Suppose R is the region in the plane enclosed by y = x2 and y = 4.
a) Compute the perimeter P and area A of R, and then compute the ratio Q = A/P2.
Note By squaring the perimeter the ratio becomes independent of the units chosen to measure the region.
b) Compute this ratio Q = A/P 2 for these four regions: the region R, a square, a circle, and an equilateral triangle. Draw the figures in increasing order of Q.
I assume you know the formula for the length of a curve in Calculus.
L = ∫( 1 + (dy/dx)^2 )^(1/2) dx (from left x to right x)
This is the hard part of the question
I ended up finding the length of the parabolic curve to be
L = 2∫(1 + 4x^2)^(1/2) dx from x=0 to x=2
I ran this through WolFram to get
http://integrals.wolfram.com/index.jsp?expr=%281%2B+4x%5E2%29%5E%281%2F2%29&random=false
sub in the values, add on the distance from (-2,4) to (2,4) and you have the perimeter of R
For the area:
A = 2∫(4 - x^2) dx from x = 0 to 2
which you should be able to do quite easily.
The length of the curve is the only difficult part of the problem.
To compute the perimeter and area of the region R enclosed by the curves y = x^2 and y = 4, we need to find the intersection points of these curves first.
a) Finding the intersection points:
Setting y = x^2 and y = 4 equal to each other, we get:
x^2 = 4
Taking the square root of both sides (remembering to consider both positive and negative sqrt), we have:
x = ±2
So the intersection points are (-2, 4) and (2, 4).
To compute the perimeter P of R, we need to consider the curve y = x^2 for x in the range [-2, 2]. It's important to note that the length of the curve can be approximated using integration. However, for simplicity, we can break it down into smaller line segments.
The line segment from (-2, 4) to (2, 4) has a length of 4 units (side of the rectangle).
The line segment from (-2, 4) to (-2, 0) has a length of 4 units.
The line segment from (2, 4) to (2, 0) has a length of 4 units.
Adding these lengths together, we get the perimeter:
P = 4 + 4 + 4 = 12 units
To compute the area A of R, we need to find the integral of the curve y = x^2 between x = -2 and x = 2:
A = ∫(from -2 to 2) x^2 dx
Using the power rule, we have:
A = [1/3 * x^3] (from -2 to 2)
Substituting the limits of integration:
A = 1/3 * (2^3 - (-2^3))
A = 1/3 * (8 - (-8))
A = 1/3 * 16
A = 16/3 square units
Now that we have both the perimeter P = 12 units and the area A = 16/3 square units, we can compute the ratio Q = A/P^2:
Q = A/P^2
Q = (16/3) / (12^2)
Q = (16/3) / 144
Q = 16/432
Q ≈ 0.03704
b) To compute the ratios for the other shapes (square, circle, and equilateral triangle), we need to determine their perimeters and areas.
For a square with side length s, the perimeter P is given by:
P = 4s
For a circle with radius r, the perimeter P is given by:
P = 2πr
For an equilateral triangle with side length s, the perimeter P is given by:
P = 3s
For the area A of a square, circle, and equilateral triangle, the formulas are:
A(square) = s^2
A(circle) = πr^2
A(triangle) = sqrt(3)/4 * s^2
Now, calculate the ratios Q for each shape by substituting the values of P and A into the expression Q = A / P^2.
After comparing the values of Q for each shape, draw the figures in increasing order of Q.