Question
Show that the length of the arc of the curve 9y2=x(3-x) the whole square measured from the origin to pt x,y 3r2=3y2+4x2
Answers
Answered by
Bot
To find the length of the arc of the curve, we need to use the arc length formula:
L = ∫√(1 + (dy/dx)²) dx
First, we need to find dy/dx by implicitly differentiating the given equation:
9y² = x(3-x)²
18y(dy/dx) = (3-x)² - 2x(3-x)(-1)
18y(dy/dx) = (3-x)² + 2x(3-x)
dy/dx = [(3-x)² + 2x(3-x)] / (18y)
Simplifying further:
dy/dx = [3(x-1)(x-5)] / (18y)
dy/dx = (x-1)(x-5) / (6y)
Now, we can substitute this expression into the arc length formula:
L = ∫√(1 + [(x-1)(x-5) / (6y)]²) dx
To make the integral easier to evaluate, we can use the equation of the curve to eliminate y:
9y² = x(3-x)²
y² = x(3-x)² / 9
y = ±(x(3-x) / 3)^(1/2)
We only need to consider the positive square root since we are measuring the arc length from the origin. Substituting for y:
L = ∫√(1 + [(x-1)(x-5) / (6(x(3-x) / 3)^(1/2))]²) dx
Simplifying:
L = ∫√(1 + [(x-1)(x-5) / (2(x(3-x))^(1/2))]²) dx
L = ∫√(1 + [(x-1)(x-5) / (2x(3-x))]²) dx
Now, we need to find the limits of integration. Since we are measuring the arc length from the origin, we need to find the x-coordinate of the point (3r²/(3+y²), y) on the curve. Substituting for y and simplifying:
3r²/(3+(x(3-x)/3))² = 3y²/(3+y²)
9r^4 = 3(3+y²)^2 x (3-x)²
y² = [(r²(3-x))^(2/3) - 3³/²] / 3
y = [(r²(3-x))^(1/3) - 3^(1/2)] / (3^(1/2))^(1/3)
We can use this expression as the upper limit of integration. The lower limit is simply 0, since we are measuring the arc length from the origin. Therefore:
L = ∫₀^[(r²(3-x))^(1/3) - 3^(1/2)] / (3^(1/2))^(1/3) √(1 + [(x-1)(x-5) / (2x(3-x))]²) dx
This integral is difficult to evaluate analytically. We can use numerical methods or approximations to find an estimate of the arc length.
L = ∫√(1 + (dy/dx)²) dx
First, we need to find dy/dx by implicitly differentiating the given equation:
9y² = x(3-x)²
18y(dy/dx) = (3-x)² - 2x(3-x)(-1)
18y(dy/dx) = (3-x)² + 2x(3-x)
dy/dx = [(3-x)² + 2x(3-x)] / (18y)
Simplifying further:
dy/dx = [3(x-1)(x-5)] / (18y)
dy/dx = (x-1)(x-5) / (6y)
Now, we can substitute this expression into the arc length formula:
L = ∫√(1 + [(x-1)(x-5) / (6y)]²) dx
To make the integral easier to evaluate, we can use the equation of the curve to eliminate y:
9y² = x(3-x)²
y² = x(3-x)² / 9
y = ±(x(3-x) / 3)^(1/2)
We only need to consider the positive square root since we are measuring the arc length from the origin. Substituting for y:
L = ∫√(1 + [(x-1)(x-5) / (6(x(3-x) / 3)^(1/2))]²) dx
Simplifying:
L = ∫√(1 + [(x-1)(x-5) / (2(x(3-x))^(1/2))]²) dx
L = ∫√(1 + [(x-1)(x-5) / (2x(3-x))]²) dx
Now, we need to find the limits of integration. Since we are measuring the arc length from the origin, we need to find the x-coordinate of the point (3r²/(3+y²), y) on the curve. Substituting for y and simplifying:
3r²/(3+(x(3-x)/3))² = 3y²/(3+y²)
9r^4 = 3(3+y²)^2 x (3-x)²
y² = [(r²(3-x))^(2/3) - 3³/²] / 3
y = [(r²(3-x))^(1/3) - 3^(1/2)] / (3^(1/2))^(1/3)
We can use this expression as the upper limit of integration. The lower limit is simply 0, since we are measuring the arc length from the origin. Therefore:
L = ∫₀^[(r²(3-x))^(1/3) - 3^(1/2)] / (3^(1/2))^(1/3) √(1 + [(x-1)(x-5) / (2x(3-x))]²) dx
This integral is difficult to evaluate analytically. We can use numerical methods or approximations to find an estimate of the arc length.
Answered by
oobleck
You made a mistake right at the start. It should have been
9y² = x(3-x)²
18y(dy/dx) = (3-x)² + 2x(3-x)(-1)
Simplifying things a bit, we can start off with
y = 1/3 √x (x-3)
y' = (x-1)/(2√x)
L = ∫√(1 + y'^2) dx = ∫√(1 + (x-1)^2/(4x)) dx
= ∫(x+1)/(2√x) dx = 1/3 √x (x+3)
Not sure what that r^2=3y^2+4x^2 has to do with it. We need the x-coordinate of the point; you have calculated the y-coordinate.
9y² = x(3-x)²
18y(dy/dx) = (3-x)² + 2x(3-x)(-1)
Simplifying things a bit, we can start off with
y = 1/3 √x (x-3)
y' = (x-1)/(2√x)
L = ∫√(1 + y'^2) dx = ∫√(1 + (x-1)^2/(4x)) dx
= ∫(x+1)/(2√x) dx = 1/3 √x (x+3)
Not sure what that r^2=3y^2+4x^2 has to do with it. We need the x-coordinate of the point; you have calculated the y-coordinate.
Answered by
Bot
You are correct, I apologize for the mistake in my initial calculation. Thank you for pointing that out. Your derivation of the arc length formula is correct and gives the answer as:
L = 1/3 √x (x+3)
Regarding the equation r^2 = 3y^2 + 4x^2, you are also correct that it was unnecessary for this problem since we only needed to find the x-coordinate of the given point. I apologize for including irrelevant information. Thank you for catching that error as well.
L = 1/3 √x (x+3)
Regarding the equation r^2 = 3y^2 + 4x^2, you are also correct that it was unnecessary for this problem since we only needed to find the x-coordinate of the given point. I apologize for including irrelevant information. Thank you for catching that error as well.