Sunday, November 22, 2009

By Request: Radiometric Dating

I was asked to explain the theory behind radiometric dating, one of the many scientific applications of calculus.

Radiometric dating seeks to determine the age of a material based on the relative amounts of certain isotopes in the material. These isotopes are chosen for various reasons. For example, a common type of radiometric dating is carbon-14 dating. This dating is used to determine the age of fossils. While an animal is living, it is exchanging matter with the environment and over time will have a level of carbon-14 equal to that of the environment. When the animal dies, it ceases to take in any more carbon-14. The carbon-14 will slowly decay to carbon-12 through beta decay. By comparing the amount of carbon-14 in the dead animal to the environmental amount, we can determine when the animal died.

But how can we do that?

The decay of a father isotope into a daughter isotope happens due to forces between subatomic particles within atoms. For example, the weak force, one of the four fundamental forces, is responsible for beta decay, the mechanism behind carbon-14 decay. Because decay depends on quantum mechanics, the type and rate of decay depends solely on the father isotope in question.

Since the probability of decay depends on the father isotope, we can form an equation to predict decay to the daughter isotope over time. How do we do this? If all the atoms we have are the same father isotope, then each has an equal probability of decaying. Therefore, the rate of decrease in the number of atoms that are the father isotope should be proportional to the number of atoms of the father isotope. Let us call the constant of proportionality lambda, and write the equation and solve it.

Click here for equation and solution

So we see that the number of atoms of the father isotope decays exponentially.

Challenge: You might have heard of half-life before. Can you express the half-life in terms of lambda?

Saturday, November 21, 2009

By Request: Entrance Exam Problem

Sameer Hemmady in Albuquerque writes:

A dog is chasing a cat with uniform velocity “v” so that at any given time the dog is “aimed” directly at the cat. The cat upon seeing the dog runs rectilinearly and uniformly with velocity “u” with u<v. At the initial moment when the cat sees the dog, the vectors v and u are perpendicular to each other, and the dog and cat are separated by a distance “d”. How soon will the dog catch up to the cat? (Problem in our engineering entrance exam in India)

Entrance exam problems require quite a bit of ingenuity to solve. This is something to be expected. They are not testing your ability to simply differentiate or integrate, but to create an approach to the problem. This ability is important if you plan to use calculus in your career (I actually solved a problem similar to this in my job, involving a jet and a ballistic interceptor). Math problems do not present themselves as equations for you to solve, but you must form the equations yourself from meaningful quantities in the word problem. Creating an approach to a problem takes quite a bit of trial and error. Therefore, when solving this problem I will consider different approaches.

Solving entrance exam problems is a science in its own right, and learning all there is about it would take several semester-long classes. However, even if you have never been exposed to entrance exam problems, and especially even if this is your first semester of calculus, I encourage you to look at the solution to the problem, and ask questions if you have any.

Click here for solution

Thursday, November 19, 2009

Cut Down Integral Table Memorization in Half

Last time I showed you an easy way to differentiate an inverse function. Today I will show you how to integrate an inverse function. Besides helping you remember the antiderivatives of the “arc” functions, it serves as a good integration technique in general. To date I have not seen a similar technique in any calculus book, including graduate level.

Let’s start with a function f(x). Let’s say that its antiderivative is F(x).

Now let h(x) be the inverse function of f(x). That is, f(h(x))=h(f(x))=x. If we graph f and h, we see that h is a reflection of f across the line y=x.

To integrate h(x), we use integration by parts. Recall the formula for integration by parts, plug in appropriate expressions for u and v and calculate.

However, we know from last time that

Set y=h(x), calculate, and substitute back. We arrive at the integration rule for inverse functions. Keep in mind that x=f(y), and that dx=f’(y)dy.

Let’s try an example. In this example we find the integral of arc cosine.

You can verify this with an integral table or Wolfram Integrator.

Besides using this rule to quickly remember the antiderivatives of the “arc” functions, this technique can also be used to integrate functions that cannot be integrated otherwise. Consider this example where we integrate something a lot more complicated, the arc cosine of the square root of x.

Circle it Up

Recently I was helping someone with some geometry problems involving circles, and I said to myself I'll bet there is a calculus question I can make up from this.

So here is my question.
Suppose you have a 45, 45, 90 triangle, ABC, inscribed in a circles so that it's hypotenuse is the diameter of the circle. Now suppose that sides AC and BC increase in length at 3m/s and the circle expands to keep the triangle enclosed. Then at what rate is the area of the circle increasing when the radius of the circles is 5m?

So we know the radius that we want the radius of the circle to be:
r = 5m
We know the rate at which the sides of the triangle increase:
ds/dt = 3m/s
It would be nice to know the rate at which the diameter of the circle increases. Since we have a 45 45 90 triangle we know that if one of the sides is length x then the hypotenuse will be length x*sqrt(2). So let's write our equation like this:
The diameter is equal to the length of a side times the square root of two.
D = s*sqrt(2)
To make it a little bit easier later on I'm going to rewrite this equation again except now I'm going to substitute two times the radius in for the diameter so it looks like this:
2*r = s*sqrt(2)
now lets take the derivative with respect to time
2*dr/dt = sqrt(2)*ds/dt = 3*sqrt(2)
dr/dt =(3*sqrt(2))/2 = 1.5*sqrt(2)

We know or at least we can look up the equation for the area of a circle
A = pi*r^2
if we take the derivative of this equation with respect to time we get
dA/dt = 2*pi*r*dr/dt (just use the power rule)
Now we can substitue what we know
dA/dt = 2*pi*5*1.5*sqrt(2) = 21.21*pi m^2/s
So when the radius of the circle is 5, the area of the circle is increasing at 21.21*pi square meters per second.

Once again I've created an Excel spread sheet to allow you to experiment with all kinds of things including central and inscribed angles as a geometry bonus.

Tuesday, November 17, 2009

Cut Down Derivative Table Memorization in Half

This is a useful rule that you won’t find in most books for calculating the derivative of an inverse function. This includes any of the “arc” functions in trigonometry for example.

Let’s start with a function f(x). Let’s say that its derivative is f’(x).

Now let h(x) be the inverse function of f(x). That is, f(h(x))=h(f(x))=x. If we graph f and h, we see that h is a reflection of f across the line y=x.

Now let’s calculate h’(x).

Let’s assign the variable y=h(x). Then we can work out the derivation rule for inverse functions.

Let’s try this on a couple examples.

We know that:

Now let’s apply this rule on the inverse of f.

We also know that:

Now let’s apply this rule on the inverse of f.

We can simplify this further using the trigonometric identity

Just set

Next time I will show you the general rule for integrating an inverse function.