5hbc: how to graph implicit relations intro packet!

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 dydx =

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 -12 ,

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 -12 ,

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 dydx 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 dydx in terms of y.

(b) Write an equation for each vertical tangent to the curve.

(c) Find d2ydx2 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!

5hbc: how to graph implicit relations intro packet!

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