Question

a particle moves on the hyperbola ( xy = 15 ) for time ( tgeq0 ) 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: Find the value of \(y\)

Given \(xy = 15\) and \(x = 3\), substitute \(x\) into the equation: \(3y=15\), so \(y = 5\).

Step2: Differentiate \(xy = 15\) with respect to \(t\)

Using the product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u = x\) and \(v = y\).
Differentiating \(xy\) with respect to \(t\) gives \(\frac{d(xy)}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}\).
Since \(\frac{d(15)}{dt}=0\), the equation becomes \(x\frac{dy}{dt}+y\frac{dx}{dt}=0\).

Step3: Substitute the known values

We know \(x = 3\), \(y = 5\), and \(\frac{dx}{dt}=6\).
Substitute these into \(x\frac{dy}{dt}+y\frac{dx}{dt}=0\): \(3\frac{dy}{dt}+5\times6 = 0\).
Simplify the equation: \(3\frac{dy}{dt}+30 = 0\).
Then \(3\frac{dy}{dt}=- 30\), and \(\frac{dy}{dt}=-10\).

Answer:

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