Find the length of the arc formed by
x^2=9y^3
from point A to point B, where
A=(0,0) and B=(81,9)
Please help, I have no clue as to where to go with this problem! Thanks!
To find the length of the arc formed by the equation x^2 = 9y^3, we can use the arc length formula for a parametric curve.
First, let's express the equation in parametric form. Let:
x = 3t^2 (taking the positive square root)
y = t^3
Now, we need to find the limits of t that correspond to the points A(0,0) and B(81,9).
For point A:
When x = 0, we get 0 = 3t^2, which simplifies to t = 0.
For point B:
When x = 81, we have 81 = 3t^2, which simplifies to t^2 = 27. Taking the positive square root, we get t = √27 = 3√3.
Now, we can find the arc length using the formula:
s = ∫[a to b] √(dx^2 + dy^2) dt,
where a and b are the limits of t, and dx/dt and dy/dt are the derivatives of x and y with respect to t, respectively.
Differentiating x = 3t^2 with respect to t, we get dx/dt = 6t.
Differentiating y = t^3 with respect to t, we get dy/dt = 3t^2.
Substituting the derivatives into the arc length formula, we have:
s = ∫[0 to 3√3] √( (dx/dt)^2 + (dy/dt)^2 ) dt
= ∫[0 to 3√3] √( (6t)^2 + (3t^2)^2 ) dt
= ∫[0 to 3√3] √( 36t^2 + 9t^4 ) dt
= ∫[0 to 3√3] t * √( 36 + 9t^2 ) dt.
This integral can be solved using a substitution.
Let u = 36 + 9t^2.
Then, du/dt = 18t, which gives us dt = du / (18t).
Substituting these values into the integral, we have:
s = ∫[0 to 3√3] ( t * √u ) * ( 1 / (18t) ) du
= (1/18) ∫[0 to 3√3] √u du
= (1/18) * (2/3) * u^(3/2) |[0 to 3√3]
= (1/27) * [ ( 36 + 9t^2 )^(3/2) ] |[0 to 3√3].
Evaluating the integral at the limits of t, we have:
s = (1/27) * ( ( 36 + 9 * (3√3)^2 )^(3/2) - ( 36 + 9 * 0^2 )^(3/2) )
= (1/27) * ( 36 + 27 * 3 * 3 - 36 )
= (1/27) * ( 36 + 81 - 36 )
= (1/27) * 81
= 3 units.
Therefore, the length of the arc formed by the equation x^2 = 9y^3, from point A(0,0) to point B(81,9), is 3 units.
To find the length of the arc formed by x^2 = 9y^3 from point A to point B, you can use calculus and the formula for arc length.
Step-by-step solution:
1. First, let's find the parametric equations for the curve x^2 = 9y^3. We can rewrite the equation as x^2 / 9 = y^3. Taking the cube root of both sides, we get y = (x^2 / 9)^(1/3).
2. The parametric equations for the curve are given by x = t and y = (t^2 / 9)^(1/3), where t is the parameter.
3. Next, we need to find the derivative of y with respect to x. Using the chain rule, we have dy/dx = (dy/dt) / (dx/dt). Differentiating the parametric equations, we get dx/dt = 1 and dy/dt = (2t/9)^(1/3) / 3*(t^2 / 9)^(-2/3).
4. Simplifying, dx/dt = 1 and dy/dt = 2^(1/3)/(3t^(2/3)).
5. The formula for arc length is given by L = ∫(a to b) √(dx/dt)^2 + (dy/dt)^2 dt.
6. Substituting the values for dx/dt and dy/dt, we have L = ∫(a to b) √(1^2 + (2^(1/3)/(3t^(2/3)))^2) dt.
7. The limits of integration are a = 0 (corresponding to point A) and b = 81 (corresponding to point B).
8. Calculate the integral: L = ∫(0 to 81) √(1 + (2^(1/3)/(3t^(2/3)))^2) dt.
9. After solving this integral, you will find the length of the arc from point A to point B.
Please note that the integral may not have a simple closed-form solution and might need to be computed numerically using numerical methods or software.
I hope this explanation helps you understand the steps to find the length of the arc formed by x^2 = 9y^3 from point A to point B.
We have a formula for the length of an arc,
I assume you have that.
If not look here
http://www.mathwords.com/a/arc_length_of_a_curve.htm
You will have to find dy/dx of your equation, then square it, add one , take the square root and then integrate that !!!