Unit tangent vector problem:
Vector r(t) =
@ganeshie8
definition : \(\large \frac{r'(t)}{|r'(t)|}\)
to take the derivative of a vector function, simply take derivative of each component
I think I am having trouble factoring the magnitude on bottom.
I have T(t) = <1, 0, -2/t^2> / ( 1 + 4/t^4 )
looks good
sorry T(t) = <1, 0, -2/t^2> / ( 1 + 4/t^4 )^1/2
oh yah
my book shows <t^2, 0, -2> (t^4 + 4)^1/2 ?
thats right
\(\large \frac{<1, 0, -2/t^2>}{\sqrt{1 + 4/t^4}} \) \(\large \frac{<1, 0, -2/t^2>}{\sqrt{(t^4 + 4)/t^4}} \)
\(\large \frac{<1, 0, -2/t^2>}{\sqrt{(t^4 + 4)}/t^2} \)
multiply top and bottom wid \(t^2\)
\(\large \frac{t^2<1, 0, -2/t^2>}{\sqrt{(t^4 + 4)}} \)
so did you first factor out t^4 ? O couldnt quite tell.
we have below in the denominator :- \(\large 1 + \frac{4}{t^2}\) right ?
to add fractions, u need to have common-denominator
\(\large 1 + \frac{4}{t^4}\) \(\large 1\times\frac{t^4}{t^4} + \frac{4}{t^4}\) \(\large \frac{t^4}{t^4} + \frac{4}{t^4}\)
got it.
now that the denominators are same, we can add the numerators directly : \(\large 1 + \frac{4}{t^4}\) \(\large 1\times\frac{t^4}{t^4} + \frac{4}{t^4}\) \(\large \frac{t^4}{t^4} + \frac{4}{t^4}\) \(\large \frac{t^4+4}{t^4} \)
I see whre your going with it (; Can I show you one more ting fro a different eq. you mght know how to simplify in a similiar way? 4sin^2(2x) + 16cos^2(2x)
thats easy, write 16 as 4 + 12
4sin^2(2x) + 16cos^2(2x) 4sin^2(2x) + 4cos^2(2x) + 12cos^2(2x) 4(sin^2(2x) + cos^2(2x) ) + 12cos^2(2x) 4(1) + 12cos^2(2x) 4 + 12cos^2(2x)
i dont think it simplifies anymore
Thats it and thank you. What time is it in India?
np :) 1:04 pm .... going for lunch.... cya must be very late over there. lol u should goto sleep :)
Its 11 36 pm (; No class tomorrow. Have a good day.
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