- 34
- 149 339
Maths Paz
Registrace 13. 12. 2023
I'm a 40 year-old father of two and I set out to create a CZcams channel such that its viewers would certainly end up benefiting from watching it. So, what better topic than Maths? I chose Maths because I have 100% confidence that my maths videos will have some degree of a positive impact on the people who watch them. And as for "Paz", I felt I should add a unique word to my channel's name to make it stand out, so I added "Paz", which means peace in Spanish. I like words with nice meanings.
So welcome. Feel free to check out my videos, and comment on them, telling everyone what you think of my work, or suggesting alternative solutions whenever you find one because I love alternative solutions.
So welcome. Feel free to check out my videos, and comment on them, telling everyone what you think of my work, or suggesting alternative solutions whenever you find one because I love alternative solutions.
How to Solve a System of Equations by Elimination and Substitution: Linear Equation in Two Variables
The tutorial should clear all the basic questions in your head about solving a system of linear equations that features a pair of equations with two unknown variables. There are two methods you can use when dealing with simultaneous equations of this kind: The elimination method and the substitution method. I have covered both of these techniques in detail to make sure everyone can understand how to find x and y. Working with equations is not only a prerequisite for getting good grades and passing the exams. It is also the best way to develop key maths and algebra skills such as removing the brackets using the distributive property, factoring, combining like terms, transposing, cross-multiplying and isolating the variables. This video demonstrates some of these skills, though not all because I have solved only one system of equations and a very easy one, too. But you can check the other videos I make on similar topics to find out more on these simple but effective techniques. I hope you find this useful!
zhlédnutí: 49
Video
Awesome Analytic Geometry Problem with Fun Solution: Finding the Areas Under the Curve with Integral
zhlédnutí 1,2KPřed měsícem
This super cool analytical geometry question is great for learning and practicing some key concepts in analytic geometry as well as calculus. I don't think this one can be considered insanely hard but it needs you to use and thus review integration as a way to find the areas of certain regions under a function on the coordinate plane. The rules of integral is one of the most advanced topics in ...
HARD Analytical Geometry Problem: Maximum Area of Rectangle Under Function Curve - incl. Derivative
zhlédnutí 1,3KPřed měsícem
This intense but super fun analytical geometry puzzle is one of the best questions I’ve created because it involves key math topics that will help improve your math and geometry skills for sure! I wanted to call this one insane but I am withholding that adjective for more special stuff I am going to present in the future. Yet, this one will also fire up the parts of your brain that deal with ma...
Crazy Analytical Geometry Problem that's HARD to solve: Circle Under Function Curve - Find the Area
zhlédnutí 15KPřed 2 měsíci
This super interesting analytical geometry question deserves to be called insane because this is the hardest maths question I solved in my entire life. But I prefer to call it crazy instead. And it is actually an extremely fun puzzle that will turbocharge your neurons with its expert-level difficulty. The solution is long and complicated but it’s a very cool solution. Overall, I think this prob...
Fun Analytical Geometry Problem that looks INSANE: Square Under the Function Curve - Find the Area
zhlédnutí 352Před 2 měsíci
This extremely interesting analytical geometry question looked insanely hard when I first created it but then I solved it and it turned out to be a very fun puzzle with a slightly above-the-average difficulty and a very cool solution that's actually very helpful for learning or reviewing important maths and geometry concepts. As you see in the figure, there is a quadratic function and it's got ...
Cool Geometry Problem to Train the Mind: Find the Area of the Excircle of the Triangle
zhlédnutí 2,8KPřed 2 měsíci
In this extremely cool and moderately difficult geometry puzzle, we've got a right triangle with a 30-degree angle and a blue inscribed circle in it, which has an area of 3 pi. Touching the triangle from outside, is a bigger, yellow circle that is the excircle of the triangle and we are asked to find the area of that yellow circle. I think this geometry problem has an average level of difficult...
