Projections of the curve onto the coordinate axes (KristaKingMath)
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- Опубліковано 7 лют 2025
- ► My Vectors course: www.kristaking...
Learn how to sketch the projections of the curve. Projections are like shadows formed by the curve on the three coordinate axes. Given components of the vector equation, you can write parametric equations of the curve. Use the parametric equations to find equations of the projections in terms of only the variables involved in the coordinate axis. Sketch the projections and then use them to draw the real curve.
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Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)
Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!”
So I started tutoring to keep other people out of the same aggravating, time-sucking cycle. Since then, I’ve recorded tons of videos and written out cheat-sheet style notes and formula sheets to help every math student-from basic middle school classes to advanced college calculus-figure out what’s going on, understand the important concepts, and pass their classes, once and for all. Interested in getting help? Learn more here: www.kristakingm...
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This is a very difficult concept to visualize, but you did an outstanding job of giving a nice, clean explanation of it. Much better than the explanation my calculus professor gave. Good Job :)
Aww thanks! :)
Wow that was great, this video is exactly what I was looking for & this channel should definitely be more popular. Thanks a lot for yur help and I hope you keep it up
Thanks!
Beautiful clear explanation of projection of curve on to coordinate planes. Exactly what I was looking for. Thanks
You're welcome, so glad it was helpful!
I believe that the formula cos(sin^{-1}(y))=sqrt{1-y^2} is only valid if we restrict the parameter t to lie in the interval [-pi/2, pi/2], which is the range of the arcsine function. The values of the arcsine are not uniquely determined if the domain of the sine function is not restricted. Moreover, if we allow, for example, t = pi, then cos(sin^{-1}(y)) would be negative.
We can avoid using the arcsine by simply noting that since x = cos(t) and z/2 = sin(t), it follows that x^2 + z^2/4 = sin^2(t) + cos^2(t) = 1, which gives us the equation of the ellipse x^2 + z^2/4 = 1.
Yo this is exactly the problem i was stuck on. thanks for the help
No problem, Freddy! So glad it helped! :)
I had this on my webassign thank you!
❤❤❤❤❤thanks, i love you
Hi, thanks Krista for the nice explanation. But what if our vector is something like this: f(x,y,z)=? Is there a formula or a more general way to find the projections of this path on the xy xz yz planes?
What if function is not defined at 0? What to do then?
Could you please provide the software name which you use to describe all the example exercises on your channel? This video is a great example of learning calc...and wished it were interactive...although that's OK as is.
Love your voice.
Thanks.
I explain here :) www.kristakingmath.com/my-videos
Could you please provide the software name you used to described the graph on 3D dimension? Thanks
+Tai H. Doan That's called "Grapher", it comes standard with every Mac.
I used Window, you know any software like that? Thaaaaaaanks
+Tai H. Doan I don't, only because I haven't used Windows in so long. But you could search online for something like "mac grapher for windows" and see what other programs come up.
Thank you. Keep doing your great job!
Is she awesome or what?
Speaking of Macs (re: graphing software), Krista is the "Mac/iPhone" of math teaching.
Yep, my exam paper will be a mess