curious_iza
- 6
- 0
Hi this is the question that I am unsure about...
Find and then sketch the level curves of the scalar field T(x,y) = (2x+y)/(x^2 -y^2) for;
T = -1
T = -0.5
T = 0
T = 0.5
T = 1
I am unsure about these answers which I got by subbing each of the values into T(x,y)
1) T = -1
After gathering x and y values on one side of the equation and completeing the square I got;
(x + 1)^2 -(y - 0.5)^2 = 3/4 ...
I am unsure about what kind of graph this is because I know that a hyperbola should be equal to 1 and this isn't.
T = -0.5
For this one I got (x + 2)^2 -(y - (0.5))^2 = (15/4)
T = 0
Finally one I could do :)
y = 2x
T=0.5
(x - 2) ^2 - (y +1)^2 = 3
T = 1
(x - 1)^2 - (y - 1/2)^2 = 3/4
I am pretty sure that I got the equations correct but I would appreciate if someone could help me with;
a) Explaining how I draw this since it is not in the form of a hyperbola ie not = 1.
b) What these mean?
Find and then sketch the level curves of the scalar field T(x,y) = (2x+y)/(x^2 -y^2) for;
T = -1
T = -0.5
T = 0
T = 0.5
T = 1
I am unsure about these answers which I got by subbing each of the values into T(x,y)
1) T = -1
After gathering x and y values on one side of the equation and completeing the square I got;
(x + 1)^2 -(y - 0.5)^2 = 3/4 ...
I am unsure about what kind of graph this is because I know that a hyperbola should be equal to 1 and this isn't.
T = -0.5
For this one I got (x + 2)^2 -(y - (0.5))^2 = (15/4)
T = 0
Finally one I could do :)
y = 2x
T=0.5
(x - 2) ^2 - (y +1)^2 = 3
T = 1
(x - 1)^2 - (y - 1/2)^2 = 3/4
I am pretty sure that I got the equations correct but I would appreciate if someone could help me with;
a) Explaining how I draw this since it is not in the form of a hyperbola ie not = 1.
b) What these mean?