This document discusses five methods for graphing implicit functions on a TI-83 graphing calculator: 1. Using function mode, programming, and Euler's method to graph solutions to a differential equation defined by the implicit function. 2. Using parametric mode and the quadratic formula to solve the implicit function for x as a parametric function of t. 3. Using function mode, solving for x as a function of y, and using DrawInv to graph the inverse relation. 4. Using function mode and the Solve() command to numerically solve the implicit equation for y as a function of x. 5. Using polar mode by rewriting the implicit equation in terms of r and θ and graphing r

1. Graphing Implicit Functions on the TI-83
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Last year, the AP Calculus exam had a question involving an implicit
function. The question was 98AB6:
Consider the curve defined by 2y3 + 6x2y - 12x2 + 6y = 1.
(a) Show that
dy
dx
=
4x - 2xy
x2 + y2 + 1
.
(b) Write an equation of each horizontal tangent line to the curve.
(c) The line through the origin with slope -1 is tangent to the curve
at point P. Find the x- and y-coordinates of point P.
All this calculus is well and good, but this isn’t the first implicit relation
or function appearing on an AP Calculus exam. The AP Calculus
Committee seems to think this a baffling question. Why, say you? Well,
notice what they don’t ask. This curve is not easy to graph! What if we
could use a graphing calculator to demystify this question? My students
and I wrestled with just this issue last May after the exam arriving at methods
1 and 5. In September, my new AP Calculus BC class tackled this problem
and came up with method 3. Methods 2 and 4 were developed with help
from the ap-calc (www.ets.org), graph-ti (www.ti.com) and calc-reform
(www.ams.org) listservs!
Let’s add a new part (d) to this question.
(new d)
Make a complete sketch of the given implicit function.

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Method 1: Function Mode and Programming (98AB6)
Consider the given differential equation in part (a) above. Could we
use the derivative of the implicit function? Why not use slope fields and/or
Euler’s Method? Setup an appropriate window such as ZOOM4, let
y1=(4X-2XY)/(X2+Y2+1)
using alpha Y and run program slopef.83p. The slope field already suggests
a family of solution curves. Part (c) shows that a point on the curve is


-1
2
,


1
2
,
so run program eulerg.83g with xi=-0.5, yi=0.5 and ∆x=0.1. Run eulerg.83g
again with xi=-0.5, yi=0.5 and ∆x=-0.1. You will get a graph through the slope
field.
Well, this is a satisfying graph as we indeed can tell that this relation
is a function with the line y = -x a likely tangent at


-1
2
,


1
2
. But, we can’t
really trace the curve or use the graphing calculator’s power to analyze it.
I’ll leave to the reader the following exercise: Modify program eulerg.83p
(graphical) to output a table of all the ordered pairs plotted above and
call the new program eulern.83p (numerical).
Of course, this method works best when
dy
dx
is defined for x in your
window.

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Method 2: Parametric Mode (98AB6)
The given relation can be solved for x. Rewriting it as
(6y-12)x2 + (2y3+6y – 1) = 0
the equation is quadratic in x and we may apply the quadratic formula.
x =
-0± 02-4(6y-12)(2y3+6y-1)
2(6y-12)
=
± -4(6y-12)(2y3+6y-1)
12y-24
A graph you can trace can then be obtained in parametric mode as
follows.
x1(t) = -4(6T-12)(2T3+6T-1) /(12T-24) y1(t) = T (pos branch)
x2(t) = -x1(t) y2(t) = T (neg branch)
x3(t) = T y3(t) = -T (y=-x tan line)
This is a much better graph as it can be traced. It is incomplete,
however, as the solution to part (b) is missing. The point of horizontal
tangency (0, 0.165) is not plotted. This is due the fact that our
parametrization is undefined when T=0! Also, the point (-0.5, 0.5) is not on
the curve.

