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babymother [125]
3 years ago
12

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?

Mathematics
1 answer:
professor190 [17]3 years ago
4 0

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.

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