killjoy1990xb9

2021-12-27

Determine the orthogonal trajectories of the hyperbola $3xy=5$ at the point $\left(2,\frac{5}{6}\right)$. Sketch the curves.

kalupunangh

Step 1
differentiate the equation $3xy=5$ with respect to x.

Step 2
Replace  and bring the x terms with dx and y terms with dy.

$\frac{{x}^{2}}{2}+C=\frac{{y}^{2}}{2}+C$
$\frac{{x}^{2}}{2}-\frac{{y}^{2}}{2}=C$
$\frac{{x}^{2}-{y}^{2}}{2}=C$
${x}^{2}-{y}^{2}=C$
To find the value of C, substitute the values of x and y.
${x}^{2}-{y}^{2}=C$
${2}^{2}-{\left(\frac{5}{6}\right)}^{2}=C$
$4-\frac{25}{36}=C$
$\frac{144-25}{36}=C$
$C=3.305$

Karen Robbins

We have to determine orthogonal trajectories of the Hyperbola $3xy=5$ at the Point (2, 5/6) and also sketch the curves
The given equation will be differentiated w.r.t. x.

So,

Now, (dets change ):
So

The orthogonal trajectories are the curves that satisfy a differential equation
We have
Integrate it:
$\frac{{x}^{2}}{2}+c=\frac{{4}^{2}}{2}+c$ (using powerrule
Now, $\frac{{x}^{2}}{2}-\frac{{4}^{2}}{2}=c$
$\frac{{x}^{2}-{4}^{2}}{2}=c$
${x}^{2}-{4}^{2}=c$
Given point $\left(x,4\right)=\left(2,\frac{5}{6}\right)$
${2}^{2}-{\left(\frac{5}{6}\right)}^{2}=c$
$c=4-\frac{25}{36}$
$c=3.305$

Vasquez

$3xy=5$ with respect to x.
$\begin{array}{}\frac{d\left(3xy\right)}{dx}=\frac{d\left(5\right)}{dx}\\ 3\left(x\frac{dy}{dx}+y\right)=0\\ \frac{dy}{dx}=\frac{-y}{x}\\ \frac{dy}{dx}=\frac{-y}{x}\\ -\frac{dx}{dy}=-\frac{y}{x}\\ \frac{dx}{dy}=\frac{y}{x}\\ xdx=ydy\\ xdx=ydy\\ \frac{{x}^{2}}{2}+C=\frac{{y}^{2}}{2}+C\\ \frac{{x}^{2}}{2}-\frac{{y}^{2}}{2}=C\\ \frac{{x}^{2}-{y}^{2}}{2}=C\\ {x}^{2}-{y}^{2}=C\\ {x}^{2}-{y}^{2}=C\\ {2}^{2}-\left(\frac{5}{6}{\right)}^{2}=C\\ 4-\frac{25}{36}=C\\ \frac{144-25}{36}=C\\ C=3.305\end{array}$

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