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Arc Length of a Parametric Curve from t0 to tsqrt8

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Find the arc length of the parametric curve given by x(t)=2+6t2x(t) = 2 + 6t^2 and y(t)=5+4t3y(t) = 5 + 4t^3 for tt in the interval [0,8][0, \sqrt{8}].

To find the arc length of a parametric curve defined by two functions, we utilize the arc length formula specific to parametric equations. This involves computing the integral of the square root of the sum of the squares of the derivatives of the x and y components of the curve with respect to the parameter t. For a parametric curve given by functions x(t)x(t) and y(t)y(t), the arc length L from t=at = a to t=bt = b is found using the integral of the square root of (dxdt)2+(dydt)2(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 with respect to t, evaluated from a to b.

In this problem, you are asked to determine the arc length of a parametric curve where x(t)=2+6t2x(t) = 2 + 6t^2 and y(t)=5+4t3y(t) = 5 + 4t^3 over a specific interval. The process starts by computing the derivatives dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}. Then, these derivatives are squared, and their sum is established under the square root. The integral is then evaluated from t=0t = 0 to t=8t = \sqrt{8}. This approach highlights a standard technique in calculus that assists in analyzing the properties of curves in parametric form, providing insights into their geometric characteristics.

Furthermore, understanding how to express and manipulate parametric equations is crucial, as they are often used to describe motion and paths in two or more dimensions. The grasp of parametric forms aids not just in computing geometric properties like arc length but is also foundational in fields such as physics and engineering, where analyzing paths, velocities, and accelerations are essential.

Posted by grwgreg 5 days ago

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