Find speed along a parametric path
A short step along the path
A point moves in the plane. At time \(t\) seconds it is at \(x = 3t\) and \(y = 4t - t^2\), in meters. At \(t = 1\) it is at \((3, 3)\), the orange point.
Move the slider to choose a short time step \(\Delta t\). The faint point shows where the point is after that time, and the orange segment joins the two positions.
For \(\Delta t = 0.1\), the point moves 0.3 m to the right and 0.19 m up. These changes are the legs of a right triangle, and both are green because both are positive.
The orange segment is the hypotenuse, so its length is \(\sqrt{0.3^2 + 0.19^2} \approx 0.355\) m. The speed is the length of a short step divided by its time, so the point moves at about 3.55 m/s.
Press the button to shrink the step. The speed approaches \(\sqrt{13} \approx 3.61\) m/s. At \(t = 1\) the rates are \(\frac{dx}{dt} = 3\) and \(\frac{dy}{dt} = 2\), and \(3^2 + 2^2 = 13\).
Two rates make one speed
This figure has no numbers. The horizontal leg is the rate \(\frac{dx}{dt}\), how fast the point moves across. The vertical leg is \(\frac{dy}{dt}\), how fast it moves up or down, and the hypotenuse is the speed.
Move the time slider. A leg is green when its rate is positive and red when its rate is negative. Press the button to reverse the horizontal motion. The horizontal leg turns red and points left, but the hypotenuse keeps its length.
The speed depends on the size of each rate, not on its sign.
Notation
The rates \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are the derivatives of \(x\) and \(y\) with respect to \(t\), and they are the legs of the triangle.
\[\cg{\dfrac{dx}{dt}},\qquad \cg{\dfrac{dy}{dt}}\]In a short time \(\Delta t\), the point moves \(\Delta x\) across and \(\Delta y\) up. Each is about the rate times \(\Delta t\), and the step is the hypotenuse of the triangle with these legs.
\[\begin{aligned} \cg{\Delta x} &\approx \dfrac{dx}{dt}\,\Delta t \\ \cg{\Delta y} &\approx \dfrac{dy}{dt}\,\Delta t \\ \text{step} &\approx \sqrt{\Delta x^2 + \Delta y^2} \end{aligned}\]Substitute \(\Delta x\) and \(\Delta y\) into the step. Each square under the root then has the factor \(\Delta t^2\), which comes out of the root as \(\Delta t\) because \(\Delta t > 0\). Divide by \(\Delta t\) and let \(\Delta t\) shrink, so that the approximations become exact.
\[\cy{\text{speed}} = \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}\]A speed is never negative, because a square root is never negative.
The formula needs both rates to exist at the time \(t\).
When it fails
The path \(x = t\), \(y = |t|\) has a corner at \(t = 0\). For \(t < 0\), \(\frac{dy}{dt} = -1\), and for \(t > 0\), \(\frac{dy}{dt} = 1\).
At \(t = 0\), the two sides give different rates, so \(\frac{dy}{dt}\) does not exist. The formula needs both rates, so it gives no speed at \(t = 0\).
Example 1: A drone
A drone flies so that \(x = t^2\) and \(y = 2t\), with distances in meters and time in seconds. Find its speed at \(t = 3\).
Differentiate each coordinate. The two rates are the legs of the triangle.
\[\dfrac{dx}{dt} = 2t,\qquad \dfrac{dy}{dt} = 2\]At \(t = 3\), the horizontal rate is 6 and the vertical rate is 2.
\[\cg{\dfrac{dx}{dt} = 6},\qquad \cg{\dfrac{dy}{dt} = 2}\]Square each rate and add the squares. Then take the square root.
\[\begin{aligned} \cy{\text{speed}} &= \sqrt{6^2 + 2^2} \\ &= \sqrt{40} = 2\sqrt{10} \end{aligned}\]As a check, \(2\sqrt{10} \approx 6.32\) m/s.
Example 2: A point given by its position
A point moves with \(x = 4t\) and \(y = t^2 - 4t\). Find its speed when it is at \((4, -3)\).
The question gives a point, not a time, so first find the time. Setting \(y = -3\) gives two times.
\[\begin{aligned} t^2 - 4t + 3 &= 0 \\ (t - 1)(t - 3) &= 0 \end{aligned}\]The equation for \(x\) decides. At \(t = 1\), \(x = 4\), which is the given point. At \(t = 3\), \(x = 12\), which is a different point.
\[4t = 4\ \Rightarrow\ t = 1\]At \(t = 1\), the horizontal rate is 4, so the horizontal leg is green. The vertical rate is \(-2\), so the vertical leg is red and points down.
\[\cg{\dfrac{dx}{dt} = 4},\qquad \cr{\dfrac{dy}{dt} = 2t - 4 = -2}\]A tempting wrong move: Add the two rates.
\[\text{speed} = 4 + (-2) = \cr{2}\]The red segment of length 2 is shorter than the horizontal leg of 4. In a right triangle, the hypotenuse is longer than either leg.
Square each rate, because a square is never negative. Add the squares and take the square root.
\[\begin{aligned} \cy{\text{speed}} &= \sqrt{4^2 + (-2)^2} \\ &= \sqrt{20} = 2\sqrt5 \end{aligned}\]As a check, \(2\sqrt5 \approx 4.47\) m/s.
Practice
A point moves with \(x = t^2\) and \(y = \dfrac{t^3}{3} - t\). Find its speed at \(t = 2\).
On the figure, move the time to 2 and read the two legs. The speed is the hypotenuse.
Solution
Differentiate each coordinate. Then substitute \(t = 2\).
\[\begin{aligned} \cg{\dfrac{dx}{dt}} &= \cg{2t = 4} \\ \cg{\dfrac{dy}{dt}} &= \cg{t^2 - 1 = 3} \end{aligned}\]The legs are 4 and 3, so the speed is 5 m/s.
\[\cy{\sqrt{4^2 + 3^2} = 5}\]At any time, the speed is \(t^2 + 1\), because \(4t^2 + (t^2 - 1)^2 = (t^2 + 1)^2\).
Questions
A point moves along \(x = 2t\), \(y = 2t - t^2\). The orange point is at \(t = 1\), and the white point is at \(t = 2\). The legs of each triangle are the two rates at that time.
The orange point, at \(t = 1\), is the highest point of the curve. What is its speed?
- 2
- 0
- 4
Answer
2
At the white point, \(t = 2\), the green leg is 2 and the red leg is 2. What is its speed?
- \(2\sqrt2\)
- 0
- 4
Answer
\(2\sqrt2\)
A point moves with \(x = 2t^3\) and \(y = 3t^2\). What is its speed at \(t = 1\)?
- \(6\sqrt2\)
- 12
- 72
- 6
Answer
\(6\sqrt2\)
Summary
You find the speed by squaring the two rates, adding the squares and taking the square root.
This works because a short step is the hypotenuse of a right triangle, and its length divided by its time is the speed.
You use it when a point moves in the plane and you know \(x(t)\) and \(y(t)\).
This is one of four skills in topic 9.3 of Which & Why: AP Calculus.