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Related Rates Problems · Tutorial 670 of 1495

Choosing Coordinates for Moving-Point Problems

Choose coordinates that expose the geometry of moving points, build the distance equation, and identify when it is valid to differentiate.

Intermediate 10 min read

What You'll Learn

  • Choose origins and axes to match the paths or fixed constraints in a motion problem.
  • Represent moving points with coordinate functions and distinguish signed position from distance.
  • Build distance equations for collinear, perpendicular, and parallel-path motion.
  • Derive the plane-distance rate equation from the Pythagorean theorem.
  • Apply the rate equation only when the points are separated, and handle coincidence separately.

Why Coordinate Choice Matters

A moving-point problem can look complicated when described only in words: two objects move, their paths are offset, and the question concerns how their separation changes. Coordinates turn that description into equations. A useful choice of axes can make the geometry visible before any differentiation is done.

The goal here is to set up positions and distances carefully. A coordinate is a signed position measured from an origin; it is not automatically a distance. In particular, a point may have a negative coordinate, and the distance between two points is nonnegative. Keeping that distinction clear prevents sign errors in rates.

For a moving point in the plane, write its position as \((x(t),y(t))\), where the coordinates may vary with time \(t\). The derivatives \(x'(t)\) and \(y'(t)\) are signed coordinate rates. Their signs record direction: a negative coordinate rate means the coordinate is decreasing.

Coordinate setup: Choose axes that match the geometry, express each point’s position using coordinate functions, and write the distance relation from those positions. Do this before differentiating.

Choosing Axes and Writing Positions

A good coordinate system usually puts a simple feature of the motion on an axis. If a point travels along a straight horizontal path, let that path be the \(x\)-axis. If two paths meet at a right angle, place one path on each coordinate axis. If two paths are parallel, put them in the same direction and record their fixed perpendicular offset in the other coordinate.

The origin can be placed wherever it makes the positions simplest: at an intersection, a fixed landmark, or one of the points’ starting references. The origin does not have to be the point whose distance is being studied. Once the axes and origin are chosen, use the same coordinate system for every point throughout the calculation.

For two points \(P=(x_1,y_1)\) and \(Q=(x_2,y_2)\), their horizontal and vertical coordinate differences are \(x_2-x_1\) and \(y_2-y_1\). The distance formula gives their separation \(r\):

$$ r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. $$

This formula contains the geometry of the setup. A fixed coordinate can be substituted into it immediately when it is truly constant, but time-varying coordinates must remain as functions of time until the relation has been differentiated.

Worked Example: Two Objects on One Straight Track

Worked Example: Two Objects on One Straight Track

Object \(A\) and object \(B\) move along the same straight track. Choose the track as the \(x\)-axis and let \(x(t)\) and \(z(t)\) be their signed positions. At a particular instant, \(x=120\text{ m}\), \(z=170\text{ m}\), \(x'=18\text{ m/s}\), and \(z'=12\text{ m/s}\). Find the rate of change of their separation at that instant.

First write the signed difference from \(A\) to \(B\):

$$ q=z-x. $$

The actual distance is \(r=|z-x|\). At this instant, \(z-x=170-120=50\text{ m}>0\), so the points are separated and \(r=z-x\) in a neighborhood of this instant. Differentiate that relation:

$$ \frac{dr}{dt}=z'-x'. $$

Substitute the given rates:

$$ \frac{dr}{dt}=12\text{ m/s}-18\text{ m/s}=-6\text{ m/s}. $$

The negative sign means the separation is decreasing. Check the calculation by substituting the rate into the differentiated relation: \(z'-x'=12-18=-6\text{ m/s}\), as obtained. The positions also give \(r=170-120=50\text{ m}>0\), which confirms that using \(r=z-x\), rather than \(r=x-z\), was correct at this instant.

For collinear motion, it is especially important not to confuse a signed coordinate difference with a distance. If the order of the points reverses, the absolute value changes its expression. At a time when the points are apart, the sign of the coordinate difference determines which expression for \(r\) applies.

Perpendicular Paths

If two paths are perpendicular, place their intersection at the origin and align each path with one axis. For example, one point on the positive \(x\)-axis has position \(P=(x,0)\), and another on the positive \(y\)-axis has position \(Q=(0,y)\). Their coordinate differences are \(-x\) and \(y\), so the Pythagorean theorem gives \(r^2=x^2+y^2\). The signs of the differences disappear when squared.