A Geometry Challenge with a Surprising Solution - Isosceles Triangle: 36, 36 and 108 Degree Angles
zhlédnutí 559Před 3 měsíci
In this minimalistic but fairly challenging geometry problem, we've got an isosceles triangle in which the top angle is 108 degrees. The only distance we know is the side across that angle, which is 10. And we are asked to find the area of the entire triangle. I think this geometry problem has an above-medium level of difficulty, so it is not impossible but most would have a hard time solving i...
A Nice Geometry Challenge to Activate the Brain: Find the Area of the Green-Shaded Trapezoid
zhlédnutí 1,2KPřed 3 měsíci
In this engaging geometry problem, we've got a rectangle with three straight lines extending from three of its vertices, and they meet at an external point, creating two separate angles. We know one of those angles is 45 degrees. We do not know any side lengths of the rectangle but we know the length of the two seeming random line segments in the diagram and we’re asked to find the area of the ...
A Lovely Geometry Puzzle for Mental Stimulation: Chords & Quadrilateral Region inside Quarter Circle
zhlédnutí 655Před 3 měsíci
In this reasonably challenging geometry problem, we've got a quarter circle with an unknown radius and it’s got two chords inside it that together form a quadrilateral region. We know the lengths of the chords. And we are asked to find the area of the green shaded region that is the quadrilateral shape. This geometry problem has a medium level of difficulty, so it is not super-difficult but als...
An Engaging Geometry Puzzle: FIND the Area of the Yellow Circle inside a Semicircle
zhlédnutí 1,6KPřed 4 měsíci
In this super interesting and fun geometry problem with a difficulty level that is somewhere between moderately hard and not so easy, we've got a big semicircle that contains a smaller semicircle and a whole circle that is squeezed into the remaining space inside the big semicircle. We are not given any side lengths whatsoever and we are asked to find the area of the yellow circle. The only thi...
Fun Geometry Challenge: Find the Area of the Yellow Circle below the Chord
zhlédnutí 2,5KPřed 4 měsíci
In this extremely fun and moderately challenging geometry problem, we've got a semicircle and a chord inside the semicircle. Underneath the chord, there is a smaller, yellow circle and we are asked to find the area of that yellow circle. This geometry puzzle has a slightly above-average level of difficulty, and I actually designed it to be a fun question that will make you have a nice and relax...
This Tricky Geometry Question Has a Super Cool Solution: Find the Area of the Purple Region
zhlédnutí 1,8KPřed 4 měsíci
In this super-exciting and fairly challenging geometry problem, we've got a quarter circle and a whole circle tightly placed inside a big square. The circles overlap and they intersect at two points but we do not know their radii. In fact, the only length we know in this problem is the length of the square. And, we are asked to find the area of an uncanny quadrilateral purple region created by ...
Crazy Geometry Problem: Find the Area of the Green Circle - Five Circles Inside One Big Circle
zhlédnutí 2,5KPřed 5 měsíci
This geometry problem is unique although at first glance, it looks like a classic circle-inside-circle question. I call it crazy because I made sure it’s a super-exciting problem with a very nice solution. There are three identical circles tightly crammed into one big circle and there is an uncanny chord separating them that also acts as a tangent to each one. There are two other smaller circle...
Thrilling Geometry Problem: Find the Shaded Area - Two Circles with a Common Tangent in a Square
zhlédnutí 3,6KPřed 5 měsíci
In this extremely fun and reasonably demanding geometry problem, two identical circles are located inside a square. The circles do not touch each other but they have a common tangent line. We are given that the radii of the circles are 5 and 12. And we are asked to find the area of the blue-colored shaded region, which is a super weird-shaped region between the circles. This geometry problem ha...
Fabulous Geometry Problem: Find the Shaded Area - Twin Circles in a Right Triangle
zhlédnutí 4,2KPřed 5 měsíci
In this extremely cool and moderately challenging geometry problem, we've got two identical circles located inside a right triangle. We do not know any sides or radii whatsoever but we are given that the hypotenuse of the right triangle is seven times the radius of the circles. We also know that a particular segment on the triangle is 2r 1 . Yes, you read right! We are not given any lengths. An...
