00:01
I think what i would do is just start with certain t values, and i remember x is secant, and then y is tangent.
00:10
I think you just have to be careful because, like, i'm just going to go at starting at zero, that secant is a reciprocal of cosine.
00:20
So on the unit circle at zero, the order pair is one, zero.
00:26
So secant is a reciprocal of one, so that's one, and tangent is sine over cosine, so 0 divided by 1.
00:34
And then maybe just go to pi over 6 next, where we're at root 3 over 2, 1 half.
00:43
So we're getting 2 over root 3, and then 1 half divided by root 3 over 2.
00:50
The 2s cancel.
00:52
And then jump to pi over 4, where we're going to get 2 over root 2, which i think if we rationalize, it's just root 2.
01:00
And then root 2 over 2 over root 2 over 2 is just 1 and then i don't know how many more that we need but pi over 3 would be exactly 2 and then tangent would be just root 3 and it may all go to pi over 2 as well which this is actually undefined and then because that pi over 2 is straight up at 0, 1 and actually tangent is undefined as well so i believe what we'll be looking at then and that's just quadrant one, is we would be starting at one, zero, and then we're getting bigger numbers for each of these.
01:49
And let me just jump to like two root three, because root three would be a little bit smaller than one, like 1 .7.
01:58
So we're up here.
01:59
I believe this graph is kind of like a hyperbola that is going in this direction, but then we're not quite done because then we're going into the second quadrant where they're both negative.
02:13
Secant and tangent are negative in quadrant two.
02:18
So that's going to take my answer and flip it over into quadrant three and go this way.
02:26
But then if i go into quadrant three, now the x -coordinate is negative and the y -coordinate is negative.
02:33
So tangent is actually positive, but secant is still negative so i believe it now jumps over in this direction like resets and goes in this direction and then resets to be in quadrant four um to finish out this cycle and then i guess we're basically done there uh but then we can find dy dx by oh wait they wanted you to find eliminate a parameter so let me do that real quick that t would equal the inverse secant of x and then i can replace that in here so y would equal tangent of the inverse secant of x that's one of the answers they wanted and i wouldn't do that for dy dx what i would do is the derivative of tangent which is secant squared and and then the derivative of secant is secant tangent, which i can definitely simplify because one of those secants will cancel.
03:44
I remember secant is 1 over cosine, and tangent is, let me write it that way, is sine over cosine, which is the same as multiplying by the reciprocal, and these would cancel, and i would rewrite dy dx as, well, cosecant.
04:12
That's what 1 over sine is.
04:13
That's one of your answers but now when they ask you for the second derivative you do the derivative of cosecant which is negative cosecant cotangent over the derivative of your x variable which is secant tangent again which again i should be able to simplify because cosecant is 1 over sine cotangent is is cosine over sine.
04:51
And instead of dividing, i'm gonna multiply by the reciprocal in secants one over cosine.
04:58
But remember, i have to do the reciprocal...