Critical points

Date: 19 Dec 1994
From: Bruce Wise
Subject: calculus question

This is a problem that I have in my calculus book:

Find all critical points (if any) in the following problem.

        k(t)=1/the square root of (t-squared +1).

Date: 19 Dec 1994 
From: Dr. Sydney
Subject: Re: calculus question

Hello there!

Thanks for writing to Dr. Math.  The critical points of a function are the
points where the derivative is 0 (at these points, functions either reach a
local maximum, local minimum, or "saddle point.").  You can see why 
this is true...the derivative is the slope of the line tangent to the function,
right?  So, when the derivative is 0 the function has "flattened out."  Take
for instance the graph y = x^2.  We know what that looks like--kind of a U
shape, right?  Well, the place where the tangent line is horizontal is where
the graph has a local minimum--at the point (0,0).  That is precisely the
point where the derivative is 0.  Neat, huh?

Anyway, to find the critical points of your function, we just need to figure
out where the derivative is 0.

I find it easier to differentiate problems like this when I write them with
exponents, so I'd write your problem as follows:

                           k(t) = (t^2 + 1)^(-1/2)

So, to differentiate, we just use power rule and chain rule to get:

                           k'(t) = -t(t^2 + 1)^(-3/2)

(If your wondering how exactly I got that, you can write back and I'll
explain in more detail)

Now, all we want to know is for what values of t is k'(t) = 0?

So, when is     -t
                ----       = 0?
             (t^2 + 1)^3/2

The denominator is always going to be positive, no matter what t you choose,
so we need not worry about it being 0 (phew!).

So, the derivative will be 0 when the numerator is 0.  And, the numerator is
0 when t = 0, right?

So, the derivative is 0 when t = 0.

For a complete answer, you usually want to say what k(t) is at the critical
point.  k(0) = 1 in this problem, so we would say that k has a critical
point at the point (0,1).

Hope this helps!  Write back if you have any more questions.

--Sydney