Posted by **Mathew** on Wednesday, October 2, 2013 at 9:36pm.

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?

- Calculus -
**Graham**, Wednesday, October 2, 2013 at 10:39pm
(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 y^{2} = 2 + xy:

Therefore: y = ±√2

Answer: (0, -√2), and (0, +√2)

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