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A Calculus Primer Part II- Applications of Derivatives

#1 macosxnerd101  Icon User is online

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Posted 25 April 2011 - 04:36 PM

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This tutorial will cover applications of derivatives, including the meaning of a derivative in context, first and second derivative tests, and L'Hospital's rule.

So What Exactly Does a Derivative Tell Me?
A derivative is simply a rate of change. If f(x) is a position function, then f'(x) tells us the velocity, or the rate at which position changes. The second derivative of f(x) (f"(x)) tells us acceleration. f"(x) is also the first derivative of velocity, and describes the rate of change of velocity. Or another way to put it is the rate of change of the rate of change.

Let's examine velocity. We know that if a function is increasing on an interval, then the velocity has to be positive on that domain. Graphically, the velocity will be above the x-axis. Conversely, if a function is decreasing, then the velocity is negative and below the x-axis.

Great, Now How Does This Help Us?
Knowing the derivative of a function gives us a lot of information about that function. For example, the first derivative tells us where the extrema of a function are, or the points at which their maximum and minimum values lie. We call this the first derivative test. So how exactly does the first derivative test work? First, start off by setting f'(x) = 0, and solving for x. These values of x are called critical values.

So how do we know where the maximum and minimum values lie on f(x)? We know that if f(x) is decreasing, then f'(x) is negative; and if f(x) is increasing, then f'(x) is positive. So where f'(x) transitions from negative to positive, we have a minimum; and when f'(x) transitions from positive to negative, we have a maximum.

The second derivative, which describes the acceleration of f(x), also provides invaluable information about the function. From the second derivative, we can determine the maximum and minimum values for the velocity, which are called points of inflection. We refer to the changes that occur at these points as changes in concavity. That is, when f"(x) < 0, the function is concave down; and when f"(x) > 0, the function is concave up.

In the image below, the hump to the left of the y-axis is concave-down, and the hump to the right axis is concave up. We can see the function initially decreasing until it reaches the maximum (think first derivative test), and then continuing to decelerate at a slower rate until it reaches its minimum, at which point it increases again.
Attached Image

Based on this information found in the second derivative, we can use the second derivative test to determine where local extrema lie on a function. Our precondition for the second derivative test is that f'(x) = 0 for our given point x. Based on this, we have the following conditions:
-f"(x) > 0: Local Minimum
-f"(x) < 0: Local Maximum
-f"(x) = 0: Inconclusive

Let's break these conditions down a little more. First, we know that f'(x) = 0. So if our acceleration is negative at x, that means there is a maximum because the function is going to change from increasing velocity to decreasing velocity at x. Similarly when f"(x) > 0, there is a minimum because the function changed from decreasing to increasing at x. Our last condition, f"(x) = 0, is inconclusive using the second derivative test because it is a possible point of inflection. You will have to test x using the first derivative test instead.

L'Hospital's Rule
L'Hospital's rule helps us evaluate limits in an indeterminant form by taking the derivative of the expression. Some indeterminant forms include 0 * infinity, +-infinity, 0^0, 1^infinity, etc. Formally, the rule is: lim(x-->a) f(x)/g(x) = lim(x-->a) f'(x)/g'(x).

Let's take a look at a couple of examples:

1) lim(x-->0) sin(x)/x

Starting off, this limit looks like infinity, which is an indeterminant form. If you have memorized this as a "special limit", you may recognize it as 1. We can prove this using L'Hospital's rule, though. So let's take the derivatives of the top and bottom functions. When we do this, we get: lim(x-->0) cos(x)/1, which evaluates to 1.


2) lim(x-->1) 2ln(x)/(x-1)

Again, we see an indeterminant form here of 0/0. So applying L'Hospital's rule to this limit, we can evaluate lim(x-->1) (2/x)/1 = lim(x-->1)2/x. This is much easier to evaluate, and again we get 2.


Conclusion
This concludes my second tutorial on derivatives. When working with particle motion, remember to always be thinking about what velocity is needed to change a particle's path, and what acceleration is needed to change the velocity. Accounting for these variables programatically and otherwise can make your life a lot easier when describing a particle.

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Replies To: A Calculus Primer Part II- Applications of Derivatives

#2 elgose  Icon User is offline

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Posted 25 April 2011 - 06:42 PM

View Postmacosxnerd101, on 25 April 2011 - 05:36 PM, said:

1) lim(x-->0) sin(x)/x

Starting off, this limit looks like infinity, which is an indeterminant form. If you have memorized this as a "special limit", you may recognize it as 1. We can prove this using L'Hospital's rule, though. So let's take the derivatives of the top and bottom functions. When we do this, we get: lim(x-->0) cos(x)/1, which evaluates to 1.

