Question

A particle is moving along a hyperbola \(xy = 8\). As it reaches the point \(\left( {4,2} \right)\), the \(y\)-coordinate is decreasing at a rate of 3cm/s. How fast is the \(x\)-coordinate of the point changing at that instant?HoH

Short answer

The \(x\)-coordinate is increasing at a rate of \(6\,{\rm{cm/s}}\).

Step-by-step solution

The product rule of differentiation

If a function\(h\left( x \right) = f\left( x \right) \cdot g\left( x \right)\)and\(f\)and\(g\)are both differentiable, then the derivative of\(h\left( x \right)\)is as follows:

\(\frac{{dh}}{{dx}} = \frac{d}{{dx}}\left( {f\left( x \right) \cdot g\left( x \right)} \right) = f\left( x \right) \cdot \frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right) \cdot \frac{d}{{dx}}\left( {f\left( x \right)} \right)\)

The derivative of the given function

Given that \(xy = 8\). Differentiate both the sides as follows:

\(\begin{aligned}{c}\frac{d}{{dt}}\left( {xy} \right) &= \frac{d}{{dt}}\left( 8 \right)\\x\frac{{dy}}{{dt}} + y\frac{{dx}}{{dt}} &= 0\end{aligned}\)

Substitute the given values

Given that \(\frac{{dy}}{{dt}} = - 3\), \(\left( {x,y} \right) = \left( {4,2} \right)\). Substitute these values in the above equation and solve as follows:

\(\begin{aligned}{c}4\left( { - 3} \right) + 2\frac{{dx}}{{dt}} &= 0\\ - 12 + 2\frac{{dx}}{{dt}} &= 0\\2\frac{{dx}}{{dt}} &= 12\\\frac{{dx}}{{dt}} &= 6\end{aligned}\)

Hence, the \(x\)-coordinate is increasing at a rate of \(6\,{\rm{cm/s}}\).