Recent content by thegoodfight
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Find T(t) at (1,0,0) when r(t)=<cos(t),sin(t),ln(cos(t))>
Ok! That makes a LOT more sense now. So t = 0 so that way the vector shows the point (1,0,0). Thanks a lot. Something small like finding this out makes it make so much more sense now.- thegoodfight
- Post #3
- Forum: Calculus and Beyond Homework Help
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Find T(t) at (1,0,0) when r(t)=<cos(t),sin(t),ln(cos(t))>
Homework Statement r(t)=<cos(t),sin(t),ln(cos(t))> Find T(t) (and N(t)) at (1,0,0) Homework Equations The Attempt at a Solution T(t)=(dr/dt)/ldr/dtl dr/dt = <-sin(t),cos(t),-tan(t)> ldr/dtl = (sqrt)((-sin(t))^2+(cos(t))^2+(-tan(t))^2) = sec(t) therefore T(t)=...- thegoodfight
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- Replies: 2
- Forum: Calculus and Beyond Homework Help
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Lim (x,y)->(0,0) (x^2*sin^2(y))/(x^2+2y^2)
Thanks a lot. I think I get it now, and the video did help some. I am going to work on using polar coordinates, even if we aren't allowed to use them. Seems to be handy even I use it just to check. Thanks again.- thegoodfight
- Post #7
- Forum: Calculus and Beyond Homework Help
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Lim (x,y)->(0,0) (x^2*sin^2(y))/(x^2+2y^2)
OK, so if I prove that the limit exist using, say, 5 or 6 different paths and they all come out to be zero, then I would be able to say that the limit exist, knowing full well that we haven't gone over every possible pathway? I keep reading that polar coordinates are supposed to be helpful with...- thegoodfight
- Post #3
- Forum: Calculus and Beyond Homework Help
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Lim (x,y)->(0,0) (x^2*sin^2(y))/(x^2+2y^2)
Homework Statement lim (x,y)->(0,0) (x^2*sin^2(y))/(x^2+2y^2) Homework Equations The Attempt at a Solution I have attempted to solve this by letting y=x giving me lim (x^2*sin^2(x))/(x^2+2x^2) = lim x^2sin^2(x)/3x^2 = sin^2(x)/3 = 0 (x,x)->(0,0)...- thegoodfight
- Thread
- Replies: 6
- Forum: Calculus and Beyond Homework Help