Question

a particle moves on the hyperbola ( xy = 18 ) for time ( tgeq0 ) seconds. at a certain instant, ( y = 6 ) and ( \frac{dy}{dt}=8 ). which of the following is true about ( x ) at this instant?
a ( x ) is decreasing by 4 units per second.
b ( x ) is increasing by 4 units per second.
c ( x ) is decreasing by 1 unit per second.
d ( x ) is increasing by 1 unit per second.

Explanation:

Step1: Find the value of \(x\)

Given \(xy = 18\) and \(y = 6\), then \(x=\frac{18}{y}=\frac{18}{6} = 3\).

Step2: Differentiate \(xy = 18\) with respect to \(t\)

Using the product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u = x\) and \(v = y\). So \(\frac{d(xy)}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}=0\) (since \(\frac{d(18)}{dt} = 0\)).

Step3: Substitute the known values

We know \(x = 3\), \(y = 6\) and \(\frac{dy}{dt}=8\). Plugging into \(x\frac{dy}{dt}+y\frac{dx}{dt}=0\) gives \(3\times8+6\times\frac{dx}{dt}=0\).

Step4: Solve for \(\frac{dx}{dt}\)

First, simplify \(24 + 6\frac{dx}{dt}=0\). Then \(6\frac{dx}{dt}=- 24\), so \(\frac{dx}{dt}=-4\). A negative value of \(\frac{dx}{dt}\) means \(x\) is decreasing. The magnitude is \(4\), so \(x\) is decreasing by \(4\) units per second.

Answer:

A. \(x\) is decreasing by \(4\) units per second.