Awesome Analytic Geometry Problem with Fun Solution: Finding the Areas Under the Curve with Integral
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- čas přidán 1. 07. 2024
- 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 maths and one that many of us, including tend to forget over time. In this problem, there is a purple parabola-shaped function curve and a blue right triangle has been placed into it in such a way as to divide the part of the parabola above the X-axis into three regions. And our job is to find the green region that is on right side of the diagram. I found this puzzle to be fun and the solution was smooth and satisfying. You could want to try this yourself. If you do, what you read upto this point contains ample clues to guide you, but if you still can't solve it don't feel bad because I showed the step-by-step solution in the video.
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.
Excellent explanation. High level math genuinely enriches my mental power. Million thanks squared^*^
Glad you enjoyed. Thanks for the fantastic comment!
Thanks for this nice problem.
Well, thanks for watching and the nice 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
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.
❤
Thanks for the 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!
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.
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