first, we must find 2z(dz/dt)=2x(dx/dt) + 2y(2y/dt) simplifying, we get: dy/dt=???(dx/dt + dy/dt)
What is that simplified?
dividing by 2, we have
zz' = xx'+yy'
y' = (zz'-xx')/y
not sure where you're going with this
the problem was: 2z dz/dt= 2x dx/dt + 2y dy/dt
dz/dt= ??? ( x dx/dt + y dy/dt)
still confused
c'mon this is algebra I, after doing the derivatives. Just divide both sides by 2z:
dz/dt = (x dx/dt + y dy/dt)/z
To simplify the given equation, let's break it down step by step.
We start with the equation: 2z(dz/dt) = 2x(dx/dt) + 2y(2y/dt)
First, distribute the coefficients: 2z(dz/dt) = 2x(dx/dt) + 4y(dy/dt)
Next, we want to isolate the term with the derivative of y, which is dy/dt. Move the term with dy/dt to one side of the equation: 2z(dz/dt) - 4y(dy/dt) = 2x(dx/dt)
Now, we factor out dy/dt on the left side of the equation: dy/dt(2z - 4y) = 2x(dx/dt)
Lastly, divide both sides by (2z - 4y) to solve for dy/dt: dy/dt = 2x(dx/dt) / (2z - 4y)
So the simplified form of the equation is dy/dt = 2x(dx/dt) / (2z - 4y).