bengaltiger14 said:dy/dx is just 2x+2y. So it does not equal correct?
bengaltiger14 said:Are you saying to solve for dy/dx in the first equation?
bengaltiger14 said:2y(dy/dx) = -2x
divide by 2y:
dy/dx = -2x/2y
bengaltiger14 said:dy/dx = -xy
That is the answer to the problem? So the DE is not a solution.
bengaltiger14 said:yeah, your right... dy/dx = -xy/y^2
But the top y would just cancel and the y^2 would become y again so why do that?
bengaltiger14 said:y^2 = 1-x^2
bengaltiger14 said:dy/dx = -xy/1-x^2
So they do equal. If you multiply by -1, it gives the DE.
The equation x^2 + y^2 = 1 is used to describe a circle with a radius of 1 centered at the origin on a graph. It is also known as the unit circle and is often used in trigonometry and geometry.
To verify if a point (x,y) is on the graph of x^2 + y^2 = 1, simply substitute the values of x and y into the equation. If the equation holds true, then the point is on the graph. If the equation does not hold true, then the point is not on the graph.
The possible solutions for x^2 + y^2 = 1 are an infinite number of points that lie on the circle described by the equation. This includes points with integer coordinates, decimal coordinates, and even irrational coordinates.
Yes, the equation x^2 + y^2 = 1 can be solved for either x or y by isolating the variable and taking the square root of both sides. However, it is important to note that there will be two possible solutions for each variable due to the nature of the equation representing a circle.
The equation x^2 + y^2 = 1 is a special case of the Pythagorean theorem, where the length of the hypotenuse of a right triangle is equal to 1. This can be seen by rearranging the terms in the equation to x^2 + y^2 = c^2, where c represents the length of the hypotenuse. This relationship is often used in geometry and physics to solve for unknown lengths or distances.