Fascinating Geometry Challenge: Find the Shaded Area - Semicircle inside a Right Triangle
zhlédnutí 6KPřed 5 měsíci
Fascinating Geometry Challenge: Find the Shaded Area - Semicircle inside a Right Triangle
Challenging Geometry Problem: Two Circles inside a Rectangle - Find the Shaded Area
zhlédnutí 6KPřed 5 měsíci
Challenging Geometry Problem: Two Circles inside a Rectangle - Find the Shaded Area
Fun Geometry Problem with a Cool Solution - Find the Shaded Area - 3 Squares in a Semicircle
zhlédnutí 6KPřed 5 měsíci
Fun Geometry Problem with a Cool Solution - Find the Shaded Area - 3 Squares in a Semicircle
Fascinating Geometry Problem: Find the Shaded Area - Two Semicircles in An Equilateral Triangle
zhlédnutí 4,3KPřed 6 měsíci
Fascinating Geometry Problem: Find the Shaded Area - Two Semicircles in An Equilateral Triangle
Geometry Challenge: Find the Area of the Blue Region - 3 Circles inside 1 Big Circle
zhlédnutí 25KPřed 6 měsíci
Geometry Challenge: Find the Area of the Blue Region - 3 Circles inside 1 Big Circle
Intriguing Geometry Problem with Fascinating Solution: Square and Circle Overlap! Find the Area.
zhlédnutí 6KPřed 6 měsíci
Intriguing Geometry Problem with Fascinating Solution: Square and Circle Overlap! Find the Area.
Circle Sectors: Geometry Challenge - Find the Sum of Angles Alpha and Beta
zhlédnutí 839Před 6 měsíci
Circle Sectors: Geometry Challenge - Find the Sum of Angles Alpha and Beta
Challenging Geometry Problem with a Cool Solution: Find the Area of the Green Region
zhlédnutí 32KPřed 6 měsíci
Challenging Geometry Problem with a Cool Solution: Find the Area of the Green Region
Complicated Geometry Problem with A Beautiful Answer: Find the Area of the Pink Circle
zhlédnutí 7KPřed 6 měsíci
Complicated Geometry Problem with A Beautiful Answer: Find the Area of the Pink Circle
4 Triangles in 1 Square - Find the Magenta Area - Nice Geometry Problem
zhlédnutí 413Před 6 měsíci
4 Triangles in 1 Square - Find the Magenta Area - Nice Geometry Problem
Find the AREA challenge! One Circle In Another - Concentric Circles - Cool Geometry Problem
zhlédnutí 10KPřed 6 měsíci
Find the AREA challenge! One Circle In Another - Concentric Circles - Cool Geometry Problem
Square inside Triangle - Find the AREA - Cool Geometry Problem
zhlédnutí 2,4KPřed 7 měsíci
Square inside Triangle - Find the AREA - Cool Geometry Problem
Find the Ratio of SHADED Regions: Cool Geometry Problem with Circles.
zhlédnutí 208Před 7 měsíci
Find the Ratio of SHADED Regions: Cool Geometry Problem with Circles.
How to Find the CIRCUMFERENCE of a Circle: C = 2πr Formula and Example Questions
zhlédnutí 4,2KPřed 7 měsíci
How to Find the CIRCUMFERENCE of a Circle: C = 2πr Formula and Example Questions
The Missing Angle: How to FIND IT in any Triangle instantly!
zhlédnutí 70Před 7 měsíci
The Missing Angle: How to FIND IT in any Triangle instantly!
Excellent Calc1-level problem. I'd almost hoped you had a different way to solve it lol.
I almost finished doing my video about the most insane parabola problem yet. I'm a bit busy nowadays so I cannot guarantee I will upload it immediately. But make sure you check back in the coming weeks.
You said "complicated", so I made it MUCH more complicated: I did a "u" substitution, ended up using quadratic eqn...and landed @ r=0.933. So close. Lol.
At the time I solved this one I thought it was super complicated. But then I kept coming up with more and more complicated problems.
Boom. Thanks again
👏
Thanks!!
