A small surveying marker follows a stipulated path on a flat floor. Its horizontal coordinate is x(t)=t²−2t and its vertical coordinate is y(t)=t+1, for 0≤t≤3. Here t is the numerical time in seconds, and x and y are the numerical coordinates in meters. Negative x means left of the chosen origin; it does not mean negative distance traveled. The coordinate formulas are a synthetic model, not measurements of a physical device.
Before drawing anything, fill the two missing coordinates at t=1. A position needs both values at the same time. The notation r(t)=(x(t),y(t)) packages that simultaneous pair; it does not add a third spatial dimension.
Time t (seconds)
x(t) (meters)
y(t) (meters)
Simultaneous position r(t) (meters)
0
0
1
(0,1)
1
−1
2
(−1,2)
2
0
3
(0,3)
3
3
4
(3,4)
Plot these paired positions with equal lengths for one meter horizontally and vertically. Join them using the stated formulas, not an arbitrary straight-segment interpolation. Draw arrows in increasing time order. The path first moves left while rising, then right while continuing to rise. A graph of x against time tells a different story: its horizontal axis is the clock, not horizontal location on the floor.
Predict
Compare t=0 with t=2. Both positions have x=0, but their y coordinates differ. Predict whether the line joining them is horizontal, vertical or neither. Then predict whether a point could leave a position and return to it while still having traveled along a nonzero-length path.
Now compare t=1 with t=3. Write the elapsed time separately from the two coordinate changes. The average horizontal change is measured in meters per second; the slope of a line drawn on the floor is measured in meters per meter. These are not interchangeable quantities. Predict the two component averages before using the display, and record what would make their ratio undefined.
Experiment
One time, three coordinated graphs
Predict the selected point and the interval's average component changes before revealing the calculation. Time is in seconds; x and y are in meters. The path uses equal x/y scales. Component panels plot time horizontally, not spatial position.
Selected simultaneous coordinates: r(0)=(0, 1) meters. The point is shared by all panels. The open circle and square on the path mark interval start and end; the dashed line joins them. Arrows show increasing time. Curves use 129 samples for drawing; the displayed formulas define the model exactly.
Move the selected time through 0, 1, 2 and 3. In each panel, locate the same selected time’s information: one point on the floor, one x value above the time axis, and one y value above the time axis. The red marker does not animate; you control its position. Read the numerical equivalent below the graphs if a curve or marker is difficult to distinguish.
Keep the quadratic path and select interval start 0 and end 2. Reveal the average component changes. Then change the interval to [1,3] and predict again. The interval endpoints are marked separately from the selected-time point. Moving that point does not change the interval calculation; changing an interval control hides the old calculation. Choosing equal or reversed times requests a different problem and is refused rather than silently swapped.
Finally select the circle and set the interval to [0,4]. This second model is x=cos(πt/2), y=sin(πt/2), with coordinates in meters and t in seconds. Inspect intermediate times before drawing a conclusion from its equal endpoints. The interval [1,3] instead produces distinct endpoints on a vertical line. Exact quarter-turn values and trigonometric parameter-symmetry identities are used for these zero tests; finite drawing samples do not establish a general zero or a theorem about the full path.
Understand
For the first path, complete the square: x(t)=(t−1)²−1. On [0,3], this gives minimum x=−1 at t=1. The square is largest at t=3, the endpoint farthest from 1, giving maximum x=3. Meanwhile y=t+1 increases from 1 to 4. The horizontal minimum and vertical minimum occur at different times; combining them into (−1,1) would invent a position. A spatial extreme is not a claim about travel speed.
For any ordered times a<b, define Δt=b−a, Δx=x(b)−x(a) and Δy=y(b)−y(a). Average component changes are Δx/Δt and Δy/Δt. On [0,2], displacement is (0,2) meters over 2 seconds, so the averages are (0,1) meters per second. The distinct endpoints determine a vertical secant: Δy/Δx is undefined because its denominator is zero. This says nothing about instantaneous velocity.
On [1,3], displacement is (4,2) meters over 2 seconds. The average components are (2,1) meters per second, whose ratio is 1/2. Directly, the Cartesian chord has slope 2/4=1/2. Elapsed time cancels in this ratio when Δx is nonzero. For a complete circle, however, both displacements vanish. The expression 0/0 determines no line direction at all. It is not the vertical-line case. Neither zero displacement nor its average implies zero traveled distance.
