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?

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

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.
CircleandTriangle.xls

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.

y = (tan x)^(1/x), Using the Log Rule

You'll have to use the Log rule find the derivative of y=(tan x)^(1/x). I wrote a very long explanation on exactly how to do this, and if you're interested in reading it, I've included a link so you can down load the word document.

The image I've included in this post gives you the basic steps for finding the derivative, with out any explanation. So if you're up to speed on taking simple derivatives you should be able to follow the picture from one line to the next.

Anyway couple of interesting things about this function. One is since it involves the tan function, it is undefined at the half pi's. Two is that it's derivative has a natural log function in it so it is undefined for all x <= 0.

Overall it's a strange function I'd say.

If you really want to get into the nitty gritty down load the word doc.
LogRuleExplanation.doc

Simple Projectile Motion

Ever get one of those problems.. that sounds something like..
"Roy hits a baseball at an angle of 30 degrees with respect to the horizontal at a velocity of 30 mph.
a. determine how high the baseball will go
b. determine how far the baseball will go"
and so on and so forth.
Well I've explained just about everything I know about simple projectile motion in this animated excel spread sheet. Including a nice little explanation of the calculus behind the equations of motion.
SimpleProjectileMotion.xls

Water Trough Practical Application

A 10ft long trough whose sides are shaped like isosceles triangles with a height of 1ft and a base of 3ft, fills up with water at a rate of 12 cubic feet per minute. How fast is the water rising when the water is 6 inches deep?

I created an excel spread sheet that illustrates whats going on here very nicely. But I'll leave one hint right here. The key is to write an equation that relates the depth of the water to the volume of the water. Follow the link to download the xls file.

WaterTroughProblem.xls

x^2y^2 + xsiny = 4, Implicit Differentiation

x^2y^2 + xsiny = 4
This problem requires you to use implicit differentiation in conjunction with the product rule and the chain rule. The thing that might scare you about this problem is the xsiny part but don't worry

x^2 + xy - y^2, Implicit Differentiation

This problem requires implicit differentiation because the equation cannot be solved for y. The fun part about this problem is that you get to use the product rule on the xy term in the middle of the equation.

Click on the image below to see a lager version of my solution.

Piecewise Function, y = |x^2-9| Plot & Find Derivative

Problem:
Find the derivative of y =|x^2-9|. Plot the function and the derivative.

Solution:
Lets say you had the graph of just x^2-9 (without the absolute value, represented by the orange line in the graph below) that would look like the graph of x^2 shifted down 9 units.

But, there is an absolute value sign so the part of the graph that goes below zero becomes positive. That’s why you have that blue bubble in the middle of the graph. This bubble is just a reflection of the orange line about the x axis. So the blue hump in the middle of the graph is y = -x^2 + 9. In order to reflect a function about the x axis just multiply the right hand side of the equation by -1.

If you redefine the function in |x^2-9| in a piece wise fashion, you can find the derivative on each part.
y = { x^2 - 9 when x <= -3
9-x^2 when -3< x < -3 (the reflection)
x^2 -9 when x >= 3

y' = {2x when x <=-3
-2x when -3>x<3
2x when x >= 3

Now every where you have a negative slope the value of the derivative is negative and everywhere you have a positive slope the value of the derivative is positive.

Notice that where the red line goes vertical you actually have jump discontinuity. So at x = -3 and x = 3 the left and right hand limits of the function are not equal and therefore the derivative of the function does not exist at these points.


if you'd like to experiment with the excel file I made for this problem here is a link PiecewiseProblem_1.xls

The Difference Quotient

Here is a link to an excel spread sheet that I put together to explain the difference quotient:
DifferenceQuotient.xls

In the spread sheet I use the graph of y = sin(x) to illustrate the application of the difference quotient. Below is a screen shot of the interactive graph which is contained in the spread sheet.