Calculus 1: Max-Min Problems (27 of 30) Find Point of Least Light
HTML-код
- Опубликовано: 30 сен 2024
- Visit ilectureonline.com for more math and science lectures!
In this video I will find the point of least light between 2 light bulbs of different intensities.
Next video in this series can be seen at:
• Calculus 1: Max-Min Pr...
Hello Michel ... I just got done watching your other video on the Two POLES with a Rope between them... where we had to find the distance "x" on the ground where a Stake would hold the Rope.. Recall that video?... well, it turns out that the distance , x, could be found with this equation: x = d(h1/(h1+h2)) where "d" is the distance between the two poles and h1 and h2 are the heights of the two poles.. Why do I even mention this????.. Great question... because the RESULT in the above video, in regards to the Two LIGHT sources, has the same type of Equation as the result for the Two Poles Video, that is.. x = d ( S1/(S1 + S2)) ... ofcourse those "S" sources require taking the Cube Root of each of them, but the Form of the two equations is the same!!.. how cool is that???.. x=d(h1/(h1+h2)) and x=d(S1/(S1+S2)) Thank you for the thorough lectures!!.. greatly appreciated... and the only way I knew of these Equations was after having watched your two videos and learned from them... I always enjoy learning how Equations are derived and proven.... thanks!
Thanks a lot, now I know that I have passed my exams (Special exams for Engineering from ULB) :D
One can see from the derivative that if the intensity of the two light bulbs are equal the least light would be the midpoint between them. Although that may seem obvious, it nice to see the derivative proves it.
There is a constraint i.e. D2=x-d
excellent once again, Michel!! I see that I replied to this video FIVE YEARS AGO!! lol, and it was a LAME reply... lol... but it's always nice to repeat watching your videos to refresh my understanding... I think I'll imagine the Earth at a point between the Moon and the sun and figure out where the Intensity of light from both the sun and the moon are Equal.. (Yeah, Right!) :) perhaps the full moon has to be within 100 miles of earth to equal the sun's Intensity... Thanks Michel !!
Hi Philip. We have plenty of new videos so you don't have to go back to the old ones. 🙂 Just kidding, glad you are enjoying them.
can you help me. what is the solution of this problem? {A lamp of 20000 lumens is suspended 6 meters above the corner of an area has a length of 6 and width of 4 meters . calculate the maximum and minimum illumination on this rectangular area.}
Use I = P / A = P / (4)(pi)(R^2) (intensity = power at the source / surface area over which it spreads. Then calculate the distance to the closest point (6 m) and the distance to the farthest point: SQRT( 6^2 + 6^2 + 4^2)
Hello. there is an error in the last step. to isolate the variable it must first subtract and divide.
The video is correct. Thanks for checking.
Thank you