Why doesn't f(t) = t/√(t² -1) have any critical point?

[selpak]

I have exam tomorrow, and while studyin i Faced many problems, two main ones are the following, please help.

why isn`t there critical points for t/sqrt(t^2 -1) ?

how can the antiderivative of (csc ρx)^2 be found? -p is pi - {i must not use Integration}


plz help

Added after 2 hours 3 minutes:

I want to say that I know i must convert the First equation so it would be (1/2)sin2x

but still i dunno how to find the Antiderivative of this, is there any specific method?

 

[kalyanram]

Re: two short questions

Q1. Why doesn't f(t) = t/√(t² -1) doen't have any critical point.

Let me define critical point first
Critical point is a point in the domain of a function where the derivative is equal
to zero or a point where the function ceases to be differentiable.

So firstly lets define the domain of the function f(t), thats R - [-1,1] right.
Now the derivative f`(t) = -1/(t²-1)√(t²-1), this function can never be zero
for any value in the domain, and it exists for all values in the domain hence
it has no critical points.

As for the second question is concerned well it a little tricky without using
integration. Let me give it a thought.

~Kalyan.

 

[kalyanram]

Re: two short questions

Q2. Well I don't know how accurately I can claim this method to be not using integration exactly but here it is any how..
lets write csc²θ = 1/sin²θ = (sin²θ + cos²θ)/sin²θ ...
Now lets rearrange the terms in the numerator
= - ( sinθ.cos'θ - cosθ.sin'θ)/sin²θ now this seems to be of the form
(u/v)' = (v.u'-u.v')/v² so u/v is its anti-derivative...
so csc²θ is the derivative of -(cosθ / sinθ) or -cotθ.
Hence -cotθ is its anti-derivative.
(ρ is upto you to take care of )

~Kalyan.