(B) You have two simultaneous equations in two variables. The original equation, and the derivative for specified values of dy/dx.
When dy/dx=1/2 :
1/2 = y/(2y-x)
Rearrange:
2y - x = 2y
Hence: x = 0
Substitute back into y2 = 2 + xy:
Therefore: y = ±√2
Answer: (0, -√2), and (0, +√2)
Consider the curve given by y^2 = 2+xy
(a) show that dy/dx= y/(2y-x)
(b) Find all points (x,y) on the curve where the line tangent to the curve has slope 1/2.
(c) Show that there are now points (x,y) on the curve where the line tangent to the curve is horizontal.
(d) Let x and y be functions of time t that are related by the equation y^2 = 2 + xy. At time t=5, the value of y is 3 and dy/dt=6. Find the value of dx/dt at time t=5.
I was able to do part A, that was easy, but then with the rest of the parts I didn't know where to start. I know I should plug in 1/2 for the slope and solve but I have two variables of x and y. How should I go about solving these problems?
2 answers
(0, -√2), and (0, +√2)
1 answer