I have to admit, the way you explained all of Chp. 12 was beautiful thank you for helping me understand through the content of calculus 3, I'm assuming you are covering using the book Thomas Calculus early transcendentals 14th edition right?
Hello ! I just stumbled apon your courses and I adore the way you explain this subject ! I'm a French student n struggle with the way my teachers explains stuff and I found that by learning from teachers that speaks in English helps a lot having another pov on the said subject. Now my question is what age student is this taught? I'm a bit lost trying to find the chapters we do in France at my age with all the algebra / calculus 1234 it's really confusing for me ... Is there a sweet soul that could help me out / explain it to me ? Thanks a lot !
Since she found the center point, and you are given two points that reference the sides of the sphere, all you need to do is find the distance from the center point to one of the given points. Distance formula (note you can use the other point): sqrt((0-1/2)^2+(2-2)^2+(3-2)^2) = sqrt(5/4) = sqrt(5)/2
That wasn't me talking, which is why the sound was so quiet. At 25:15, a student in my class was answering. The main point is that the inequality shown is a sphere with the inside filled in.
I’m confused on why for example six, R2 has a horizontal line. The y axis isn’t up and down, its on the left and right. Wouldn’t you graph the line vertically?
i think The lecturer assumed the center point to be (-1/2, 2, 1), which is incorrect. They simply changed the x-coordinate of one endpoint, but the y and z coordinates remained the same. This does not represent the midpoint of the diameter.
I calculated the distance between the center and one of the endpoints of the diameter. For example: r = sqrt[(-1 - -1/2)^2 + (2 - 2)^2 + (1 - 2)^2] = sqrt[(-1/2)^2 + (-1)^2] = sqrt(5/4). Either endpoint will work.
Since she found the center point, and you are given two points that reference the sides of the sphere, all you need to do is find the distance from the center point to one of the given points. Distance formula (note you can use the other point): sqrt((0-1/2)^2+(2-2)^2+(3-2)^2) = sqrt(5/4) = sqrt(5)/2