Proud of myself for arriving @ "r=162/25". Probably took me like an hour and a half. I solved everything in terms of "X", ended up with some nasty fractions. Thanks for this, Paz
That looks good kudos. And thanks for the comment. I really meant for the 9s and 5s to cancel out but I misplaced the numbers somewhere when creating the problem and ended up with this ugly. I was so frustrated 😢
You blazed through the solution so fast and changed the graphics so quickly, its all but impossible to learn. Not a fan. I attacked it a different way: [⅓Big Circle]-2*[Small Circle Segments]-[⅓Triangle] [⅓Pi(2+Sqrt(3))²]-[2*(150/360)Pi(Sqrt(3)²]-[⅓•½•2Sqrt(3)•3]
Hi Nandi. Thanks for the feedback. It's hard to pinpoint shat my viewers need and that looks like some important feedback I need to consider seriously. I didn't assume most people would be patient enough to watch a math video for more than 5 minutes so I tried to get to the point as soon as possible. But I guess I need to pay more attention to the needs of people who are prepared to spend dome time to learn stuff. As for the solution. It's a great solution. Actually I realized after I made the video that the problem could be solved more quickly like that.
Where do you find questions like this
I create them myself. I try to create the coolest possible questions. I hope you're enjoying.
@@Maths-Paz huh wow that's amazing. You have a really high order of learning. I been trying to find more questions like this.
Just hang on for a few days more. The most insane parabola area problem yet is on the way!
Great choice of fun math problem! Excellent graphics with a clear explanation. If you want to use a free, browser-based modeling tool to draw and/or solve math problems, see how I solved a somewhat similar problem at czcams.com/video/BLDpTQ5ediM/video.html.
found a simple way to solve this instead, just think about the max where x is on the parabola (x=10), then we know that (x-10)^2+y^2=r^2, but we can otherwise make it so the y offset is away from the x - axis, so the equation you need to work with is (x-10)^2 + (y-r)^2=r^2, then plug in y= 10x - 1/2x^2 - 18 into our circle equation, now go find the minimum for r. which will be (x-10)^2 + ( [10x - 1/2x^2 - 18] - r)^2 = r^2. although thinking about this requires a graphing calculator, but is otherwise another way to solve through it
Never knew that geometry can be so fun
How did forget to reply this concise and awesome comment! Thank you for this!
AC = (4 + 2√2)√2 = 4√2 + 4 = 𝒓√2 + 𝒓 hence 𝒓 = 4 The point of intersection to the left of B is called E. The tangent point below B is called F. Centre of circle O. 2(AE)² = 8² = 64 ∴ AE = 4√2 So EB = 4 + 2√2 - 4√2 = 4 - 2√2 BF = 4 + 2√2 - 4 = 2√2 Therefore, by Pythagoras, (EF)² = 32 - 16√2 Apply the law of cosines to △EOF to obtain ∠EOF = 45° Area of circular segment created by chord EF = [sector EOF] - [△EOF] = (4²)π·(45°/360°) - (4·4sin45°)/2 = 2π - 4√2 Red shaded area = [△EBF] - [segment] = ((4 - 2√2)2√2)/2 - (2π - 4√2) = 8√2 - 4 - 2π
f(x) = 8x - x^2 - 1 = 0 → x1,x2 = Q,P = 4 ± √17 ∎ABCD → AB = BC = CD = CM + DM = a + a = 2a → AM = a√5 → AMB = δ → sin(δ) = 2√5/5 → cos(δ) = √5/5 → tan(δ) = 2 PM = 4 - √17; QM = 4 + √17 → PM + QM = 8 → PM = QM = 4 → g(x) = 2x - 8 = f(x) → x1, x2 = 3 ± 4 → x = 7 → a = DM = 7 - 4 = 3 → 2a = 6 → area ∎ABCD = 36
Great vid, great explanation, straight forwards, many thanks, Sir. Never stop learning 🙂
Thank you for your fantastic comments Murdock. Great commenta like this motivate me and I need them to keep making more and more interesting videos! I appreciate it!
