Question

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?

Answer 1

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.

Answer 2

The value of $x$ at the instant when particle moves on the hyperbola $xy=18$ from time $t\geq 0\;\rm seconds$ is $3$.

Differentiating the given equation  $xy=18$ with respect to the $'t'$ to get an addition equation by  implicit differentiation,

$x\dfrac{dy}{dt}+y\dfrac{dx}{dt}=0----(i)$

Also, from the given equation, we can write $x=\dfrac{18}{y}$

Substitute the value of $x$ in the equation (i), we get

$\begin{aligned}\dfrac{18}{y}\dfrac{dy}{dt}+y\dfrac{dx}{dt}&=0\\\dfrac{dx}{dt}&=-\dfrac{-18}{y^2}\dfrac{dy}{dt}\end{aligned}$

Substitute the value of $y=6$ and $\dfrac{dy}{dt}=8$ in the above equation, we get

$\begin{aligned}\dfrac{dx}{dt}&=-\dfrac{-18}{y^2}\dfrac{dy}{dt}\\&=\dfrac{-18}{6^2}\times 8\\&=-4\end{aligned}$

Now, from equation (i), we can re-write the expression to evaluate the value of $x$ at that instant as-

$x=-y\dfrac{\frac{dx}{dt}}{\frac{dy}{dt}}\\x=-6(\dfrac{-4}{8})\\x=3$

Hence, the value of $x$ at the instant when particle moves on the hyperbola $xy=18$ from time $t\geq 0\;\rm seconds$ is $3$.

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