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Method 3: Function Mode and DrawInv (98AB6)
Use the same solution for x = f(y) as in Method 2 and graph it as y =
f(x). In other words let y1 = -4(6X-12)(X3+6X-1) /(12X-24) then use DrawInv
from the draw menu. What y1 is graphing is the inverse relation. If you
deselect y1, then DrawInv y1 and DrawInv –y1 will draw the given relation.
This is, however, another method which does not lend itself to trace
mode.

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Method 4: Function Mode and Solve()
As we are really solving the equation 2y3 + 6x2y – 12x2 + 6y – 1 = 0, let
y1=solve(2Z3 + 6x2Z – 12x2 + 6Z – 1,Z,0.5) and graph in ZOOM4. Be careful to
use alpha Z as X and Y are reserved for home screen output. What this
does is solve (from the catalog menu) for Z for each value of x in the
window using Newton’s Method near Z=0.5 and graphs the results as y
values. 0.5 is a good ‘guess’ for Newton’s Method as we know that y=0.5
is in the range of the relation.
This is by far the best method, if a little slow, for trace mode and the
calc menu as all ordered pairs in the implicit function are graphed for a
given window. We can even verify that the points (-0.5, 0.5) and (0, 0.165)
are part of the function. Also, f ’(0) = 0 and f ’(-0.5) = -1.
Please note, however, that this method works best when graphing a
function.

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Method 5: Polar Mode (94AB3)
What if the relation is quadratic in both x and y? We can then
discuss conic sections (see activity sheets). Graphing is also easy with a
change in coordinate system from cartesian to polar.
Consider the curve defined by x2 + xy + y2 = 27.
(a) Write an expression for the slope of the curve at any point (x, y).
(b) Determine whether the line tangent to the curve at the x-
intercepts of the curve are parallel. Show the analysis that
leads to your conclusion.
(c) Find the points on the curve where the lines tangent to the
curve are vertical.
(new d)
Make a complete sketch of x2 + xy + y2 = 27 including the
tangent lines found in parts (b) and (c).
Let x=rcos(θ) and y=rsin(θ), substitute, solve for r as an explicit function
of θ and let r1=f(θ) in polar mode and ZOOM6. Part (b) above is especially
easy to confirm with trace mode and the calc menu!
r2cos2(θ) + rcos(θ)rsin(θ) + r2sin2(θ) = 27
r2(cos2(θ) + cos(θ)sin(θ) + sin2(θ) = 27
r2(cos(θ)sin(θ) + 1) = 27
r2 =
27
cos(θ)sin(θ) + 1
r =
27
cos(θ)sin(θ) + 1

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You pick the Method: (92AB4)
Consider the curve defined by the equation y + cos y = x + 1
for 0 ≤≤≤≤ y ≤≤≤≤ 2ππππ.
(a) Find
dy
dx
in terms of y.
(b) Write an equation for each vertical tangent to the curve.
(c) Find
d2y
dx2 in terms of y.
(new d)
Make a complete sketch of y + cos y = x + 1 including the
vertical tangent.
This one is very easy to solve for x, so DrawInv or Parametric Mode
come to mind. What do you think? Use your graph to confirm your results
in parts (a) and (c). Where is the function increasing or decreasing?
Where is the function concave up or down?

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Activity Sheets: Exploring Implicit Functions and Conic Sections
In the attached activity sheets, we explore the world of conic
sections using questions similar to 94AB3. A conic section can be written
implicitly as:
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.
We will use Function Mode and Polar Mode methods here. By
letting some of these coefficients vary while others are zero, the student
can get a feel for what each coefficient affects in the shape of the conic.
When B=0, the student will see the effect of D and E by completing
the square and solving for y as an explicit function in x.
By letting D=E=0 and varying B, the student will learn to predict the
behavior of the implicit graph based on the value of B2 – 4AC in polar
mode.
We will investigate what happens when A and C are of equal
value, unequal value, same sign and differing sign.
All conic sections will be generated from circles and ellipses to
parabolas and hyperbolas to even the degenerate black sheep of the
family!