This is really awesome, many thanks, Sir! Way beyond most other YT math videos. Go ahead! 😊
Thanks for this encouraging comment Murdock. Make sure you check back in the coming weeks because I've got a problem which is insanely hard and it's coming soon.
❤
Thanks for the comment!
This is the method that I used: I did b=6a-a². So what I did is [integrate y=6x-x² - ½(ab)=32/3]. After I got what's a and what's b which a=4 and b=8, I subtract the whole area under the curve from (32/3) and the triangle to get the answer of 28/3.
That looks perfectly correct! Congrats and thanks for the comment!
@@Maths-Paz thank you
Excellent explanation. High level math genuinely enriches my mental power. Million thanks squared^*^
Glad you enjoyed. Thanks for the fantastic comment!
Don't forget the +C
Oh I know about the C in the integral. I guess they show the C when explaining integral in theoretical manner in a sense that integral is an inverse derivative so we should think maybe the original function had a constant term of some sort. Well, I have to yet to remember many things in calculus 😊
@@Maths-Paz if you solve a definite integral (with interval) no need the +C since the answer will be a constant
Oh that's some good news 😄 so I was right? C is just a theoretical something?
@@Maths-Paz C is there to prevent contradictions like 0=1
That made things more clear.
Sir, you wrote in the text that the purple area was 32/2 and I solved with that value getting different solution. Moreover if you want you can use the archimedean method in which the area under a parabola is 4/3 the area of the triagle whose vertexes are the intersection with x-axis and the the vertex of parabola. This area is 6*9*1/2 =27 then the area under the curve = 4/3*27 = 36 Finally thank you for the simple integral😅
I must be rushing really bad not to see that blatant mistake. I guess I'll have to literally live with that because there is no way to change videos after upload and it's going to stay there forever. The only good thing is it's very small for most people to care. Thanks for realizing and telling me, that'll let me be more careful in the future. As for the parabola 4/3 formula. I am shocked to learn this because nobody taught me and it's not something I would imagine on my own. I think that's a super handy rule. Thanks a lot!
Let y = mx,find intersection and then find the area substrating mx from the curve equation between the intersections and equate to area value to find the value of m ( slope) which will be 2. Then find the intersection ( which was previous in terms of m) as (8, 4) other than (0, 0) of course and now integrate the curve integral from x=4 to x=6
Looks like a solution by a math expert. I think after subtracting mx, you need integral to find the area. Am I right?
@@Maths-Paz Yes which will give ( 6-m) ^3 = 64
OK. (6-m)^3=64 is some sort of rule for the area of the parabola? Or is it integral?
@@Maths-Paz Hahaha good way to increase comments... Equating integral area which is ((6-m) ^3) /6 to 32/3( Area of the parabola above the line with slope m, gives the above result...
Finally! It took me some time to figure out x = 6 - m ... That does save some algebra. And coukd even help with more complicated problems that otherwise seem impossible to solve.
I solved it slightly differently but I think your method is better. -- Find the point of intersection of the curve and the line (y = mx): 6x-x² = mx -> [ (6 - m), m(6 - m) -- Integrate the difference between the curve and the line ∫(6x - x² - mx)dx, between 0 and (6 - m) = 32/3 which gives m =2 which in turn gives the abscissa of the intersection at x = 4. -- Finally integrate the curve between 4 and 6 to find the answer. P.S. I wonder if the curve had different coefficients, would we have to solve a cubic equation to find a?
Thanks for this alternate solution. I guess your and my solutions would take approximately the same time to complete. It's always good to look from a different perspective. And about the cubic equation. Yes it is possible. More likely you need a cubic function to get a cubic equation. Actually my initial plan was a cubic equation function with a super cool-looking graph but I tried to solve it and realized I was doomed no matter how I adjust the coefficients. Sometimes I create problems I can't solve.
@@Maths-Paz Ok that’s hard to believe. You seem to find the most inventive way to solve the unsolvable. When does your son come back? Find myself missing his contribution.
@@Maths-Paz lol...ur not alone...
@@Ron_DeForest ..hes the best 😁 .. always comes in and solves everything nicly in the end...
