Evaluate vector derivatives at an instant
One time, two vectors
A point has position \(\mathbf r(t) = \langle 2t^2,\ t^3 - 3t\rangle\). Its velocity is \(\mathbf r'(t) = \langle 4t,\ 3t^2 - 3\rangle\).
Choose \(t = 2\) and put it into \(\mathbf r\). The point is at \((8, 2)\), because \(2\cdot 2^2 = 8\) and \(2^3 - 3\cdot 2 = 2\).
Put the same \(t = 2\) into \(\mathbf r'\). The velocity is \(\langle 8,\ 9\rangle\), because \(4\cdot 2 = 8\) and \(3\cdot 2^2 - 3 = 9\).
The dashed arrow from the origin ends at this point. The orange arrow \(\langle 8,\ 9\rangle\) starts there and shows how the point moves at the instant \(t = 2\).
Drag the time slider. The point and the arrow change together, because both use the same \(t\).
One value of t for every part
Drag the slider. It sets one moment \(t_0\), and the point and both parts of the arrow move together.
The point is \(\big(x(t_0),\ y(t_0)\big)\), and the arrow starts there.
The arrow is \(\langle x'(t_0),\ y'(t_0)\rangle\), so its parts show how fast the point moves horizontally and vertically.
Every part uses the same \(t_0\). If the parts come from different times, the arrow describes no moment of the motion.
Notation
Differentiate first. Find the formula for \(\mathbf r'(t)\).
\[\mathbf r'(t) = \langle x'(t),\ y'(t)\rangle\]If you put a number into \(\mathbf r\) first, the result is a vector of constants, and its derivative is \(\langle 0,\ 0\rangle\).
Then put the moment \(t_0\) into every component of \(\mathbf r'\).
\[\mathbf r'(t_0) = \cy{\langle x'(t_0),\ y'(t_0)\rangle}\]Put the same \(t_0\) into \(\mathbf r\) to find where the arrow starts.
\[\mathbf r(t_0) = \big(x(t_0),\ y(t_0)\big)\]Both components of \(\mathbf r'\) must exist at \(t_0\). If \(\mathbf r'(t_0) = \langle 0,\ 0\rangle\), the value is still correct, but the arrow shows no direction.
When it fails
For \(\mathbf r(t) = \langle t^3,\ t^2\rangle\), the velocity is \(\langle 3t^2,\ 2t\rangle\). At \(t = 0\), it is \(\langle 0,\ 0\rangle\).
The arrow has length 0, so it shows no direction. The point stops for an instant, and the path has a sharp point at the origin.
Example 1: A velocity that points up
A point has position \(\mathbf r(t) = \langle 4\sin 2t,\ t^2\rangle\). Find \(\mathbf r'\!\left(\frac{\pi}{4}\right)\) and the point where the arrow starts.
Differentiate each component. The first one needs the chain rule.
\[\mathbf r'(t) = \langle 8\cos 2t,\ 2t\rangle\]Put \(t = \frac{\pi}{4}\) into both components. The horizontal part is 0, so the arrow points straight up.
\[\mathbf r'\!\left(\tfrac{\pi}{4}\right) = \cy{\left\langle 8\cos\tfrac{\pi}{2},\ \tfrac{\pi}{2}\right\rangle = \left\langle 0,\ \tfrac{\pi}{2}\right\rangle}\]The figure draws the arrow at half its length.
Put the same \(t\) into \(\mathbf r\) to find where the arrow starts.
\[\mathbf r\!\left(\tfrac{\pi}{4}\right) = \left(4,\ \tfrac{\pi^2}{16}\right)\]
Example 2: Substitute at the right time
A point has position \(\mathbf r(t) = \langle t^2 - 4t,\ \frac{6}{t}\rangle\), with \(t > 0\). Find its velocity at \(t = 3\) and the point where the arrow starts.
Put \(t = 3\) into \(\mathbf r\) to find the point.
\[\mathbf r(3) = \langle 9 - 12,\ 2\rangle = \langle -3,\ 2\rangle\]A tempting wrong move: Differentiate \(\langle -3,\ 2\rangle\). It is a constant vector, so the derivative is zero.
\[\mathbf r'(3) = \langle 0,\ 0\rangle\]The position at any one instant is a constant vector, so this method gives \(\langle 0,\ 0\rangle\) at every time, as if the point never moved. The faint points show it at \((-4, 3)\) at \(t = 2\) and at \((0, 1.5)\) at \(t = 4\).
Differentiate first, as a function of \(t\).
\[\mathbf r'(t) = \left\langle 2t - 4,\ -\dfrac{6}{t^2}\right\rangle\]Then put \(t = 3\) into both components. The arrow starts at \((-3, 2)\), and it points to the right and slightly down.
\[\mathbf r'(3) = \cy{\left\langle 2,\ -\dfrac{2}{3}\right\rangle}\]
Practice
A point has position \(\mathbf r(t) = \langle 2t^3 - 3t,\ \sin\pi t\rangle\). Find \(\mathbf r'\!\left(\frac12\right)\) and the point where the arrow starts.
On the figure, set the time to 1/2. The arrow starts at the point and points to the left.
Solution
Differentiate each component.
\[\mathbf r'(t) = \langle 6t^2 - 3,\ \pi\cos\pi t\rangle\]Put \(t = \frac12\) into both components.
\[\begin{aligned} \mathbf r'\!\left(\tfrac12\right) &= \left\langle \tfrac32 - 3,\ \pi\cos\tfrac{\pi}{2}\right\rangle \\ &= \cy{\left\langle -\tfrac32,\ 0\right\rangle} \end{aligned}\]The figure draws the arrow at 0.3 of its length.
Put the same \(t\) into \(\mathbf r\) to find where the arrow starts.
\[\mathbf r\!\left(\tfrac12\right) = \left\langle \tfrac14 - \tfrac32,\ 1\right\rangle = \left\langle -\tfrac54,\ 1\right\rangle\]
Questions
Questions 1 and 2 use the curve \(\mathbf r(t) = \langle 4 - t^2,\ 2t\rangle\). The orange arrow is the velocity at the chosen time, drawn from the point. You can drag the point.
The orange arrow is \(\mathbf r'(2)\). Which vector is it?
- \(\langle -4,\ 2\rangle\)
- \(\langle 0,\ 4\rangle\)
- \(\langle -2,\ 2\rangle\)
Answer
\(\langle -4,\ 2\rangle\)
At which time does the orange arrow point straight up?
- \(t = 0\)
- \(t = 2\)
- \(t = 1\)
Answer
\(t = 0\)
A point has position \(\mathbf r(t) = \langle \ln(t^2 + 1),\ 3t^2\rangle\). What is \(\mathbf r'(1)\)?
- \(\langle 1,\ 6\rangle\)
- \(\langle \ln 2,\ 3\rangle\)
- \(\langle 2,\ 6\rangle\)
- \(\langle 1,\ 3\rangle\)
Answer
\(\langle 1,\ 6\rangle\)
Summary
To find a velocity at an instant, you differentiate each component first and then put the same value of \(t\) into every component.
This works because one moment gives both the point and the arrow that starts at it.
You use it when a question asks for the velocity or acceleration at a given time.
This is one of four skills in topic 9.4 of Which & Why: AP Calculus.