A particle starts at the origin and moves along the curve $y = \dfrac{2x^{3/2}}{3}$ in the positive $x$-direction at a speed of $4$ cm/sec, where $x$ and $y$ are in measured in cm. Find the $x$-coordinate of the particle at $t = 5$.
I honestly have very limited ideas on how to approach this problem. I set $\dfrac{dx}{dt} = 4$, to get $\int dx = \int 4dt$, and ultimately that $x = 4t + C$. But I haven't done anything with integrals with parametric equations before and I don't know how to continue. Could someone help me?
If $f'(x)$ is continuous then the distance along the graph of $f$ from $(0,f(0))$ to $(p,f(p))$ is $$D(0,p)=\int_0^p (1+f'(x)^2)^{1/2}dx.$$ At $4$ cm/sec, after $5$ seconds we have $D(0,p)=20.$ We have $f'(x)=(2x^{3/2}/3)'=x^{1/2}.$ So we seek $p$ such that $$20=\int_0^p (1+x)^{1/2}dx.$$