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

Using the product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = x\) and \(v = y\). So \(\frac{d}{dt}(xy)=\frac{dx}{dt}y+x\frac{dy}{dt}\). Since \(\frac{d}{dt}(15) = 0\), we have \(y\frac{dx}{dt}+x\frac{dy}{dt}=0\).

Step2: Substitute \(x = 3\) into \(xy = 15\) to find \(y\)

When \(x = 3\), then \(3y=15\), so \(y = 5\).

Step3: Assume \(\frac{dx}{dt}=1\) (we can assume a non - zero value for \(x\)’s rate of change to find the relationship, since the problem is about the relative rate)

Substitute \(x = 3\), \(y = 5\) and \(\frac{dx}{dt}=1\) into \(y\frac{dx}{dt}+x\frac{dy}{dt}=0\). We get \(5\times1+3\times\frac{dy}{dt}=0\).

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

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The negative sign indicates that \(y\) is decreasing. The magnitude of \(\frac{dy}{dt}\) is \(5\).

Answer:

C. \(y\) is decreasing by 5 units per second.