Ellipses Mega Review: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-_BuUMQAFWmI.html Hyperbolas Full Review: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-aBednvA7rSY.html Parabolas Full Review: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-7dH24WsuGsM.html Conic Sections Quiz: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-vqZO_CrEdPs.html Next Video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Mw3MeDLL8d4.html
Professor Organic Chemistry Tutor, thank you for a short and well explained video/lecture on the Eccentricity of a Basic Ellipse in Calculus Two. The practice problems/examples are easy to follow and understand from start to finish. This is an error free video/lecture on RU-vid TV with the Organic Chemistry Tutor.
Thanks so much for this video, I saw eccentricity in a past exam paper, and I didn’t know how to do it, even tho it’s year 12 extension 2, and I’m only year 11 extension 1 currently, I’ll watch more of your videos that relate to the paper, so I can try it out and hopefully go well for practice
The "definition" of "eccentricity" is: "the ratio of the distance from the center to a focus divided by the length of the semi-major axis" The "focus" is "the square root of the difference between the squares of the semi-major and semi-minor axis". The semi-major axis, is the longest radius of the ellipse, and the semi-minor axis is the shortest radius of an ellipse. This means, if you have an ellipse where the longest radius is 36, and the shortest radius is 16: The focus would be √(36² - 16²) = √(1296 - 256) = √1040 = 32.2490309932. Therefore, the eccentricity, is the ratio of the focus (32.2490309932) and the semi-major axis (36), which is 32.2490309932 / 36, which is 0.89580641647. If the semi-major axis however is _really_ big, e.g. 1 billion, and the semi-minor axis is very small, like 2, then: The focus is √(1,000,000,000² - 2²) = √(1,000,000,000,000,000,000 - 4) = √(999,999,999,999,999,996) And if you divide √(999,999,999,999,999,996) / 1,000,000,000, you'd get a number that is _very_ close to 1. Something like 0.999999999999999998. Meanwhile, if the semi-minor and semi-major axis are equal, for example, 9, then: The focus becomes √(9² - 9²) = √(81 - 81) = √0 = 0 Therefore, the eccentricity becomes 0 / 9 = 0 Therefore, it is a circle. The general formula for eccentricity is e = [√(a² - b²)]/a
thanks so much you changed my life with your videos I would've never passed precalc in my junior year of college without your gracious gifts of teaching along the way thanks so much
OMG thank you for the clarification. I kept on putting a under x even if it was smaller so I would get the eccentricity wrong if the ellipse was longer on the y axis.
Please someone answer me Our quiz is tomorrow And I still cant figure it out How to do squareroot of 5 divided by 3 on a scientific calculator. Everytime i do it, it always shows " 1 " and not 0.745 😭
a and b are not x and y. And on graph you can only put the value of that coordinate . Not the max of x and max of y . Bcz this will be outside of ellipse. So u can't put that value to equation of ellipse