Question

How does -cost-sint =1?

Answer 1

?

Answer 2

i'm going over a calculus problem which had to find the unit tangent vector of <cost,sint,t> deriving it we get <-sint, cost,1>. taking the modulus of that derivative we get \[\sqrt{1^{2}+1^{2}}=\sqrt{2} \]
therefore putting this all into the tangent vector equation: \[r'(t)/|r'(t)|\]= [-sin(t) + cos(t) + 1]/√2
but having to solve its derivative is what confuses me...
T'(t)=[-cos(t)-sin(t)]/√2
How does |T'(t)|= 1/√2?

Answer 3

are you sure the rest is right?

Answer 4

\[\vec r(t)=\langle\cos t,\sin t,t\rangle\]\[\vec r'(t)=\langle-\sin t,\cos t,1\rangle\]\[\|\vec r'(t)\|=\sqrt2\]\[\vec T(t)={\vec r'(t)\over\|\vec r'(t)\|}=\frac1{\sqrt2}\langle-\sin t,\cos t,1\rangle\]\[\vec T'(t)=\frac1{\sqrt2}\langle-\cos t,-\sin t,0\rangle\]\[\|\vec T'(t)\|=\frac1{\sqrt2}\sqrt{(-\cos t)^2+(-\sin t)^2}=\frac1{\sqrt2}\]make sense?

Answer 5

how is \[\sqrt{\left( -\cos t \right)^{2}+\left( -\sin t \right)^{2}}=1?\]
i don't understand that part.

Answer 6

(-cost)^2+(-sint)^2 = cos^2(t) + sint^2(t) = 1 = sqrt(1)

Answer 7

is the cos or sin 0 while the other is 1? that is my problem i don't know which is which

Answer 8

google pythagorean theorem

Answer 9

@mellylol do you not agree that\[\sin^2x+\cos^2x=1\]?

Answer 10

but yes that is true, cos(pi/2) = 0 sin(pi/2) = 1

Answer 11

but of course this is not always true.

Answer 12

but if you sum their squares they always equal 1:)

Answer 13

yes...how does this help me in this problem @TuringTest ?

Answer 14

do you also agree that\[(-a)^2=a^2\]?

Answer 15

it helps because (-cost)^2+(-sint)^2 = cos^2(t) + sint^2(t) = 1

Answer 16

you might be thinking about -(cos(t))^2 and not (-cos(t))^2. big difference

Answer 17

indeed

Answer 18

\[(-a)^2=a^2\neq-(a)^2\]

Answer 19

that is not my problem...i know that. it's the sine and cosine that confuses me.

Answer 20

ok let's take this step by step....
you agree that\[\sin^2x+\cos^2x=1\]no matter what x is, and you agree that since\[(-a)^2=a^2\]we have that\[(-\sin x)^2=\sin^2x\]and\[(-\cos x)^2=\cos^2x\]so that gives\[(-\sin x)^2+(-\cos x)^2=\sin^2x+\cos^2x=1\]now what part exactly confuses you here?

Answer 21

ok...

Answer 22

do you know the theorem I posted?

Answer 23

so you get it then?

Answer 24

thank you both :)