..had the same approach... when i first did see it .. its just... wait whats the slope..lol
Thanks for this nice problem.
Well, thanks for watching and the nice comment!
green area=(28/3)u^2
I don't want to give spoilers for my super exciting video 🤣😅 but it looks good
I gave you my answer before waching your answer ..👍@@Maths-Paz
I knew that 😊 congrats!
Trying again... Here's a Desmos graph for this problem. You can resize the rectangle and see the area as function of the x-coordinate of the left hand edge (description in the formula panel). Because comments with links keep getting removed automatically: 1) Go the Desmos graphing calculator and pick any graph from the examples list via the menu (icon with three horizontal lines) 2) Replace the end of the URL with: g4eblvofjr
Hi there! CZcams has removed my comment with a Desmos link again. Feel free to make it publicly viewable if you like it (only the latest attempt with the full URL). Cheers!
Hi Fran! Nice to see your comment. I think your link comment is completely deleted. I don't see it in awaiting review box or anywhere. Just try typing the end part of the URL in the comment and say something like "put this at the end of the desmos url." Most people would figure that out.
GREAT VIDEO !!! SUBS FROM INDIA !!!❤❤❤
Welcome to my channel Arawindh. Thanks for subscribing!
Worked it out in my head, glad to see answer was correct.
Dude I completely imagined from scratch and solved a few of the problems on this channel while I took kids for walk but this one is way too hard for that. I am starting to feel like the world's highest iq people are watching me.
@@Maths-Paz Haha, well it maybe comes from having done similar problems a lot with students (even if not quite at this level), but it was essentially a quadratic equation and a derivative.
Nice to have maths teachers watch my channel. Hope you like the way I explain things.
Bro wake up , Maths Paz uploaded
Thanks for the comment bro! I am so proud to have loyal fans like you!
@@Maths-Paz no problem also I missed your son's voice haha
Oh don't worry. he'll be back!
At first glance this seems to be a very difficult problem, but after watching your video, I have seen that your reasoning works very well and in a simple way. Thanks for making me learn this new technique to solve problems like this.
Thanks for this awesome comment Max! I am so energized to hear that my video was helpful to you!
That was awesome. The only thing you didn’t do was rationalize the denominator. Missed your little guy. Hope he’s having an amazing time.
Thanks a lot Ron. Every time I release a video I look forward to seeing your comment! I am trying to explain everything as simply as possible. I hope it's helpful for everyone. My son's having a good time. He's got a dozen or so cousins over there. But he'll be back eventually.
@@Maths-Paz one thing.. tha max of a quadratic of the form y=ax²+bx +c is x= -b/2a [when y'=0]-> for s=12u² - 48u + 32 -> -(-48)/2(12) =2 [s'=0 when u=2] √c=u -> c=u² -> c=4 the intersection points of u is -b/2a±Δu [Δu= √( (b²-4ac)/2a)] or written as the quadratic standard form solution (-b ± √(b²-4ac))/2a
Why you add the childs voice to this tutorial, I don't know, compleatly ruined it
Introducing the parameter 'c 'complicates matters. It is simpler if solve the problem in terms of b. The upper two points of the rectangle are intersections of the line y = b and the curve. The difference of the two solutions of b = 4 √x - x , is 'a' = 8√(4 - b), And the Area of the rectangle A = 8b√(4 - b). Differentiate A with respect to b to find the maximum at b = 8/3
That was an excellent alternate solution Andreas! Thank you for that! You see I am trying hard to explain in a way so that nobody would end up like "hold on, where did that come from?" Of course, that poses the risk of overthinking and overcomplicating things on my part. So these different approaches that I am seeing in the comments are really helping me to stay on the course and optimize my future solutions. So everyone! Keep sending feedback.
I did in this way too, but confess I didn’t remember how to derivative a radical
And by the way, integration question is coming soon.
@@Maths-Paz Ahi!!! 😅
Nice simplication! It turns out that “a” is just the square root of the determinant. It also suggests that the max value of b is 4 which is the max height of the curve, and the min value of b is 0 which is the nonzero x intercept of the curve.
