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