# Related rates problem with bead sliding down a curve.

I keep getting the wrong answer on this problem.

A bead slides down the curve $xy=10$. Find the bead's horizontal velocity at time $t=2$ if its height at time $t$ seconds is $y=400-16t^2$.

I do $$\frac{dy}{dt}=-32t=-\frac{10}{x^2}\frac{dx}{dt}.$$

At $t=2$, $y=336$, so $x=\frac{10}{336}$. Plugging in everything above I get $\frac{dx}{dt}=\frac{5}{882}$ when $t=2$. The right answer is supposed to be $\frac{16}{9}$. How did I mess up?

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I see nothing wrong with your answer. What’s more, the supposed correct answer is unreasonable. At $t=0$, $x=\frac1{40}$, and at $t=2$, $x=\frac5{168}$, so in $2$ seconds the object has travelled
$$\frac5{168}-\frac1{40}=\frac{25-21}{840}=\frac1{210}$$
units horizontally. That’s an average horizontal speed of $\frac1{420}$. Granted, it’s accelerating, so its instantaneous horizontal speed should be a bit more than this, but that description fits $\frac5{882}$ much better than $\frac{16}9$. $\frac1{210}$ is about $0.004762$, while $\frac5{882}$ is about $0.005669$.