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
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
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.
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.
@@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))
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
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
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!
@@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 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
@@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
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 ^^
@@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)
@@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
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 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)
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.
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!
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 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
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.
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.
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
@@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.
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
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))
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 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.
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!
I haven't done the math yet, but my idea is to calculate the intersection of the parabola and the function representing the half circle above the x-axis to figure out the point to be rotated, and then use the cosine law.
I did so but I don't know what is cosine law so I just calculated f'(x1) where f(x) is x^2 and x1 is the point of intersection and tan(θ) = f'(x1) because of the geometrical interpretation of the derivative
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.
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°.
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
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.
I did it another way, without polar coords: You can draw a right triangle where the hypotenuse is 5,so you knoe that y² +x²=5². Furthermore, y =x² so y+y²-25=0 so sin(beta) = y/5 so beta = arcsin (sqrt(1.01)-0.1)
Okay this is probably overkill. But I used rotation transformation. Since it's a clockwise rotation, so the matrix will be: cos(-t) -sin(-t) sin(-t) cos(-t) cost sint A= -sint cost The vector v=(x,x^2) Now, Av=(5,0) xcost+x^2sint=5.....(1) -xsint+x^2cost=0 x^2cost=xsint x=tant tantcost+tan^2tsint=5 sint+sint*tan^2t=5 sint=5cos^2t sint=5-5sin^2t 5sin^2t+sint-5=0
If you know how to write equation for any graph rotated about origin by an angle θ anticlockwise, we just substitute: X = xcosθ + ysinθ Y = -xsinθ + ycosθ Since we are rotating clockwise, we take (2π - θ) as the angle. So X = xcosθ - ysinθ Y = xsinθ + ycosθ Now equation of curve: Y = X² And we need to find angle when it intersects (5,0), so we put in (5,0) in the substitution and equation satisfies: X = 5cosθ Y = 5sinθ =⟩ 5sinθ = (5cosθ)² =⟩ sinθ = 5cos²θ =⟩ tanθ secθ = 5 =⟩ tanθ(√(1 + tan²θ)) = 5 =⟩ tan²θ(1+tan²θ) = 25 =⟩ tan²θ = (√(101)-1)/2 =⟩ tanθ = √[(√101 - 1)/2]
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!
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..
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).
r=sec(theta)tan(theta) does help solve the problem. If we have the unknown angle of rotation be phi, then the equation becomes r = sec(theta + phi)tan(theta + phi) for clockwise rotations. Now just solve for phi where r=5 when theta=0. Reduces to 5 = sec(phi)tan(phi).
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 think mine is simpler, We have to equation 1) y=x^2 2) x^2+y^2=5^2 Subtitute 1 into 2 we get y^2+y-25=0 Solve it we get y=4.52494 y=-5.52494 Take the positive y and by using cosin We get Sin-1(4.52494÷5)= 64.8 degree
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.
3:02 i think it can help... We put r=5 and get theta from it. Then that would be the answer for the question as theta is angle of (5,0);origin and x axis. So in rotated graph, theta is zero and in first graph theta is what we found. So the difference, which is the same, is the answer
Looked at the question and saw that (5,0) is a point which is 5 distant to the vertex, of which there are only two, so to find that, plugging in y^2 + x^2 = 25 gives you x^2 = (sqrt(101)-1)/2 noting that x > 0. from there tan(theta) is x^2/x = x the diagram helped a lot, but in the contest itself, sketching would definitely be the first step.
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
I used complex numbers; take the original graph as x + i*x^2, then the new graph as e^-i(theta)(x+i*x^2) (this is saying the rotation of the graph by theta in the clockwise direction). Then pretty much the same thing follows
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 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!
@@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.
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
One other method we can use is one that is frequently used in our physics class. We can rotate the entire plane so that the parabola is where we want it to be. then all we need to do is find the angle between the primary and secondary field. we can use the polar form to do so.
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
I was able to solve this on my own, but in a different way. First I created a system of equations: x^2 + y^2 = 25 x^2 = y Then rewrote the first equation y=(25-x^2)^1/2 Plugged that into the second equation and rewrote it x^4+x^2-25=0 Let x^2 =u and do the quadratic formula and you get u=-1/2+sqrt(25.25) x= (-1/2+sqrt(25.25))^1/2 Plugged the solution back into the first equation and you get: y=(25+1/2-sqrt(25.25))^1/2 Now solve for the angle 5cos(x)= (-1/2+sqrt(25.25))^1/2 x = cos^-1(((-1/2+sqrt(25.25))^1/2)/5) 5sin(x)=(25+1/2-sqrt(25.25))^1/2 x = sin^-1(((25+1/2-sqrt(25.25))^1/2)/5) And I got x = approximately 1.131 radians.
made a line that intersects the parabola at the origin and at a point (x, y) in such a way that the distance from that point to the origin is 5, that is, x² + y² = 25, and using y = x², I calculated the coordinates (x, y) of the point and then I calculated the slope of the line, to then apply the concept that the slope of the line is the tangent of the angle with respect to the X axis, and voila.
