Find the length of the curve. Select the correct answer.
$$x = 3 t ^ { 2 } + 8 , y = 2 t ^ { 3 } + 8,0 \leq t \leq 1$$
A) $2 \sqrt { 2 } - 2$
B) $4 \sqrt { 2 }$
C) $4 \sqrt { 2 } - 1$
D) $4 \sqrt { 2 } - 2$
E) None of these

Question

Quizplus · Accepted Answer

The length of the curve can be found using the formula:
$$L = \int_{t_1}^{t_2}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$$
where $t_1$ and $t_2$ are the lower and upper limits of $t$.

Taking the derivatives of $x$ and $y$ with respect to $t$, we get:
$$\frac{dx}{dt} = 6t$$
$$\frac{dy}{dt} = 6t^2$$

Substituting these values into the formula for $L$, we get:
$$L = \int_0^1 \sqrt{(6t)^2 + (6t^2)^2}dt = \int_0^1 6t\sqrt{1+36t^2}dt$$
To evaluate this integral, we can use the substitution $u = 1 + 36t^2$, $du = 72t dt$, so the integral becomes:
$$L = \int_1^{37} \sqrt{\frac{u-1}{4}}\frac{du}{72} = \frac{1}{18}\int_1^{37} \sqrt{u-1}du$$
We can evaluate this integral using the substitution $v = u - 1$, $dv = du$, so the integral becomes:
$$L = \frac{1}{18}\int_0^{36} \sqrt{v}dv = \frac{2}{9}\cdot\frac{v^{3/2}}{3/2}\bigg|_0^{36} = \frac{4\sqrt{2}-2}{3}$$
Therefore, the answer is (D) $4 \sqrt { 2 } - 2$.

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

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