Both your values of c are relevant. By subtracting them you get the value of a = 16/rt3, and by using the formula y=4rtx-x with either of them you get b = 8/3. Then Amax = 8.16/3rt3= your answer. There is no need to use the horrible rectangle formula again.
Oh you're absolutely right Ken! I was aware the whole time, that once you get the c values you can simply calculate the sides and multiply. It's only that I feel compelled to show that things are really the way I say they are. So I thought maybe, if I went on to calculate the side lengths some couldn't follow. Thanks for the comment by the way.
This was a nice question to solve. Funnily enough Newton's iterative method doesn't seem to work for finding the other value of x, which left me scratching my head a bit. I found that you can define the area using only y as A = 8*y*sqrt(4-y), which leads to the same solution. Thank you
It's great to find out other people out there enjoy this stuff like I do Ralph. Thanks for the comment! I don't know much about Newton's method but I know it's basically a series of approximations. I am not sure if I am qualified to say why it didn't work but I guess it's got something to do with the second value not exactly being a minimum value in a broad sense. I believe (though I didn't think very hard on that) it's possible to create blue rectangles that have areas thst go all the way to minus infinity by moving its upper corners to below the x-axis. I might be wrong. Just my guess 😊
There is a simple solution .You can write the parabola in the form y = c - (x-xo)^2 and the circle as (x-xo)^2+ (y-r)^2 = r^2. Now we set x-xo =z . In order to determine the radius r we consider the equation z^2 + (y-r)^2 - r^2 = 0 ,where y = c - z^2 . This is a quadratic equation for z ^2. The discriminant of this quadratic equation must be zero , since there should be only one solution for z^2 . This leads immediately to the radius r.
Hi Rene. Thanks for this very cool solution! Though I would call it advanced rather than simple 🙂 I think I'll study it in more detail but there is a particular term in your solution that I am not sure if I'm grasping or not. It's Xo . Can you tell me the name of that Xo, the word that describes it like discriminant or whatever.
xo is the abscissa of the axis of the parabola (= abscissa of the center of the circle) The discriminant of a second order equation a x^2 + b*x + c= 0 is the expression D =b^2- 4 a c . If D=0 then there is only one solution . This is what I used.
A very cool word 😁 googled it and found out about ordinate too. Thanks Rene. That helped. I learned this stuff back in 2000 or something and I can still solve the problems somehow but I don't remember some topics even existed. and reading comments is really helping.
You could move the objects to the origin than y=-1/2×^2÷32-R R ist the radius, the intersection point xo fullfill the equation y=mr × x0 and the for the slope mr we have mr*mt=-1 mt is the slope of then tangent but mt is -xo therfore mr=1/xo and then yo=mr*xo=1 so the intersection point is P(xo|1) but this point must fullfill two equstions it is on the parabel so 1=-1/2(xo^2)+32-R and on the circle R^2=xo^2+1 and last eleminate xo from this both equations you get R^2+2R-63=0 solution is R=7
Please 🙏 try to make as many videos in Mathematics as you can. GOD bless you. 🙌 You are an excellent teacher/professor. 👏 ❤.
Thanks for this beautiful comment bro. May God bless you too. I am not a maths teacher but I'm doing my best to be useful to people in my channel.
49π with intersections being at y = 8.
It was difficult for me until I realized the axis of symmetry of the parabola could be used to find the x-coordinate of the center of the circle.
Hi tyger. Thanks for the comment. Good news you found this one challenging because I tried to make this one hard enough to be a good piece of math practice. And good to know it works as I intended. Just make sure you stay around. I have some more analytical geometry questions coming soon!
Whilst I am all in favour of using tangent secant theorem, (or intersecting chord properties as we called them,) I used Pythagoras twice to solve this, and I do not think it took any longer to do. Good problem!
I love pythagoras theorem too. Straightforward, all-purpose formula. I thought I would use something different in my next problem so I came with this one. Thanks for the comment by the way!
