Give a definite integral representing the length of the parametric curve $x = t ^ { 3 } , y = t ^ { 4 } , 0 \leq t \leq 1$ .
A) $\int _ { 0 } ^ { 1 } \left( t ^ { 3 } + t ^ { 4 } \right) d t$
B) $\int _ { 0 } ^ { 1 } \sqrt { t ^ { 3 } + t ^ { 4 } } d t$
C) $\int _ { 0 } ^ { 1 } \sqrt { 1 + 3 t ^ { 2 } } d t$
D) $\int _ { 0 } ^ { 1 } \sqrt { t ^ { 2 } + 4 t ^ { 3 } } d t$
E) $\int _ { 0 } ^ { 1 } \sqrt { 9 t ^ { 4 } + 16 t ^ { 6 } } d t$
F) $\int _ { 0 } ^ { 1 } \sqrt { 4 t ^ { 4 } + 9 t ^ { 6 } } d t$
G) $\int _ { 0 } ^ { 1 } \sqrt { t ^ { 5 } + t ^ { 7 } } d t$
H) $\int _ { 0 } ^ { 1 } \sqrt { 8 t ^ { 6 } + 6 t ^ { 8 } } d t$

Question

Give a definite integral representing the length of the parametric curve $x = t ^ { 3 } , y = t ^ { 4 } , 0 \leq t \leq 1$ .
A) $\int _ { 0 } ^ { 1 } \left( t ^ { 3 } + t ^ { 4 } \right) d t$ 
B) $\int _ { 0 } ^ { 1 } \sqrt { t ^ { 3 } + t ^ { 4 } } d t$ 
C) $\int _ { 0 } ^ { 1 } \sqrt { 1 + 3 t ^ { 2 } } d t$ 
D) $\int _ { 0 } ^ { 1 } \sqrt { t ^ { 2 } + 4 t ^ { 3 } } d t$ 
E) $\int _ { 0 } ^ { 1 } \sqrt { 9 t ^ { 4 } + 16 t ^ { 6 } } d t$ 
F) $\int _ { 0 } ^ { 1 } \sqrt { 4 t ^ { 4 } + 9 t ^ { 6 } } d t$ 
G) $\int _ { 0 } ^ { 1 } \sqrt { t ^ { 5 } + t ^ { 7 } } d t$ 
H) $\int _ { 0 } ^ { 1 } \sqrt { 8 t ^ { 6 } + 6 t ^ { 8 } } d t$

Quizplus · Accepted Answer

The length of a parametric curve given by $x = f(t)$ and $y = g(t)$ from $t = a$ to $t = b$ is calculated using the formula $\int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$. For $x = t^3$ and $y = t^4$, $\frac{dx}{dt} = 3t^2$ and $\frac{dy}{dt} = 4t^3$. Substituting these into the formula gives $\int_{0}^{1} \sqrt{(3t^2)^2 + (4t^3)^2} dt = \int_{0}^{1} \sqrt{9t^4 + 16t^6} dt$.

Give a Definite Integral Representing the Length of the Parametric x=t3,y=t4,0≤t≤1x = t ^ { 3 } , y = t ^ { 4 } , 0 \leq t \leq 1x=t3,y=t4,0≤t≤1

Give a Definite Integral Representing the Length of the Parametric $x = t ^ { 3 } , y = t ^ { 4 } , 0 \leq t \leq 1$