Question

a particle moves on the hyperbola ( xy = 15 ) for time ( t geq 0 ) seconds. at a certain instant, ( x = 3 ) and ( \frac{dx}{dt}=6 ). which of the following is true about ( y ) at this instant?
a ( y ) is decreasing by 10 units per second.
b ( y ) is increasing by 10 units per second.
c ( y ) is decreasing by 5 units per second.
d ( y ) is increasing by 5 units per second.

Explanation:

Step1: Differentiate the equation \(xy = 15\) with respect to \(t\)

Using the product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = x\) and \(v = y\).
We get \(x\frac{dy}{dt}+y\frac{dx}{dt}=0\).

Step2: Find the value of \(y\) when \(x = 3\)

Since \(xy=15\) and \(x = 3\), then \(y=\frac{15}{3}=5\).

Step3: Substitute \(x = 3\), \(y = 5\) and \(\frac{dx}{dt}=6\) into \(x\frac{dy}{dt}+y\frac{dx}{dt}=0\)

We have \(3\frac{dy}{dt}+5\times6 = 0\).

$$3\frac{dy}{dt}+30 = 0$$

$$3\frac{dy}{dt}=- 30$$

\(\frac{dy}{dt}=-10\).

The negative sign of \(\frac{dy}{dt}\) indicates that \(y\) is decreasing. And \(\vert\frac{dy}{dt}\vert = 10\), so \(y\) is decreasing by 10 units per second.

Answer:

A. \(y\) is decreasing by 10 units per second.