Critical pointsDate: 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 |
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