implicit differentiation...?

Question

1. Consider the expression
(x^2)/9 + (y^2)/4 = 1

whose graph is an ellipse. We wish to find the slopes of the tangent lines to the graph of this ellipse at the point x = 1.

a) Solve the above expression for the variable y. How many solutions do you obtain? Graph each of these solutions separately. What do you notice when these graphs are viewed together on the same set of coordinate axes?
b) While the ellipse itself is not the graph of a function (why?), each of the solutions for y you obtained in (a) does represent a function (why?). In order to obtain an answer to our original question regarding the slopes of the tangent lines at x = 1, describe what you would do with the functions obtained in (a), and carry out this operation.

Answer 1

Hello

A) In order to graph we need to solve for y.

(x^2)/9 + (y^2)/4 = 1
(y^2)/4 = 1 - (x^2)/9
(y^2) = 4 - 4(x^2)/9
y = +- square root (4 - 4(x^2)/9)

When we graph we see that sqr rt +(4 - 4(x^2)/9) is the top portion of our ellipse and - sqr rt(4 - 4(x^2)/9) is the bottom portion of our ellipse.

B) In order to be a function there has to be 1 y value with only 1 x value. Meaning a graph with pts (1,2) & (1,8) is not a function.

If we look at y = sqr rt(4 - 4(x^2)/9) is a function since each x value corresponds to 1 y value.

If we look at y = -sqr rt(4 - 4(x^2)/9) is a function since each x value corresponds to 1 y value.

But together the equation is not a function: example, we have the pts (0,2) and (0,-2). Here we have 2 y values for the 1 x value. Thus would fail the vertical line test. Not a function.

C). Lets go ahead and find the slope at x=1. So lets Implicit Differentiate this.

Here is the way I teach my students (the Saxon Method).

d = (x^2)/9 + (y^2)/4 = 1
d' = (2x/9)dx + (1y/2) dy = 0
Since we are wanting to find dy - we divide everything by dx.
d' = 2x/9 + y/2 dy/dx = 0
Now lets solve for dy/dx.
dy/dx = (-2x/9)(2/y)
dy/dx = -4x/9y

So you want the tangent line at x = 1, and y = +- sqrt rt 32/9

Note*: Using +-sqr rt(4 - 4(x^2)/9) -- lets plug in x=1 and we get y= +- sqrt rt 32/9. And this is about +-1.8856181.

So lets plug in the points (1,1.8856181) and (1,-1.8856181) into dy/dx = -4x/9y

Thus the we get dy/dx = -.2357022 & .2357022604.

So the 2 slopes at x=1 is -.2357022 & .2357022604. We have 2 values here since our an ellipse is not a function.

Hope this helps

Answer 2

2x/9+y*y´/2 = 0 so y´=4x/9y
given x= 1 we have 1/9+y^2/4 = 1 so y^2 = 32/9 and y =+-4/3(2)^1/2
For each x we get 2 values of y so this is not a function
slopes =+-1/3(2)^1/2= +-1/6 sqrt(2)so the tangents are symmetrical with respect to the x axis
The equation of each tangent is a linear function.