Expert: Scotto - 5/19/2011

Question
QUESTION: I can't for the life of me see how the sume of the infintesimal lines on the curve translates into the integral of that function.

http://en.wikipedia.org/wiki/Arc_length#Another_way_to_obtain_the_integral_formu

if you look at the proof given there, I don't get how they jump from the sum of the very small lines along the curve to getting the integral of the formula for line length? There is a jump there from the limit as the change in x approaches 0 to the integral. How does an "area under the curve" come into play, exactly what would this look like on a graph, is there any way to show it visually? Any way to make this more clear?

thanks alot

ANSWER: When the length of a curve is looked at, first consider finding the hypotenuse of a right triangle.  Given one side is x and the other is y, and the hypotenuse is s, then
s^2 = x^2 + y^2.

We could make each of these a function of t that changes, which gives
[s(t)]^2 = [x(t)]^2 + [y(t)]^2.

Just as in finding area we start with a summation and let the change be considered a function,
thus finding the area, we can let s(t) goto 0 and do an infinite sum, which is an integral.


To look at the way they jump, consider finding the area under f(x) = x from 6 down to 0.
It is known that this is a triangle with width 6 and height 6, so the area is 6*6/2, which is 18.

If we approximate with a difference of 6 between the x values, then that would give x0 = 0 and x1 = 6.  From here we can use 6 as the base and a height that is between 0 and 6.
This would say the area was between 6*0= 0 and 6*6 = 36.

If the difference was taken to be 3, the lower value of the curve would be
3*f(0) + 3*f(3), which is 3*0 + 3*3 = 9.  The upper limit would be 3*f(3) + 3*F(6) = 3*3 + 3*6 = 9 + 18 = 27.  This would put the area in the interval (9,27).

If the difference was 2, the area would be between 2*f(0) + 2*f(2) + 2*f(4) =
2*0 + 2*2+ 2*4 = 12 and 2*f(2) + 2*f(4) + 2*f(6) = 4 + 8 + 12 = 24.
Thus, the area is in (12,34).

For difference 1, we would get f(0) + f(1) + f(2) + f(3) + f(4) + f(5) = 0+1+2+3+4+5 = 15 and
f(1) + f(2) + f(3) + f(4) + f(5) + f(6) = 1+2+3+4+5+6 = 21.  Now we can say the ares is betwen 15 and 21.

If we took difference 0.1, the lower limit would be 0.1(f(0) + f(0.1) + f(0.2) + ... f(5.9)) and
the upper limit ofg 0.1(f(0.1) + f(0.2) + ... f(5.9) + f(6)) .  This puts the area in
the interval (17.7,18.28).

If the difference was 0.01, the lower limit would be 17.97 and the upper would be 18.03.
This puts it in (17.97, 18.03).

The smaller the difference taken, the closer the two numbers get to 18,
and that is the value of the integral of x from 6 downto 0.


I don't have time to put it all here, but take finding the length of a circle that has center 0 and radius R that is contained in the I quadrant.  This problem of integration could be converted to polar.  It would give pi*R/2, which is correct.  This is because the length of the whole circle on the outside is known as 2*pi*R.  Since we have a quarter of the circle with length pi*R/2, this gives the total length as 4(pi*R/2) = 2*pi*R, so it can be seen to be correct.


To show it visually, draw the curve and pick points along the curve, putting a line between each set of successive point.  As the difference grows to be infitessimally small, the summation approaches the integral.


---------- FOLLOW-UP ----------

QUESTION: Well that still doesn't answer my question Scotto. I get everything up until the point when we say that the sum of the hypoteneuses on the line equals the integral. Yeah I understand that an integral is a sum, but it is the sume of infinitely small rectangles under a curve, an integral is not the sum of infinitely small hypoteneuses along a curve, so how do we jump to the integral of that function? I do not get how it can be interpreted as an area under the curve.

ANSWER: When finding the value of an integral, it finds the area of an infinite number of rectangle below the curve.  The width is taken care of by the dx and the height is the function itself.

When finding the length of a line, it is the squareroot of the sum of the squares.
Thus, the length of y = x from 0 to 1 is the √(1+1) = √2.

If integration was used on the line y = x, we need to find the value of the integral from
1 downto 0 of √(1+(y')²)/  Since y' = 1, that is the integral from 1 downto 0 of √2 dx.
The value of this is (√2)x, from 1 downto 0.  That is √2(1-0) = √2.

This could be applied to a quarter circle, y = √(1-x²), in quadrant I.
Since y' = -x/(1-x²), (y')² = x²/(1-x²).  From this, When finding the value of an integral, it finds the area of an infinite number of rectangle below the curve.  The width is taken care of by the dx and the height is the function itself.

Since y' = -x/(1-x²), (y')² = x²/(1-x²).  Now √[1+(y')²] = √[1 + x²/(1-x²)].
That reduces to √[1/(1-x²)].

Using a triangle with A as the angle, 1 as the hypotenuse, and x as the far side,
that makes √(1-x²) the near side.  When x = 0, A = pi/2.  When x = 1, A = 0.
It can be seen that √(1-x²) = cosA.  Since x = sinA, dx = cosA dA.
We are now integrating cos²A from 0 down to pi/2.
This is the integral of -cos²A from pi/2 down to 0.
It is a trig identity that cos²A = [1 + cos(2A)]/2, and that can be integrated.
The integral is A/2 + sin(2A)/4.

Since the sin of 2A goes from pi to 0, the sin of both of these is 0.
When A/2 is looked at, we get pi/2.
Now it is known that the circumference of a circle is 2pi,
so the length in one quadrante is pi/2, since that is 1/4 of the entire circle.

So, using two example of lengths that are known,
the method has been applied and shown to be correctn.


---------- FOLLOW-UP ----------

QUESTION: Well now were gettin somewhere thanks alot Scotto, but ya know what? I'm not entirely sure what the input(s) look like for the equation (square root of 1 + dy/dx^2)dx. What exactly do we input? What goes into dx? a given x value that is as far over on the x axis as we want to go? and what about for (dy/dx) Do we just plut the x value into the equation for the derivitive and get the derivitive at that point? I think I'm kind of lost here about how the inputs work.

Answer
If you have a question asking for and integral of squareroot(1 + d^2y/dx^2) dx,
that means to find the second derivative of y with respect to x for d^2y/dx^2.

Once this has been done, add it to 1, take the squareroot, and integegrate with respect to x.

The dx at the end is used to designate what variable the integral is being done with respect to.