are relative to each other the angle of rotation would remain the same..
so basically we'll get a triangle with hypotenuse as 5, one side say x² and the third side being x..
OK, I was just doing a complex analysis problem where I was asked if the graph of a complex function is a parabola and I thought it is a tilted version. And this showed up!
I didn't know it was possible to rotate functions. Neat~
The general formula for getting a point (r,0) by rotation is rotating the angle arctan( sqrt [ -1 + sqrt(4r^2 + 1) / 2]) degrees. (though i didn't find it out by myself, i referred to your video, and then used a general value x^2 + y^2 = r^2
I ACTUALLY DID IT WITHOUT WATCHING THE SOLUTION!!! Im so happy :)))
Great job 👍!!
Yesss, I managed to solve it using the pythagoras method!
I solved it by setting up the triangle you showed later instead of the circle, then using pythagoras to get the same equation as the circle formula, 25 = x^2 + x^4 and then solved from there
i solved by this way too, high 5!)
Just rotated the number 5+0i anticlockwise by angle θ like so:
( 5+0i )( cosθ + isinθ) = 5cosθ + 5isinθ
Now y = x² so 5sinθ = (5cosθ)²
So, 5sin²θ + sinθ - 5 = 0 and solving by the quadratic equation(and ruling out the one wrong root) we get:
θ = arcsin( (√(101)-1)/10) = 1.131 rad =64.82° (approx)
This is the reason I *LOVE* Complex Numbers.
i don't understand a thing but i love these kinds of videos.
I SOLVED IT!!
Simply I take a parametric point on the parabola (t/2 , t^2/4) So , tan@=t/2.
Using distance formula, equate distance of parametric point from the origin equal to 5. This will give us value of t.
using t tan@ can be calculated.
To do this we can first find how to rotate a point(using a right angled triangle) then its darn simple , plug in the result in the equation to rotate the parabola by any arbitrary angle(however take care of anticlockwise and clockwise rotation),then we can put in (5,0) in our equation then find the angle satisfying our equation
I assumed that y=i*x^2 by setting the y-axis as the axis of an imaginary number.
A point on the function is an imaginary number x+i*x^2, so
I tried to find the real solution x of (x+i*x^2)(cosQ+i*sinQ) = 5+0i.
x=2.12719 and Q=1.1313rad, same result.
This will be the same method as the rotation matrix.
I learned a lot from this lecture and thank you.
At 2:45 I totally disagree, you totally could've used r = sec(theta)tan(theta) to solve the problem by setting r =5, solving for theta and a little minor adjustment... nice idea though!
Yeah, I solved this way too
I totally messed up there.
@@blackpenredpen lol no worries! I appreciate your humility
Another great method is by axis rotation. For an observer on the parabola, if I were to rotate the parabola clockwise by theta then it would appear that we're rotating the axis anticlockwise by theta. As we know, in case of axis rotation, the new y coordinate
x(n) = x(old) cos t + y(old) sin t and y(n) = -x(old) sin t + y(old) cos t
Plugging in x(old) and y(old) as 5, 0 we get:
x(n) = 5 cos t, y(n) = - 5 sin t
now, all this has to obey is the parabolic equation if it is to touch the parabola (ie: y = x^2)
25 cos^2 t = 5 sin t
solving for t we get 1.131 radians!
At a glance. I would find the point on the graph y = x^2 that is of distance 5 from the origin. Then you will get a point (a,b) and have a sort of circle of radius 5. Then you have a triangle with base a and height b, and you can use inverse tangent to get the angle
The problem becomes really trivial if you also visualize rotating the axis of symmetry of the parabola. That right angle triangle pops right out.
I am a Japanese university student.
At first, I thought that theta = 45° just by looking at it, but when I saw the video, I was impressed. Thank you for your easy to understand English!
Now I'm curious. What would the x- and y-intercepts be if we rotated the parabola by exactly 45 degrees?
@@BillyBob-wh4sq
x = r cos(theta)
y = r cos(theta)
put that in y = x^2
r sin(theta) = r^2 cos^2(theta)
sin(theta) = r (1-sin^2(theta))
r sin^2(theta) - r + sin(theta) = 0
set theta = 45°
r/2 - r + 1/sqrt(2) = 0
r/2 = 1/sqrt(2)
r = sqrt(2) so after rotating y = x^2 by 45° we will get (sqrt(2), 0) point.
Another way is to calculate intersection point of y = x^2 and y = x, so we have (0,0) and (1,1) (lets ignore the first one).
If we rotate y = x by 45°, then every point (a, b) will become (sqrt(a^2+b^2), 0), so by puting a=1 and b=1 we are getting the answer.
@@hejkanaklejka1021 this makes sense, thank you very much!
I know that to do a clockwise rotation, you replace x with xsinθ + ycosθ and replace y with xcosθ - ysinθ. Then I plugged in 5 for x, 0 for y, and solved. My answer was θ=arcsin[(-1+sqrt(101))/10] which is also about 64.8°.
I just solved for the intersection between the parabola and a circle of radius five with x^2=sqrt(25-x^2) and then solved for x by subbing t=x^2 and using the quadratic formula. Then I just used inverse tan on the y over x values from the previous solution to find theta.
great stuff.
I also thought about finding the point of intersection with circle with radius 5. Then I thought dealing with sin/cos/etc. is too much for me, so it would be easier to find a ratio between the perimeter of the entire circle and the part of the perimeter between the point of intersection with parabola and x-axis. It would result in a ratio between 360 degree and the angle in question. Then the question transforms into finding the length of the part of the perimeter (I think there is a word for it, maybe it's called sector or segment) between 2 points on the perimeter. At this point I couldn't see an easy way to calculate it, so I gave up.
Sir please do a video on relations and function..please
I used 5 as a triangular hypotenuse, then found a data point that gave 5 on the function y=x^2. Roughly 2.125 and 4.5156. Found the angle between them. Then 90-phi gave our theta as 64.8
This funny dude is doing cool videos 👍
The way I solved it was representing only the right side of y=x^2 with x=sqrt(y). I then found a line that starts at (0, 0) and intersects a point on x=sqrt(y) so that it has a length of 5. I found the x and y points by starting with x^2 + y^2 = 25 and sqrt(y) = x. I then derived y + y^2 = 25 which can be represented as y^2 + y - 25 = 0. I used the quadratic formula to get y = (sqrt(101)-1)/2. From there I found x by plugging it into x = sqrt(25-y^2). Then the answer is y/x and the angle is arctan(y/x).