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Consider a ramp modeled by the function $y = \frac{1}{x}, x>0$. A ball slides down the ramp so that the x-coordinate of its position at any time $t$ seconds is increasing by a rate of $f(x)$ units per second. If its y-coordinate is decreasing at a constant rate of $1$ unit per second, find $f(x)$.

The answer is $x^2$. By "ramp modeled by the function $y = \frac{1}{x}$, did they mean height at a given point is $\frac{1}{x}$. How would you go about solving this question.

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We have (from the model)

$\large y=\frac{1}{x}$ (Eq. 1)

Let us denote the time variable as $t$. We are given that

$\large\frac{dx}{dt}=+f(x)$

$\large \frac{dy}{dt}=-1$

Using the chain rule we have

$\large \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ (Eq. 2)

Differentiating $y$ w.r.t $x$ results in

$\large \frac{dy}{dx}=-\frac{1}{x^2}$

Substituting this into (Eq. 2), along with the given rates of change w.r.t. time, we obtain

$\large -1=-\frac{1}{x^2}f(x)$

This leads to

$\large f(x)=x^2$

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