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Parametrized curve

by ILoveBaseball
Tags: curve, parametrized
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ILoveBaseball
#1
Apr7-05, 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*t-24
y' = -12*t^2+24*t

[tex]\int_{0}^{1} \sqrt{(24*t-24)^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?
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xanthym
#2
Apr7-05, 06:04 AM
Sci Advisor
P: 412
Quote Quote by ILoveBaseball
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*t-24
y' = -12*t^2+24*t

[tex]\int_{0}^{1} \sqrt{(24*t-24)^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?
Note that:

[tex] 1: \ \ \ \ \int_{0}^{1} \sqrt{(24*t-24)^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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