Question

A particle moves on the hyperbola xy=18 for time t≥0 seconds. At a certain instant, y=6 and dydt=8. What is x that this instant?

Answer 1

Answer:

The value of x at this instant is 3.

Step-by-step explanation:

Let $x\cdot y = 18$, we get an additional equation by implicit differentiation:

$x\cdot \frac{dy}{dt}+y\cdot \frac{dx}{dt} = 0$ (1)

From the first equation we find that:

$x = \frac{18}{y}$ (2)

By applying (2) in (1), we get the resulting expression:

$\frac{18}{y}\cdot \frac{dy}{dt}+y\cdot \frac{dx}{dt} = 0$ (3)

$y\cdot \frac{dx}{dt}=-\frac{18}{y}\cdot \frac{dy}{dt}$

$\frac{dx}{dt} = -\frac{18}{y^{2}} \cdot \frac{dy}{dt}$

If we know that $y = 6$ and $\frac{dy}{dt} = 8$, then the first derivative of x in time is:

$\frac{dx}{dt} = -\frac{18}{6^{2}} \cdot (8)$

$\frac{dx}{dt} = -4$

From (1) we determine the value of x at this instant:

$x\cdot \frac{dy}{dt} = -y\cdot \frac{dx}{dt}$

$x = -y\cdot \left(\frac{\frac{dx}{dt} }{\frac{dy}{dt} } \right)$

$x = -6\cdot \left(\frac{-4}{8} \right)$

$x = 3$

The value of x at this instant is 3.