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.
You could make it even simpler. With the equation R = sec(teta)tan(teta), R is simply sqrt(x^2 + y^2), where x and y are the coordinates of a point on the rotated function. Because (5;0) will be on our function, just plug them in. You get 5=sec(teta)tan(teta). You also get 2 answers which is more correct, see as though there can be either « branch » of the parabola that cuts through (5;0). The second answer is ~115.2 degrees. Of course, you’d need an advanced calculator for this method…
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).
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.
Omg wow I got it right My approach was to find a point cordinate which was exactly 5units away from the origin in y=x2 graphy then find tantheta= y/x to find the angle x2+y2= 25 y2+y-25=0 Quadratic formula y= -1+sqrt(101)/2 x = sqrt(-1+sqrt(101)/2) Tan theta = y/x = sqrt(-1+sqrt(101)/2)
You could've used the polar form to begin with. Just consider a phase shift and solve for the phase shift such that it goes through the polar point (5, 0). As he showed in the end, a phase shift in polar form rotates the plot rather than shifting left and right like rectangular coordinates.
My initial thought - rather than rotate the parabola, rotate the x axis. Or more properly, find the line y=mx where sqrt(x^2+y^2)=5. That gets me to x^4+x^2-25=0.
What I did: knowing that you can rotate any function θrads if you define y:=xsin(θ)+ycos(θ); x:=xcos(θ)-ysen(θ) you can define f(x, y): y=x^2 as f𝝑(x, y, θ): xsin(θ)+ycos(θ)=(xcos(θ)-ysen(θ))^2 and fix x=5 and y=0 => 5sin(θ)=(5cos(θ))^2; Solving for θ: tan(θ)sec(θ)=5 and by approximations (dunno if you can solve that by paper and pen, I couldn't) you get 1.13135rads.
This was so suspiciously easy I thought there was gonna be some catch, I was like "well you just draw a circle, where's the trap?" And there was no catch in the end.
Instead of rotating the parabola, rotate the x axis line. Parametric equations: (tcos(theta),tsin(theta)) and solve for theta where it intersects y=x^2 at t=5. Gives same result and less of a headache
Solved it using the rotation matrix and the vector valued function r(t)= representing the parabola. [cos(θ) -sin(θ)][t^2] =[5] [sin(θ) cos(θ)][t ] [0] From this I got = Using the magnitude, I then derived the equation t^4+t^2-25=0 Solving t for real numbers I got t = +-sqrt( (-1+sqrt(101))/2 ) Then from the vector I looked at (t^2)cos(θ)+tsin(θ) = 0 And got that tan(θ)=-t Hence θ = arctan(-t) θ = arctan( sqrt( (-1+sqrt(101))/2 ) )
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.
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
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.
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.
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...
I would probably just use a generic rotation matrix (a 2x2 matrix with cos(a) and sin(a) in) to transform the vector (x, x^2) then substitute in the coords for x and y and solve for a which should be fairly simple and just simplify to tan(a) = b
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!
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
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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
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)
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
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
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 : )
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!
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
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)
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.
At first i wondered why he's holding a pokeball, then i remembered it is a Mic.
Nice problem and a great explanation!!! 😍
Thanks, SyberMath!
@@blackpenredpen No problem, man! I love your teaching style!
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!
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 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 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
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.
You could also use the polar form and set "r = 5" and solve for theta
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.
Lmao this was what I thought when first learning about quadratic equation "I wonder if I can rotate this graph freely?"
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.
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
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))
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
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
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.
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!
I haven't done the math yet, but my idea is to calculate the intersection of the parabola and the function representing the half circle above the x-axis to figure out the point to be rotated, and then use the cosine law.
I did so but I don't know what is cosine law so I just calculated f'(x1) where f(x) is x^2 and x1 is the point of intersection and tan(θ) = f'(x1) because of the geometrical interpretation of the derivative
I'll try this out
@@lukandrate9866 can u explain eloperately
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
I never pause to do the problem, but today I did, and I got it right!!
Nice!!!!
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.
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°.
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
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.
I did it another way, without polar coords:
You can draw a right triangle where the hypotenuse is 5,so you knoe that y² +x²=5². Furthermore, y =x² so y+y²-25=0 so sin(beta) = y/5 so beta = arcsin (sqrt(1.01)-0.1)
Okay this is probably overkill. But I used rotation transformation.
