Determine the orthogonal trajectories of the hyperbola 3xy = 5

killjoy1990xb9

killjoy1990xb9

Answered question

2021-12-27

Determine the orthogonal trajectories of the hyperbola 3xy=5 at the point (2,56). Sketch the curves.

Answer & Explanation

kalupunangh

kalupunangh

Beginner2021-12-28Added 29 answers

Step 1 
differentiate the equation 3xy=5 with respect to x. 
d(3xy) dx =d(5) dx  
3(x dy  dx +y)=0 
 dy  dx =yx 
Step 2 
Replace  dy  dx  with dx  dy  and bring the x terms with dx and y terms with dy. 
 dy  dx =yx 
 dx  dy =yx 
 dx  dy =yx 
x dx =y dy  
x dx =y dy  
x22+C=y22+C 
x22y22=C 
x2y22=C 
x2y2=C 
To find the value of C, substitute the values of x and y.
x2y2=C 
22(56)2=C 
42536=C 
1442536=C 
C=3.305

Karen Robbins

Karen Robbins

Beginner2021-12-29Added 49 answers

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, 3(x dy  dx +y)=0 
x dx  dy =4 
[ dy  dx =4x] 
Now, (dets change  dy  dx  with dx  dy ): 
So  dx  dy =4x 
 dx  dy =4x 
x dx =y dy  
The orthogonal trajectories are the curves that satisfy a differential equation
We have x dx =y dy  
Integrate it: 
x22+c=422+c (using powerrule xm dx =xn+1n+1+c) 
Now, x22422=c 
x2422=c 
x242=c 
Given point (x,4)=(2,56) 
22(56)2=c 
c=42536 
c=3.305

Vasquez

Vasquez

Expert2022-01-09Added 669 answers

3xy=5 with respect to x.
d(3xy)dx=d(5)dx3(xdydx+y)=0dydx=yxdydx=yxdxdy=yxdxdy=yxxdx=ydyxdx=ydyx22+C=y22+Cx22y22=Cx2y22=Cx2y2=Cx2y2=C22(56)2=C42536=C1442536=CC=3.305

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