Paramentric eq. of tangent line

[Calcgeek123]

Homework Statement

Find the equation of the tangent line to f(t)=<cot(t),csc(t)> at the point (1/sq.rt of 3, 2/sq.rt of 3)

Homework Equations

n/a

The Attempt at a Solution

I started by finding the slope, y'/x', so I got csc(t)cot(t)/csc^2(t). I then used the equation of a line, y=mx+b. Doesn't a tangent line have the opposite slope? So I think i'd plug in 2/sq.rt of 3 for y, csc^2(t)/csc(t)cot(t) for the slope, 1/sq.rt of 3 for x, and then solve for b. Do I need to plug in my x and y points though into the slope? Then that gets tough because I don't know what csc^2(1/sq. rt of 3) is. Any suggestions?

 

[LCKurtz]

Homework Statement

Find the equation of the tangent line to f(t)=<cot(t),csc(t)> at the point (1/sq.rt of 3, 2/sq.rt of 3)

Homework Equations

n/a

The Attempt at a Solution

I started by finding the slope, y'/x', so I got csc(t)cot(t)/csc^2(t).

Any law against simplifying that?

I then used the equation of a line, y=mx+b.

Why use that form instead of the form y - y₀ = m(x - x₀)? After all, you are given a point on the curve through which the line must pass.

Doesn't a tangent line have the opposite slope?

Opposite? Isn't the slope of the tangent line defined to be the slope of the curve?

So I think i'd plug in 2/sq.rt of 3 for y, csc^2(t)/csc(t)cot(t) for the slope, 1/sq.rt of 3 for x, and then solve for b. Do I need to plug in my x and y points though into the slope? Then that gets tough because I don't know what csc^2(1/sq. rt of 3) is. Any suggestions?

Can't you figure out what t gives the point in question and just use it in your calculations?

 

[ramb]

I say you should try to figure out what value of t makes that point. Once you do that you can derive that vector equation, then with the derivative you can find the slope at any given point. With that slope you can find the equation of the line.

 

[Calcgeek123]

Thank you! I took the advice from the first post, and I've ended up with y-csc(t)=cos(t) (x-csc(t)). Now I'm just lost as to what to do with the point I've been given, (1/sq. rt of 3, 2/sq. rt of 3). I could plug in those numbers into the equation i arrived at for x and y, but then what would i solve for?

 

[ramb]

do what i said on my post after that, because if you have the derived vector equation, you need to plug in a t value