Since it's a clockwise rotation, so the matrix will be:
cos(-t) -sin(-t)
sin(-t) cos(-t)
cost sint
A=
-sint cost
The vector v=(x,x^2)
Now, Av=(5,0)
xcost+x^2sint=5.....(1)
-xsint+x^2cost=0
x^2cost=xsint
x=tant
tantcost+tan^2tsint=5
sint+sint*tan^2t=5
sint=5cos^2t
sint=5-5sin^2t
5sin^2t+sint-5=0
If you know how to write equation for any graph rotated about origin by an angle θ anticlockwise, we just substitute:
X = xcosθ + ysinθ
Y = -xsinθ + ycosθ
Since we are rotating clockwise, we take (2π - θ) as the angle. So
X = xcosθ - ysinθ
Y = xsinθ + ycosθ
Now equation of curve:
Y = X²
And we need to find angle when it intersects (5,0), so we put in (5,0) in the substitution and equation satisfies:
X = 5cosθ Y = 5sinθ
=⟩ 5sinθ = (5cosθ)²
=⟩ sinθ = 5cos²θ
=⟩ tanθ secθ = 5
=⟩ tanθ(√(1 + tan²θ)) = 5
=⟩ tan²θ(1+tan²θ) = 25
=⟩ tan²θ = (√(101)-1)/2
=⟩ tanθ = √[(√101 - 1)/2]
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
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..
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).
r=sec(theta)tan(theta) does help solve the problem. If we have the unknown angle of rotation be phi, then the equation becomes r = sec(theta + phi)tan(theta + phi) for clockwise rotations. Now just solve for phi where r=5 when theta=0. Reduces to 5 = sec(phi)tan(phi).
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 think mine is simpler,
We have to equation
1) y=x^2
2) x^2+y^2=5^2
Subtitute 1 into 2 we get
y^2+y-25=0
Solve it we get
y=4.52494
y=-5.52494
Take the positive y and by using cosin We get
Sin-1(4.52494÷5)= 64.8 degree
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.
3:02 i think it can help...
We put r=5 and get theta from it. Then that would be the answer for the question as theta is angle of (5,0);origin and x axis.
So in rotated graph, theta is zero and in first graph theta is what we found. So the difference, which is the same, is the answer
I understood about 3/4 of what was explained, but I still enjoyed watching this video
wow a bprp video I was actually able to solve using the same method as the one shown in the video!
Looked at the question and saw that (5,0) is a point which is 5 distant to the vertex, of which there are only two, so to find that, plugging in y^2 + x^2 = 25 gives you x^2 = (sqrt(101)-1)/2 noting that x > 0.
from there tan(theta) is x^2/x = x
the diagram helped a lot, but in the contest itself, sketching would definitely be the first step.
4:04 Thanks for pointing that out :)
😂
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
I used complex numbers; take the original graph as x + i*x^2, then the new graph as e^-i(theta)(x+i*x^2) (this is saying the rotation of the graph by theta in the clockwise direction). Then pretty much the same thing follows
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!)
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
I really like problems like this, where there is a visual trick/strategy to come up with the correct answer.
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!
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
One other method we can use is one that is frequently used in our physics class. We can rotate the entire plane so that the parabola is where we want it to be. then all we need to do is find the angle between the primary and secondary field. we can use the polar form to do so.
What a beautiful question
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
I was able to solve this on my own, but in a different way.
First I created a system of equations:
x^2 + y^2 = 25
x^2 = y
Then rewrote the first equation
y=(25-x^2)^1/2
Plugged that into the second equation and rewrote it
x^4+x^2-25=0
Let x^2 =u and do the quadratic formula and you get
u=-1/2+sqrt(25.25)
x= (-1/2+sqrt(25.25))^1/2
Plugged the solution back into the first equation and you get:
y=(25+1/2-sqrt(25.25))^1/2
Now solve for the angle
5cos(x)= (-1/2+sqrt(25.25))^1/2
x = cos^-1(((-1/2+sqrt(25.25))^1/2)/5)
5sin(x)=(25+1/2-sqrt(25.25))^1/2
x = sin^-1(((25+1/2-sqrt(25.25))^1/2)/5)
And I got x = approximately 1.131 radians.
made a line that intersects the parabola at the origin and at a point (x, y) in such a way that the distance from that point to the origin is 5, that is, x² + y² = 25, and using y = x², I calculated the coordinates (x, y) of the point and then I calculated the slope of the line, to then apply the concept that the slope of the line is the tangent of the angle with respect to the X axis, and voila.
Wow, I figured out the same way after he said to pause and try
I have no idea what he said but I love it!
Your english is really, really good for being (I assume) from China. And your enthusiasm is contagious.
JEE method that I used.
Rotation of Axes ( Clock ) formula.