This could be a prime example for showing squeeze theorem as well, since this is assuming we "know" that d(sin(x))/dx = cos(x). If we find the derivative with f'(x) = lim(h->0) (f(x+h)-f(x))/h, it offers an interesting hole:

d(sin(x))/dx = lim(h->0) (sin(x+h) - sin(x))/h
(trig identity) = lim(h->0) [(sin(x)cos(h) + cos(x)sin(h) - sin(x))/h]
(combining like terms) = lim(h->0) [(sin(x)(cos(h)-1) + cos(x)sin(h))/h]
(separating terms in the numerator) = lim(h->0) [sin(x)(cos(h)-1)/h + cos(x)sin(h)/h]
(applying limit to inner terms) = sin(x) * lim(h->0)[(cos(h)-1)/h] + cos(x) * lim(h->0)[sin(h)/h]

Now, the next leap assumes we already know that lim(x->0) sin(x)/x = 1 (and, almost as popularly, lim(x->0)(cos(x)-1)/x = 0), but by using that assumption it defeats the purpose of this exercise. Nonetheless, we see that our derivative is:
(applying known rules) = sin(x) * 0 + cos (x) * 1
(simplifying) = cos (x)

Of course, plugging this back in, we get
lim (x->0) sin(x)/x = (using L'Hospital's Rule) lim (x->0) cos(x)/1 = cos(0) = 1

This kind of confirms our work, but using the ways of finding derivatives that are usually taught first (the difference quotient) we find an obvious loop - we know d(sin(x)/dx = cos(x) because we know that lim(x->0)sin(x)/x = 1, and vice versa. "A is true because of B, and B is true because of A" is less than satisfactory, but an awesome segway to show derivatives via squeeze theorem or via graphical means. Maybe in Part III?

Overall, good primer.
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#3 macosxnerd101  Icon User is online

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Posted 25 April 2011 - 07:08 PM

I've never heard of the squeeze theorem before today. It looks simple enough, though. I could add it in as a miscellaneous topic when I get to integrals, before I hit area between two curves. That's a good suggestion! :)
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#4 cfoley  Icon User is offline

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Posted 26 April 2011 - 12:43 AM

f'''(x) is also known as jerk or jolt.

This post has been edited by cfoley: 26 April 2011 - 04:44 AM

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#5 Dormilich  Icon User is offline

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Posted 26 April 2011 - 02:48 AM

View Postmacosxnerd101, on 26 April 2011 - 12:36 AM, said:

Based on this information found in the second derivative, we can use the second derivative test to determine where local extrema lie on a function. Our precondition for the second derivative test is that f'(x) = 0 for our given point x. Based on this, we have the following conditions:
-f"(x) > 0: Local Minimum
-f"(x) < 0: Local Maximum
-f"(x) = 0: Inconclusive

you can conlude that if you compute further derivatives. If the (2n)th derivative differs from zero, you got an extremum, otherwise ((2n+1)th derivative) you have an inflection point (higer order root).

The question is whether it is worth doing all those derivatives. Sometimes sheer logic can solve that as well (between a minimum and a maximum can only be an even number of extrema, so an odd number of candidates implies at least one inflection point)

This post has been edited by Dormilich: 26 April 2011 - 02:51 AM

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#6 macosxnerd101  Icon User is online

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Posted 26 April 2011 - 04:33 AM

@cfoley: I thought that was the third derivative. :)
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#7 cfoley  Icon User is offline

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Posted 26 April 2011 - 04:44 AM

Yes, that's what I meant. I thought it would be worth including since it gives a physical feel for the maths.
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#8 macosxnerd101  Icon User is online

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Posted 26 April 2011 - 10:42 AM

@elgose: Actually, we could just call the Squeeze Theorem an extension of the Intermediate Value Theorem. :)
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#9 ahamedirshad123  Icon User is offline

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Posted 27 April 2011 - 01:50 AM

Quote

2)lim(x-->1) 2ln(x)/(x-1)

Again, we see an indeterminant form here of 0/0. So applying L'Hospital's rule to this limit, we can evaluate lim(x-->1) (2/x)/1 = lim(x-->1)2/x. This is much easier to evaluate, and again we get 1.



For this example answer should be 2 as the limit x tends to 1
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#10 macosxnerd101  Icon User is online

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Posted 27 April 2011 - 04:14 AM

Yup. My bad on the division. Fixed. :)
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