Elimination needs restrictions as well as algebra. From y=t+1, recover t=y−1 and obtain x=(y−2)²−1 with 1≤y≤4. Every permitted t produces a point satisfying this equation and interval. Conversely, take any point satisfying the equation with 1≤y≤4. Choosing t=y−1 puts t in [0,3] and reproduces both coordinates. These two implications prove equality of the reached set and the restricted Cartesian description. The unrestricted parabola includes (8,5), which requires t=4 and is not reached here.
Another branch, with a different restriction
For x=t² and y=t with t≥0, the reached set is x=y² with y≥0. Substitution proves one inclusion. Conversely, for any point with x=y² and y≥0, choose t=y to recover the original formulas. The point (1,−1) satisfies x=y² but fails y≥0. Replacing the pair by y=±√x would incorrectly add a second branch. This proof uses the parameter domain rather than the apparent shape of a finite plot.
Cover both hyperbola branches, not just a few points
A parametrization must give every requested point as well as only valid points. For (x−h)²/a²−(y−k)²/b²=1 with a,b>0, set x=h+a sec t and y=k+b tan t, on 0≤t<2π excluding π/2 and 3π/2. The identity sec²t−tan²t=1 proves all outputs lie on the curve. For the reverse direction, put X=(x−h)/a and Y=(y−k)/b. Since X²−Y²=1, X≠0 and (1/X,Y/X) lies on the unit circle. Its angle t gives cos t=1/X and sin t=Y/X, so sec t=X and tan t=Y. Both branches are therefore reached. Omitting the angles with negative cosine would lose the left branch.
For the vertical hyperbola (y−k)²/b²−(x−h)²/a²=1, interchange the roles: x=h+a tan t, y=k+b sec t with the same exclusions. For a parabola or any explicit graph y=f(x), the simple parametrization (t,f(t)) uses the actual domain of f. If f is one-to-one, (f(t),t) parametrizes its inverse graph on that same parameter domain; the first coordinate now ranges over the original range. It need not be the time variable of a physical motion.
A supplied velocity is not a supplied position
A problem may give v(t) directly, without asking you to derive it from position. For example v(t)=(2−t,t+1) meters per second on 0≤t≤3 specifies rightward motion before t=2, zero horizontal motion at t=2, and leftward motion afterward. Vertical motion is always upward. At t=1, velocity is (1,2), speed is √5 meters per second, and the same-direction unit vector is (1/√5,2/√5), equivalently (1/√5)i+(2/√5)j. A nonzero vector w is normalized by w/||w||; this has magnitude one because scalar multiplication scales magnitude by the scalar’s absolute value. The zero vector has no direction to normalize.
The magnitude of a position vector is distance from the origin; the magnitude of a velocity vector is speed. Substituting a time into a supplied velocity formula gives velocity at that time, not the particle’s coordinates. Without an initial position, translating the entire motion by any constant vector leaves its velocity unchanged, so position cannot be determined. No differentiation or integration is needed for these interpretations. The changed task below uses v(t)=(3−t,2t−2), includes component-zero cases, and separately asks for a function/inverse parametrization to ensure the two meanings are not confused.
Read co-variation from an implicit curve
The equation y²=x, restricted to 0≤x≤4, is not one function of x: it consists of y=√x and y=−√x, meeting at the origin. On the upper branch, if 0<x₁<x₂≤4, then √x₂−√x₁=(x₂−x₁)/(√x₂+√x₁)>0, so Δy/Δx is positive. On the lower branch it is negative. This relates changes on a branch without differentiating or assuming the implicit equation has a globally single-valued output.
On a vertical segment, distinct points have Δx=0 and Δy≠0: Δx/Δy is zero and Δy/Δx is undefined. On a horizontal segment the roles reverse. Calling both quotients zero would erase the difference between a fixed coordinate and an invalid division. A curved graph’s finite nearby secant is an interval rate, not automatically an exact instantaneous one.
An axis crossing similarly requires choosing the correct component. For p(t)=(t−1,t²−4), 0≤t≤3, x(t)=0 at t=1 gives the y-axis point (0,−3); y(t)=0 at the allowed t=2 gives the x-axis point (1,0). The other algebraic root −2 is outside the parameter domain. Constructing the branches or listing exact coordinate-and-curve descriptions is valid graphical evidence; testing a few arbitrary points is not a full implicit locus.
