What is the arc length of $f (t) = (3 t - 4, t^{3} - 2 t)$ $f(t)=(3t-4,t^3-2t)$ over $t \in [- 1, 2]$ $t in [-1,2]$ ?

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Answer 1 · 2016-10-23T21:27:00

Integration by WolframAlpha
$L = 28.7405$ $L = 28.7405$

$\frac{d x}{d t} = 3$ $dx/dt = 3$

$\frac{d y}{d t} = 3 t^{2} - 2$ $dy/dt = 3t^2 - 2$

Let L = the arclength

$L = \int_{a}^{b} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t$ $L = int_a^bsqrt((dx/dt)^2 + (dy/dt)^2)dt$

Substituting in our values:

$L = \int_{- 1}^{2} \sqrt{{(9)}^{2} + {(3 t^{2} - 2)}^{2}} d t$ $L = int_-1^2sqrt((9)^2 + (3t^2 - 2)^2)dt$

$L = 28.7405$ $L = 28.7405$

What is the arc length of f(t)=(3t-4,t^3-2t) f(t)=(3t−4,t3−2t)f(t)=(3t-4,t^3-2t) over t in [-1,2]t∈[−1,2]t in [-1,2]?