For y=x²,
Formula:
X= xcos - ysin
Y = - ( xsin + ycos )
now new rotated Parabola equation,
Y = X²,
But (5,0) is Old system i.e (x,y) system,
so x=5, y=0.
Substituting above in Formula and substituting that in new Parabola equation, we get,
5 sin = 25 cos²
sin = 5-5sin²
Solving for sin, we get,
sin =( -1 + √101 ) / 10
tan = ( -1 + √101)/ √2 * √( √101 - 1)
Rationalizing,
tan© = √ ( √101 -1) /√2 ~ [ 3/√2 ]
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.
You could make it even simpler. With the equation R = sec(teta)tan(teta), R is simply sqrt(x^2 + y^2), where x and y are the coordinates of a point on the rotated function. Because (5;0) will be on our function, just plug them in. You get 5=sec(teta)tan(teta).
You also get 2 answers which is more correct, see as though there can be either « branch » of the parabola that cuts through (5;0). The second answer is ~115.2 degrees.
Of course, you’d need an advanced calculator for this method…
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).
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.
Omg wow I got it right
My approach was to find a point cordinate which was exactly 5units away from the origin in y=x2 graphy then find tantheta= y/x to find the angle
x2+y2= 25
y2+y-25=0
Quadratic formula
y= -1+sqrt(101)/2
x = sqrt(-1+sqrt(101)/2)
Tan theta = y/x
= sqrt(-1+sqrt(101)/2)
I ACTUALLY DID IT WITHOUT WATCHING THE SOLUTION!!! Im so happy :)))
Great job 👍!!
I could somewhat follow the reasoning for this. I’m learning!
You could've used the polar form to begin with. Just consider a phase shift and solve for the phase shift such that it goes through the polar point (5, 0). As he showed in the end, a phase shift in polar form rotates the plot rather than shifting left and right like rectangular coordinates.
I never pause to do the problem, but today I did, and I got it right!!
I worked it out by hand up until taking the inverse tangent. Turns out I used just about the same method. Yay!
I solved it by instead rotating the axes and using the distance (5) from the origin to solve the equations tan(theta) = x and x^4 + x^2 = 25
this is the first problem on this channel i was able to solve! (out of the ones i have watched)
wish they teached these cool tricks in my school
My initial thought - rather than rotate the parabola, rotate the x axis. Or more properly, find the line y=mx where sqrt(x^2+y^2)=5. That gets me to x^4+x^2-25=0.
What I did: knowing that you can rotate any function θrads if you define y:=xsin(θ)+ycos(θ); x:=xcos(θ)-ysen(θ) you can define f(x, y): y=x^2 as f𝝑(x, y, θ): xsin(θ)+ycos(θ)=(xcos(θ)-ysen(θ))^2 and fix x=5 and y=0 => 5sin(θ)=(5cos(θ))^2; Solving for θ: tan(θ)sec(θ)=5 and by approximations (dunno if you can solve that by paper and pen, I couldn't) you get 1.13135rads.
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.
This was so suspiciously easy I thought there was gonna be some catch, I was like "well you just draw a circle, where's the trap?" And there was no catch in the end.
Yea. For some reasons I thought the question was hard and tricky too. But it wasn’t 😆
Instead of rotating the parabola, rotate the x axis line. Parametric equations: (tcos(theta),tsin(theta)) and solve for theta where it intersects y=x^2 at t=5. Gives same result and less of a headache
This is one of the first problems on your channel that I actually managed to solve
Solved it using the rotation matrix and the vector valued function r(t)= representing the parabola.
[cos(θ) -sin(θ)][t^2] =[5]
[sin(θ) cos(θ)][t ] [0]
From this I got
=
Using the magnitude, I then derived the equation
t^4+t^2-25=0
Solving t for real numbers I got
t = +-sqrt( (-1+sqrt(101))/2 )
Then from the vector I looked at
(t^2)cos(θ)+tsin(θ) = 0
And got that
tan(θ)=-t
Hence
θ = arctan(-t)
θ = arctan( sqrt( (-1+sqrt(101))/2 ) )
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.
0:46 I'll have a link in the descsoop
Finding the angle is slightly simpler if you notice that the hypotenuse is 5, so theta = arcsin((sqrt(101)-1)/10).
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
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.
i don't understand a thing but i love these kinds of videos.
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.
What a great question! I paused the video, solved it, got off the toilet, and let me say: it was definitely satisfying!
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...
I would probably just use a generic rotation matrix (a 2x2 matrix with cos(a) and sin(a) in) to transform the vector (x, x^2) then substitute in the coords for x and y and solve for a which should be fairly simple and just simplify to tan(a) = b
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!
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
Him: "You did?"
Me: Not even trying.
Him: "Cool!"
Not cool
Oh, that was nice!