Find the arc length of the curve on the given interval. $x = t ^ { 2 } + 3 , y = 8 t ^ { 3 } + 9 , - 1 \leq t \leq 0$
A) $\frac { ( 770 ) ^ { \frac { 3 } { 2 } } - 8 } { 864 }$
B) $\frac { ( 145 ) ^ { \frac { 3 } { 2 } } - 1 } { 216 }$
C) $\frac { ( 2,308 ) ^ { \frac { 9 } { 4 } } - 8 } { 1728 }$
D) $\frac { ( 2,308 ) ^ { \frac { 9 } { 4 } } - 1 } { 864 }$
E)$\frac { ( 769 ) ^ { \frac { 3 } { 2 } } - 1 } { 1728 }$

Question

Quizplus · Accepted Answer

The arc length $L$ of a curve defined by parametric equations $x = f(t)$ and $y = g(t)$ over an interval $a \leq t \leq b$ is given by the integral $L = \int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. For $x = t^2 + 3$ and $y = 8t^3 + 9$, we find $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 24t^2$. Substituting these into the formula gives $L = \int_{-1}^{0} \sqrt{(2t)^2 + (24t^2)^2} dt$, which simplifies and evaluates to the expression in choice B.

Find the Arc Length of the Curve on the Given x=t2+3,y=8t3+9,−1≤t≤0x = t ^ { 2 } + 3 , y = 8 t ^ { 3 } + 9 , - 1 \leq t \leq 0x=t2+3,y=8t3+9,−1≤t≤0

Find the Arc Length of the Curve on the Given $x = t ^ { 2 } + 3 , y = 8 t ^ { 3 } + 9 , - 1 \leq t \leq 0$