Note: If we also wanted the circle below the x axis and touching the x axis at (10,0) we would use radius R = 9 (with center (10,-9)) implies A = 81 * pi or we could use two circles one above and one below the x - axis implies A(above) = 49 * pi(center (10,7)) and A(below) = 81 * pi(center (10,-9)). Here two inscribed circles are tangent at (10,0).
You're right Rocco. There is another circle touching the upper circle at y=0 and its radius is 9. It actually satisfies the conditions of the upper circle, it's only that it's under the x-axis.
a not just, u usually dont nee the drivate to find the midpoint of a quadratic midpoint is -b/2a .. wich lie in the middle of [x]=[-b/2a±(√Δ/2a)] wich is the quadratic formula simply, x is in the middle of the two itersection points wityh the x-axis just... u can also say |y'| when k=0 is equal to [x] = -b/2a for quadratics on standard form...
You're right Patrick. But I used the derivative again which saved me some time. It's funny that I didn't realize I could find the vertex without derivative until I read the comments. You see it's been 20 somethinh years since I stopped doing maths on a regular basis. It's taking some time for my brain to power up fully 😅
@@Maths-Paz well, now u know..
🙏
@@Maths-Paz 😊 .. so what r the coordinates for the intersection points between the circle and the parabola? ...
Oh I don't remember the exact numbers right now but I know one of them had a square root of something in it...
Yes, very nice!
Thanks!
Once it's watched, it's so easy !
Nice to know my video helped 😊
Please do videos on vector calculus. Amazing video friend 👍🏾
I have only just begun to warm up my friend! So make sure you stay around for much more insane and complicated problems. And thanks a lot for the support!
This was a good one!
Thanks! Glad you enjoyed it!!
I found three solotion of R -9;7;16 but i took 7
16 looks like the radius of the circle that rests on the x-axis and its top touches the parabola's vertex. The vertex has a Y value of 32. No clue how you found it though. -9 must be the radius of the circle that touches the parabola at 2 points but it's under the X-axis. It's expectable that you find it through quadratic equation solution. Taking 7 was the right thing to do...
@@Maths-Paz Algebric expresion gives you all slotion .
I guess it was a cubic equation that gave you 3 solutions. If you don't mind the time, can you write that cubic equation and show how you solved it?
@@Maths-Paz I got two variables R and Y , and you can wach my slotion in my chanel below
czcams.com/video/vrcGScYgsvg/video.htmlsi=mAwvAkCqZMiAm_Wt
y = -0.5x²+10x-18 ...(1) dy/dx = -x+10 = 0 at turning point, so x=10 is locus followed by possible centre of circle circle: (x-10)²+(y-r)² = r² ...(2) d/dx of above gives 2(x-10)+2(y-r)dy/dx = 0 dy/dx = (-x+10)/(y-r) At point of intersection of circle and parabola, the tangents are equal (they touch rather than cross) -x+10 = (-x+10)/(y-r) so y-r = 1 or y = r+1 sub into both ...(1) and ...(2) r+1 = -0.5x²+10x-18 or -0.5x²+10x-19 = r so -x²+20x-38 = 2r ...(3) and (x-10)²+(r+1-r)² = r² x²-20x+101 = r² ...(4) add ...(3) and ...(4) 63=r²+2r=r(r+2) so r=7 area = 49π
Hi Francis, thanks! It's great to see this detailed solution because it tells me you found my question fun and engaging and I am learning new things every time I read a comment like this. Just make sure you check back in the coming weeks because I have plans for similar analytical geometry problems. Only, they will be even crazier!
Ouch!
It was hard?
I actually did it like this. I find out like this: 0=8x-x²-1 and used the quadratic formula Then after that i found the midpoint of two x's(x=4+√15 and x=4-√15) which was (4,0) Then found the gradient which was 2 and -2 to find the line of equation. Used those line of equations to find the the points (7,6) and (1,6). And because 7-1=6 so if I go down (4+3=7 and 4-3=1) Making the area 6²=36
That looks like a great alternative method sir! Thank yiu for taking the time to post this comment!
Well presented, excellent graphics.
Thanks sir!