another math question

[SparklyUnicorn]

find dy/dx and find the slope at the indicated value of the parameter

 

x=2+sec t

y=1+2 tan t

t+pi/6

 

I get as far as 2 sec t (all over) tan t

 

and I do get the correct slope (I'm guessing she'd be ok with my answer as is)

 

the answer says 2 csc t

 

how does one get from what I got to 2 csc t?

 

 

[Caroline]

2sec(t)/tan(t)=

 

(2/cos(t))/(sin(t)/cos(t))=

 

(2/cos(t))*(cos(t)/sin(t))=

 

(2/sin(t))=

 

2csc(t)

[SparklyUnicorn]

wow..thank you...that's pretty convoluted

 

LOL

 

 

[luuknam]

wow..thank you...that's pretty convoluted

 

 

Not really. You just want to go back to basics with these things. So, you know (or can look it up, but you'd probably want to memorize it) that the sec is 1/cos, and that the tan is sin/cos, and that csc is 1/sin. So, you just plug in those things where appropriate, and then simplify. You probably want to make a habit of trying to simplify stuff whenever you have some equation with one of those things over another one of those things, or one of those multiplied by another one of those... often you can simplify it. 

 

(not a fan of sec and csc etc either, which is one reason I'd try to simplify them away if possible, lol)

 

ETA: also, this probably looks a bit less convoluted on paper than typed in a forum post, since on paper you can just cross out certain things etc.

Edited November 25, 2017 by luuknam

[regentrude]

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Edited November 26, 2017 by regentrude