Find the length of the curve.
-$$x = \frac { 1 } { 3 } \left( t ^ { 3 } - 3 t \right) , y = t ^ { 2 } + 6,0 \leq t \leq 2$$
A) $\frac { 14 } { 3 }$
B) 5
C) $\frac { 8 } { 3 }$
D) $\frac { 10 } { 3 }$

Question

Quizplus · Accepted Answer

The length of the curve can be found using the formula $\int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } dt$, where $x(t)$ and $y(t)$ are given, and $t$ ranges from 0 to 2. First, find the derivatives: $\frac{dx}{dt} = t^2 - 1$ and $\frac{dy}{dt} = 2t$. Then, plug these into the formula and integrate from 0 to 2: $\int_{0}^{2} \sqrt{ (t^2 - 1)^2 + (2t)^2 } dt = \int_{0}^{2} \sqrt{ t^4 - 2t^2 + 1 + 4t^2 } dt = \int_{0}^{2} \sqrt{ t^4 + 2t^2 + 1 } dt = \int_{0}^{2} \sqrt{ (t^2 + 1)^2 } dt = \int_{0}^{2} (t^2 + 1) dt$, which evaluates to $\frac{t^3}{3} + t \Big|_0^2 = \frac{8}{3} + 2 = \frac{14}{3}$.

Find the Length of the Curve x=13(t3−3t),y=t2+6,0≤t≤2x = \frac { 1 } { 3 } \left( t ^ { 3 } - 3 t \right) , y = t ^ { 2 } + 6,0 \leq t \leq 2x=31​(t3−3t),y=t2+6,0≤t≤2

Find the Length of the Curve $x = \frac { 1 } { 3 } \left( t ^ { 3 } - 3 t \right) , y = t ^ { 2 } + 6,0 \leq t \leq 2$