Implicit Differentiation

Equations involving x and y define curves in the xy-plane. For example, the graph of the equation

x²

3²

y²

4²

= 1 is the set of all points (s, t) in the plane such that substituting s for x and t for y in the equation produces a true statement. In the applet, make a = 3, b =4, and c = 0 to see what this curve looks like. By varying the parameters, we can see a family of curves, all of which go through the four points (±a, 0) and (0, ±b). Given a curve, how can we find the slope of a line tangent to a point on the curve?

Explore

Circle. Make a = 5, b = 5, and c = 0. Notice that the point (4,3) is on the circle. What is the slope of the line tangent to the circle at (4,3)?
Ellipse. Make a = 4, b = 4, and c = −0.1 (you can fine-tune c by clicking the dot on the slider then using the left and right arrow keys).
1. What is the slope of the line through (0, 4)? (The applet shows the answer to be
  4
  5
  , but use implicit differentiation to determine whether or not this is the exact answer or an approximation.)
2. Find the two points where the tangent line is horizontal.
Parallel(?) Lines. The graph of the equation x² − xy +
1
4
y² = 1 appears to be a pair of parallel lines. See this by setting a = 1, b = 2, and c = −1. Can you prove that these lines are really parallel, or show that they are not?

Challenge

Use implicit differentiation to find the slope of the curve

x²

a²

+ cxy +

y²

b²

= 1 at any point (x, y) on the curve. Keep a, b, and c as parameters, so you'll treat them like numbers as you differentiate.

Enter your answer, in terms of a, b, c, x, and y:
When you click submit, you'll see the value of your expression in the bottom left corner of the applet. The x and y values are the x and y values of the point on the curve. See how the value of your expression compares to the value of the slope computed by the applet.