
#1
Apr705, 04:41 AM

P: 30

Find the length of parametrized curve given by
[tex]x(t) = 0t^3 +12t^2  24t[/tex], [tex]y(t) = 4t^3 +12t^2+0t[/tex], where t goes from zero to one. Hint: The speed is a quadratic polynomial with integer coefficients. it's an arclength question right? x' = 24*t24 y' = 12*t^2+24*t [tex]\int_{0}^{1} \sqrt{(24*t24)^2 + (12*t^2+24*t)^2}[/tex] i get 6.49 when i use a math program to integrate it which is incorrect. anyone know where i went wrong? 



#2
Apr705, 06:04 AM

Sci Advisor
P: 412

[tex] 1: \ \ \ \ \int_{0}^{1} \sqrt{(24*t24)^2 + (12*t^2+24*t)^2} \, dt \ = \ \int_{0}^{1} \sqrt{ \left ( 12t^{2} \  \ 24t \ + \ 24 \right )^{2} } \, dt \ = [/tex] [tex] 2: \ \ \ \ = \ \int_{0}^{1} \left  \left ( 12t^{2} \  \ 24t \ + \ 24 \right ) \right  \, dt \ \color{red} = \ \mathbf{\left (16 \right )} [/tex] ~~ 


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