Arc Length and Speed Can anyone help me with this calculus problem?
Calculate the length of the path over the given interval:
$c(t)=(5t^2, 10t^3), 1≤t≤3$
I'm not sure how to figure out the equation to graph it.
Any help is appreciated! 
 A: Set $x=5t^2,y=10t^3$. Solve for $x$ in terms of $t$ and substitute it into $y$. To find the length of the path, however, you can use the following formula:
$$s=\int^b_a\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}dt$$
A: Note: I initially evaluated the integral from $0$ to $3$; it has been amended.
What do we know about $V(D\alpha)$ where $\alpha(t) = (5t^2, 10t^3)$ is a parameterisation of a $1$-manifold in $\mathbb{R^2}$?
$V(D\alpha) = \sqrt{x'(t)^2 + y'(t)^2} = \sqrt{100t^2 + 900t^4} = 10t\sqrt{1 + 9t^2}$.
So we compute $$\int_{t=1}^{3} 10t\sqrt{1+9t^2} \\ = 5/9 \int_{t=1}^{3} 18t\sqrt{1+9t^2} dt \\ = 5/9 \int_{u=10}^{82} \sqrt{u} \\ = 5/9\cdot 2/3\cdot (82^{3/2} - 10^{3/2}) \\ = 10/27(82^{3/2} - 10^{3/2}).$$
A: to find $x$ in terms of $y$, try finding $t^6$ in terms of both $x$ and $y$, then set them equal to each other.
$$x=5t^2$$
$$x/5=t^2$$
$$x^3/5^3=t^6$$

$$y=10t^3$$
$$y/10=t^3$$
$$y^2/10^2=t^6$$

$$\therefore x^3/5^3=y^2/10^2=t^6$$
$$ y^2=4/5x^3$$
$$ y=\pm 2x\sqrt{x/5}$$