Apply
Without changing the tool, investigate a fresh model on paper: u(s)=(s²−4s,2s+1), for 0≤s≤4 seconds. Construct simultaneous positions at s=0,1,2,3,4. Draw a spatial path and two separate component graphs. Locate the horizontal extreme and calculate average component changes over [1,4]. Explain why no finite endpoint calculation alone supplies the traveled distance along a general curve.
Worked feedback for the changed model
The positions are (0,1),(−3,3),(−4,5),(−3,7),(0,9) meters. Completing the square gives x=(s−2)²−4, so horizontal minimum −4 occurs at s=2; maximum 0 occurs at both endpoints. Vertical position increases from 1 to 9. From s=1 to s=4, displacement (3,6) over 3 seconds gives average components (1,2) meters per second and Cartesian secant slope 2. That slope does not measure meters traveled per second. Eliminating s gives x=((y−5)/2)²−4, with 1≤y≤9; the inverse choice s=(y−1)/2 proves the reverse inclusion.
In the practice questions, another pair of component graphs is stipulated by straight segments and their full vertex lists. That is an exact piecewise-linear model, so interpolation along its segments is justified. Sparse measurements of an unknown trajectory would not grant the same guarantee. Likewise a repeated circle equation describes a locus, while the order and number of traversals come from its parameter and time interval.
Continue
Keep your paired table, three sketches and interval calculation together. A correct Cartesian equation without its reached branch is incomplete evidence; a correct displacement without units or a time interval is also incomplete. The written tasks below include changed formulas and require your constructions and explanations, not reproduction of the displayed numbers.
Return to parametric curves and restrictions for a short recap, or explore vectors and dot products when you want more ways to compare directions. These are optional routes, not prerequisites for using this investigation. This Unit 4 bridge is course enrichment; it is not part of the AP Precalculus examination weighting.
Apply · explain · revise
Practice and reasoning
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For r(t)=(2t²−4t,3−t), 0≤t≤3 seconds, construct a simultaneous table at t=0,1,2,3 and sketch the directed Cartesian path with meter axes. State r(3/2) in vector notation.
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Two component graphs are defined exactly by joined straight segments: x(t) joins (0,2),(1,−1),(3,3),(4,0); y(t) joins (0,0),(2,4),(4,2), on 0≤t≤4 seconds. Coordinates are meters. Draw both time graphs, identify all global extrema with their times, and recover the simultaneous spatial position at t=1.
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For x=t²−t and y=2t+1 on 0≤t≤4 seconds, calculate average component changes on [1/2,5/2]. Give the Cartesian secant slope and explain why neither that slope nor the magnitude of average displacement per second is automatically average travel speed.
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A path has x=2cos(πt/2), y=2sin(πt/2), 0≤t≤4 seconds. Compare the secants on [1,3] and [0,4]. Compute average component changes and diagnose vertical versus 0/0. Can either case justify saying the point never moved?
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Let x=t²+2 and y=3−t for 1≤t≤3. Eliminate t and prove exactly which Cartesian points are reached, including the reverse implication. Exhibit one point on the unrestricted equation that is not reached.
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Compare P(t)=(cos t,sin t), Q(t)=(cos(−t),sin(−t)), and R(t)=(cos(2t),sin(2t)), each for 0≤t≤2π. Show that they have the same spatial locus but different traversal information. Use ordered landmarks; do not differentiate.
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Parametrize the entire hyperbola (x−1)²/9−(y+2)²/4=1 using secant and tangent. State excluded parameter values in 0≤t<2π and prove both that your points lie on the hyperbola and that every point is reached. Give one parametrization for the corresponding vertical hyperbola (y+2)²/4−(x−1)²/9=1.
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A model directly supplies velocity v(t)=(3−t,2t−2) m/s on 0≤t≤4 s; it does not supply position. Find every interval of right/left/up/down motion and all times when a component is zero. Compute speed and a same-direction unit vector at t=2. Can these data alone give the particle position at t=2? Separately parametrize the graph of f(x)=√(x−1) and its inverse using the same parameter, giving the parameter domain.
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Construct the full restricted implicit locus y²=x for 0≤x≤4 as a graph or exact two-branch description. Using pairs with 0<x1<x2≤4, determine the sign of Δy/Δx on each branch without derivatives. Compare with the vertical segment x=2,−1≤y≤1 and horizontal segment y=2,0≤x≤4: identify which rate is zero and which quotient is undefined. For p(t)=(t−1,t²−4),0≤t≤3, find every axis crossing, naming which component must vanish.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