Worked Example: Two Walkers Moving Away from a Corner

Two walkers start at a right-angle corner and move along perpendicular paths. Let their distances from the corner be \(x(t)\) and \(y(t)\), so their positions are \(P=(x,0)\) and \(Q=(0,y)\). At an instant, \(x=6\text{ m}\), \(y=8\text{ m}\), \(x'=2\text{ m/s}\), and \(y'=3\text{ m/s}\). Find the rate at which the distance between them is changing.

The distance equation follows from the perpendicular coordinate differences:

$$ r^2=x^2+y^2. $$

At the specified instant, the separation is \(r=\sqrt{6^2+8^2}=\sqrt{36+64}=10\text{ m}\). Since \(r>0\), differentiate the equation with respect to time:

$$ 2r\frac{dr}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}. $$

Divide by \(2\), then substitute the values and rates:

$$ r\frac{dr}{dt} =x\frac{dx}{dt}+y\frac{dy}{dt} =(6\text{ m})(2\text{ m/s})+(8\text{ m})(3\text{ m/s}) =12\text{ m}^2/\text{s}+24\text{ m}^2/\text{s} =36\text{ m}^2/\text{s}. $$

Using \(r=10\text{ m}\), solve for the separation rate:

$$ \frac{dr}{dt} =\frac{36\text{ m}^2/\text{s}}{10\text{ m}} =3.6\text{ m/s}. $$

Check by substituting back into the differentiated equation: \(r\,dr/dt=(10\text{ m})(3.6\text{ m/s})=36\text{ m}^2/\text{s}\), which equals \(x x'+y y'=36\text{ m}^2/\text{s}\). The positive rate agrees with the geometry: both walkers are moving away from the corner, so their separation is increasing at this instant.

Parallel Paths in the Plane

When two points move along parallel paths, choose the \(x\)-axis in the direction of those paths. If their paths are a fixed perpendicular distance \(h\) apart, their positions can be written \(P=(x,0)\) and \(Q=(u,h)\). The horizontal separation is \(u-x\), while the vertical separation is always \(h\). Thus their distance satisfies \(r^2=(u-x)^2+h^2\).

Worked Example: Two Boats on Parallel Channels

Two boats move along parallel straight channels that are \(3\text{ m}\) apart. Choose the direction of travel as the \(x\)-axis and write their positions as \(P=(x,0)\) and \(Q=(u,3\text{ m})\). At an instant, \(x=4\text{ m}\), \(u=9\text{ m}\), \(x'=5\text{ m/s}\), and \(u'=2\text{ m/s}\). Find the instantaneous rate of change of their separation.

The coordinate differences give the distance relation:

$$ r^2=(u-x)^2+(3\text{ m})^2. $$

At this instant, \(u-x=9\text{ m}-4\text{ m}=5\text{ m}\). Hence \(r^2=(5\text{ m})^2+(3\text{ m})^2=25\text{ m}^2+9\text{ m}^2=34\text{ m}^2\), so \(r=\sqrt{34}\text{ m}>0\).

Differentiate the distance relation. The offset \(3\text{ m}\) is constant, so its derivative is zero:

$$ 2r\frac{dr}{dt} =2(u-x)\left(\frac{du}{dt}-\frac{dx}{dt}\right). $$

Substitute the values and rates:

$$ 2(\sqrt{34}\text{ m})\frac{dr}{dt} =2(5\text{ m})\bigl(2\text{ m/s}-5\text{ m/s}\bigr) =-30\text{ m}^2/\text{s}. $$

Therefore,

$$ \frac{dr}{dt} =\frac{-30\text{ m}^2/\text{s}}{2\sqrt{34}\text{ m}} =-\frac{15}{\sqrt{34}}\text{ m/s}. $$

To check, substitute into the differentiated equation: \(2r\,dr/dt=2(\sqrt{34}\text{ m})(-15/\sqrt{34}\text{ m/s})=-30\text{ m}^2/\text{s}\). The right side is \(2(5\text{ m})(2-5)\text{ m/s}=-30\text{ m}^2/\text{s}\), so the equation is satisfied. The negative rate means the separation is decreasing: the first boat is gaining on the second in the horizontal direction, while the vertical offset stays fixed.

Why the Plane Distance Equation Gives the Rate Relation

The rate equation is not a separate geometric rule to memorize. It follows by differentiating the distance equation. The conditions matter: at a time when the points are separated, \(r>0\). If the coordinate functions are differentiable, the squared distance is differentiable, and the positive square root defining \(r\) is differentiable there as well. The chain rule can therefore be applied at such a time.

