Find the length of the loop of the curve x=t^2 y=t-t^3/3
Answers
Answer :
Length of the loop curve L = 4√3 units
what is loop curve ?
A loop curve is a type of curve in which the path of the curve encloses a bounded region in the plane and returns to its starting point without intersecting itself.
Explanation :
To find the length of the loop of the curve x = t^2 and y = t - t^3/3, we can use the arc length formula:
L = ∫√(1 + (dy/dx)^2) dx from t = a to t = b
First, we need to find dy/dx:
dy/dx = (dy/dt) / (dx/dt) = (1 - t^2) / (2t)
Now, we can substitute this into the arc length formula:
L = ∫√(1 + ((1 - t^2) / (2t))^2) dt from t = a to t = b
Simplifying the integrand, we get:
L = ∫√(1 + (1 - 2t^2 + t^4) / (4t^2)) dt
L = ∫√((5t^4 - 2t^2 + 4) / (4t^2)) dt
L = (1/2)∫√((5t^2 - 2 + 4/t^2)) dt
Using a trigonometric substitution of u = √(5t^2 - 2), we can simplify the integrand to:
L = (1/2)∫(u^2 - 3/4)^(-1/2) du
Using another substitution of v = u^2 - 3/4 and simplifying, we get:
L = ∫(v^(-1/2)) du = 2√v
Substituting back in for v and u, we get:
L = 2√(5t^2 - 11/4) from t = a to t = b
To find the length of the loop, we need to find the values of t where the curve intersects itself. Solving for x = t^2 and y = t - t^3/3, we get:
t^3 - 3t + 2 = 0
Factoring, we get:
(t - 1)(t + 1)^2 = 0
So the curve intersects itself at t = 1 and t = -1. Therefore, the length of the loop is:
L = 2√(5t^2 - 11/4) from t = -1 to t = 1
L = 4√3 units
To learn more about the loop curve follow the given link :
https://brainly.in/question/7609848
https://brainly.in/question/33335962
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