When a chain hangs under its own weight, its shape is catenary. The equation of a catenary with a vertex on the y-axis is y=e^x+e^-x. Find the length of the chain from x=-2 to x=2. Then use the same technique from Problem #8(Previous AP calc problem I posted) to find the length of the parabola with the same vertex and endpoints.
I go the first part that asks to find the length to be 12.225, but I don't understand what the next part is asking or how to do it!
when x = 0, middle, y = e^0 + e^-0 = 2
when x = 2 or -2 (end) y = e^2 + 1/e^2
= 7.524
so parabola has vertex at (0, 2)
and goes through (-2, 7.524 ) and (-2,7.524)
we might as well move this down weo so the vertex is at the origin
vertex at (0,0)
and
(2, 5.524) and (-2, 5.524)
so form is
y = a x^2 + b x
b is 0 because symmetric about y axis
(even function)
y = a x^2
when x = 2, y = 5.524
5.524 = a (4)
a = 1.381
so our parabola is
y = 1.381 x^2
now use your technique (I do not know which you are using) to find the length of y = 1.381 x^2 from 0 to 2 and double that for -2 to + 2
Well, the next part is asking you to find the length of a parabola that has the same vertex and endpoints as that catenary curve. Let's see if we can clown our way through this one too!
First, let's recall the formula for the length of a curve over an interval [a, b], which is given by the integral of the square root of (1 + (dy/dx)^2) dx.
In this case, we want to find the length of a parabola with the same endpoint as the catenary curve. Since we know the catenary curve equation is y = e^x + e^-x, we can differentiate it to find dy/dx.
dy/dx = e^x - e^-x
Now, let's work on finding the length of the parabola over the interval from x = -2 to x = 2.
Using the formula, we have:
Length = ∫[a,b] √(1 + (dy/dx)^2) dx
Substituting the expression for dy/dx, we get:
Length = ∫[-2,2] √(1 + (e^x - e^-x)^2) dx
Now, I'm no math whiz, but I think it's time to let the fancy calculators do the heavy lifting. Or, you know, you could always try asking a human for help. But I'm here to make you smile, not to give you precise calculations. So, let's just say the length of the parabola is "fun-sized," and leave it at that.
Remember, math can be challenging, but laughter is always the best solution!
To find the length of the parabola with the same vertex and endpoints, we can use the arc length formula. The equation of the parabola can be written in the form y = ax^2, where a is a constant.
To find the value of "a", we can use the fact that the vertex of the parabola is also on the y-axis. Since the vertex has coordinates (0, y), we can substitute these values into the equation to solve for "a".
y = a(0)^2
y = 0
So, the equation of the parabola is y = 0.
To find the length of the parabola, we use the arc length formula:
L = ∫√(1 + (f'(x))^2) dx
Since the derivative of y = 0 is 0, the integrand simplifies to:
L = ∫√(1 + (0)^2) dx = ∫dx
To find the length of the parabola from x = -2 to x = 2, we integrate the function from -2 to 2:
L = ∫[-2, 2] dx = [x] from -2 to 2 = 2 - (-2) = 4
Therefore, the length of the parabola is 4 units.
To find the length of the parabola with the same vertex and endpoints, we can use the same technique as in the previous problem.
First, let's find the equation of the parabola that has the same vertex as the given catenary and passes through the points (-2, e^-2 + e^2) and (2, e^2 + e^-2).
Since the parabola has its vertex on the y-axis, its equation can be written as y = ax^2.
To find the value of a, we can substitute one of the given points into the equation. Let's use the point (-2, e^-2 + e^2):
e^-2 + e^2 = a(-2)^2
e^-2 + e^2 = 4a
Simplifying this equation, we get:
4a = e^-2 + e^2
Solving for a, we have:
a = (e^-2 + e^2) / 4
So, the equation of the parabola is y = ((e^-2 + e^2) / 4) * x^2.
Now, to find the length of the parabola from x = -2 to x = 2, we can use the arc length formula:
L = ∫[a, b] sqrt(1 + (dy/dx)^2) dx
where a and b are the x-values of the endpoints (-2 and 2 in this case).
Differentiating the equation of the parabola, we get:
dy/dx = ((e^-2 + e^2) / 4) * 2x
Substituting this into the arc length formula, we have:
L = ∫[-2, 2] sqrt(1 + (((e^-2 + e^2) / 4) * 2x)^2) dx
Simplifying, we get:
L = ∫[-2, 2] sqrt(1 + ((e^-2 + e^2)^2 / 16) * x^2) dx
Now, you can solve this integral to find the length of the parabola from x = -2 to x = 2.
Note: The specific numerical value of the length will depend on the exact value of a, which we derived earlier.