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.
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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)=...
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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.
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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...
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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)...