Nice. Two application points mentioned in the vid. 1) To maximize something, take the 1st derivative. 2) The critical point of the derivative (equation) is zero to find the maximum value of the interested variable, the angle theta in this case.
Nice idea for a problem However that calculation looks unnecessary to me. If T=55/x and B=45/x the tangents of top and bottom angles then by angle difference formula for tangent tan(theta)=Tan(top_angle-bot_angle)=(T-B)/(1+TB)=10x/(x^2+9*11*25) that is 55 or 45 but i just factored it out this thing is much nicer to differentiate and knowing the functions is uniformly increasing the min/max will not be changed. tan'(theta)=(10*(x^2+9*11*25)-10x*2x)/(x^2+9*11*25)^2=(10*45*55-10x^2)/(x^2+9*11*25)^2 so we got the extrema points of x=+-15sqrt(11) I don't think the difference is major but i just didn't feel like working with the arctangent lol
Just PowerPoint. The secret to create movement is the morph transition. Create (duplicate) two slides, and create some changes in the second slide. When you run those two slides with morph transitions, any changes from the first slide to the second slide will appear a motion, so with that as the idea, you can create the animation you want fairly easily. Here is how I usually do the animation. How to Animate Math Equations #math th-cam.com/video/S_Ey0AKdzfM/w-d-xo.html
Interesting how close yet different the intuitive answer of 50ft is from the analytic result.
That's a great observation!
Nice. Two application points mentioned in the vid. 1) To maximize something, take the 1st derivative. 2) The critical point of the derivative (equation) is zero to find the maximum value of the interested variable, the angle theta in this case.
Great video!
Thanks!
Nice problem! One can also use theta = tan^-1(x/45)-tan^-1(x/55)
Nice idea for a problem
However that calculation looks unnecessary to me. If T=55/x and B=45/x the tangents of top and bottom angles then by angle difference formula for tangent
tan(theta)=Tan(top_angle-bot_angle)=(T-B)/(1+TB)=10x/(x^2+9*11*25) that is 55 or 45 but i just factored it out
this thing is much nicer to differentiate and knowing the functions is uniformly increasing the min/max will not be changed.
tan'(theta)=(10*(x^2+9*11*25)-10x*2x)/(x^2+9*11*25)^2=(10*45*55-10x^2)/(x^2+9*11*25)^2
so we got the extrema points of x=+-15sqrt(11)
I don't think the difference is major but i just didn't feel like working with the arctangent lol
Just discovered your channel. Great stuff! Keep it up. Subbbed.
Awesome, thank you!
What tool do you use for your animations? Thanks
Just PowerPoint. The secret to create movement is the morph transition. Create (duplicate) two slides, and create some changes in the second slide. When you run those two slides with morph transitions, any changes from the first slide to the second slide will appear a motion, so with that as the idea, you can create the animation you want fairly easily.
Here is how I usually do the animation. How to Animate Math Equations #math
th-cam.com/video/S_Ey0AKdzfM/w-d-xo.html
@learnmathbydoing Thank you for the detailed explanation.
And I calculate this critical angle in my head when watching live performance.
Ama phoking genius!!
Well that small guy could solve it by walking