Derivation: Let \(P(t)=(x_1(t),y_1(t))\) and \(Q(t)=(x_2(t),y_2(t))\) have differentiable coordinate functions. Define \(\Delta x=x_2-x_1\) and \(\Delta y=y_2-y_1\). At any time when \(r>0\), the distance relation is \(r^2=(\Delta x)^2+(\Delta y)^2\). Differentiating both sides gives \(2r r'=2\Delta x\,\Delta x'+2\Delta y\,\Delta y'\). Dividing by \(2r\), which is permitted because \(r>0\), yields \(r'=(\Delta x\,\Delta x'+\Delta y\,\Delta y')/r\). Since \(\Delta x'=x_2'-x_1'\) and \(\Delta y'=y_2'-y_1'\), this expresses the change in separation using coordinate differences and their rates.

This derivation justifies the common form

$$ r\frac{dr}{dt} =(x_2-x_1)(x_2'-x_1') +(y_2-y_1)(y_2'-y_1'), \qquad r>0. $$

For special geometries, some terms are zero or combine into a simpler expression. For instance, the perpendicular-path equation follows by setting \(P=(x,0)\) and \(Q=(0,y)\); the parallel-path equation follows by setting \(P=(x,0)\) and \(Q=(u,h)\) with constant \(h\).

Coincidence Requires Separate Care

At a time when the points coincide, \(r=0\), so the derivation above cannot be used: dividing by \(r\) is not allowed. More importantly, differentiability of the coordinate functions alone does not guarantee that the distance itself is differentiable at coincidence. For example, if \(P(t)=(t,0)\) and \(Q(t)=(0,0)\), then both coordinate functions are differentiable, but the distance is \(r(t)=|t|\), which has no derivative at \(t=0\).

At coincidence, inspect the original distance function and determine whether its derivative exists. If it does exist, the rate can be found from that function. The displayed rate equation applies at coincidence only if \(r\) is differentiable there; even then, its undivided form has \(r\,r'=0\) on the left and does not, by itself, determine \(r'\). The useful divided equation is specifically the equation for \(r>0\).

A Common Coordinate Mistake

A frequent error is to use a distance where a signed coordinate difference is required. In the straight-track example, the coordinate difference is \(z-x\), but the distance is \(|z-x|\). If \(z-x>0\), then \(r=z-x\) and \(r'=z'-x'\). If \(z-x<0\), then \(r=x-z\) and \(r'=x'-z'\). Writing \(r=z-x\) without checking the order can produce a negative “distance,” or a rate with the wrong sign.

Correction: For collinear points, first write \(r=|z-x|\). At a time when \(z\ne x\), determine the sign of \(z-x\), choose the corresponding expression for \(r\), and then differentiate that expression locally.

Another setup error is to treat a coordinate as constant merely because it is not changing in one direction. In the parallel-channel example, the vertical offset is constant, but the horizontal difference \(u-x\) changes. The full distance equation keeps both components visible, making it clear which one can be treated as fixed.

A Reliable Setup Sequence

1
Match axes to the paths.
Put straight paths, their intersection, or their shared direction on coordinate axes when this simplifies the geometry.
2
Assign position functions.
Represent each moving point by its signed coordinates, and mark which coordinates are constant and which depend on time.
3
Write the distance relation.
Use the coordinate differences and the distance formula; use absolute value for a collinear distance.
4
Check the instant of interest.
Determine whether the points are separated. The differentiated distance-rate equation derived above requires \(r>0\).
5
Differentiate the relation.
Use the chain rule on the time-dependent quantities, then substitute the values and rates for the specified instant.

These choices matter because the coordinate equation records the actual geometry of the motion. Once that equation is correct, differentiation translates the changing positions into a changing separation. A carefully chosen origin and axes do not alter the physical situation; they make its constraints easier to express and its signs easier to interpret.

Check Your Understanding

Use coordinate differences and the distance relation to explain each answer.

  1. Two points on a line have coordinates \(a(t)\) and \(b(t)\). What expression gives their distance, and why must its sign be checked before differentiating?
  2. How would you place the axes for two paths that meet at a right angle?
  3. Two parallel paths are a fixed distance \(h\) apart. If the points on them have coordinates \((x,0)\) and \((u,h)\), write their distance equation.
  4. State the condition under which the derived plane-distance rate equation can be divided by \(r\).
  5. Why does differentiability of the coordinate functions not guarantee differentiability of the distance at a time when the points coincide?

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