Calculus problem help, please?
next, consider curve given by y^2=2+xy
B)find all points (x,y) on the curve where the line tangent to the curve has slope 1/2
C)show there are points (x,y) on the curve where the line tangent to the curve is horizontal
D)let x and y be continuous functions of time t related by the equation y^2=2+xy. At any time t=5, the value of y is 3 and dy/dt=6. Find the value of dx/dt at time t=5.
I have no idea what I'm doing. I missed the last week. All i could do is the derivative, which is y/2y-x
Comments
using implicit derivative
2*y*y'=y+x*y' so y'= y/(2y-x) =1/2 so 2y= 2y-x and x=0 and y =+-sqrt2
y'=0 provides y=0 and 0=2 absurd there are no points with horizontal tangent
if it were verticla tangent x=2y and y^2= 2+2y^2 and y^=-2 impossible
dy/dx=dy/dt*dt/dx so dt/dx= dy/dx/dy/dt dx/dt= dy/dt/dy/dx At t=5 y=3 so 9=2+3x so x= 7/3
dy/dx=3/(6-7/3)=9/11 so dx/dt= 6/(9/11)=66/9=22/3
check my numbers please