Question

Find the exact length of the curve.
x = 1 + 9t2
y = 2 + 6t3
0 ≤ t ≤ 2

Answer 1

Answer:

6*( 5^1.5 - 1 )        

Step-by-step explanation:

Given:-

- The curve is defined by the following two parametric equations:

                                   x = 1 + 9t^2

                                   y = 2 + 6t^3

- The interval for which the required length of curve is defined is [ 0 , 2 ].

Find:-

Find the exact length of the curve

Solution:-

- The length of the curve expressed in the form of time domain parametric equation can be determined from the following integral for length " L ":

                                   $L = \int\limits^a_b {\sqrt{(\frac{dx}{dt} )^2 + (\frac{dy}{dt} )^2 } } \, dt$

- Where the limits of integral are domain for the parameter t = [ 0 , 2 ].

- Compute the first derivative of both parametric coordinates ( x & y ) with respect to parameter ( t ).

                                  $\frac{dx}{dt} = 18*t\\\\\frac{dy}{dt} = 18*t^2$

- Substitute the respective derivatives in the integral for calculating the length " L " of the curve:

                        $L = \int\limits^2_0 {\sqrt{(18t )^2 + (18t^2 )^2 } } \, dt\\\\L = \int\limits^2_0 {\sqrt{(18^2t^2 + 18^2t^4) } } \, dt\\\\L = 18 \int\limits^2_0 {\sqrt{t^2 + t^4 } } \, dt\\\\L = 18 \int\limits^2_0 {t \sqrt{ 1 + t^2 } } \, dt\\\\L = 9 \int\limits^2_0 {2t (1 + t^2)^0^.^5 } \, dt$

- Apply the general power rule of integration and evaluate the limits:

                            $L = 9 * \frac{2}{3} (1 + t^2)^\frac{3}{2} /^2_0 \\\\L = 6* (1 + t^2)^\frac{3}{2} /^2_0 \\\\L = 6* (1 + 4)^\frac{3}{2} - 6* (1 + 0)^\frac{3}{2} \\\\L = 6*( 5^\frac{3}{2} - 1 )$    

 

Answer: The exact length of the curve is L = 6*( 5^1.5 - 1 )