Differentiate each vector component
A short change in the position vector
A point moves with position vector \(\mathbf r(t) = \langle t^2,\ t^3\rangle\), the dashed arrow from the origin to the point. This is a vector function: it has one component for each coordinate.
At \(t = 1\), the point is at \((1, 1)\). Set the time step \(h\) to 0.1. The faint dot marks \(\mathbf r(1.1)\), and the line to it is the change \(\Delta\mathbf r = \langle 0.21,\ 0.331\rangle\), with green legs because both parts are positive.
Dividing the change by \(h = 0.1\) makes it ten times longer, so the orange arrow is \(\langle 2.1,\ 3.31\rangle\). This is the average rate of change of the position during the step.
Shrink \(h\). The orange arrow approaches \(\langle 2,\ 3\rangle\), whose parts are the derivatives of \(t^2\) and \(t^3\) at \(t = 1\): \(2t = 2\) and \(3t^2 = 3\).
Each part has its own rate
The position vector \(\mathbf r(t)\) has a horizontal component \(x(t)\) and a vertical component \(y(t)\).
The orange arrow is the derivative \(\mathbf r'(t)\), the limit of \(\frac{\Delta\mathbf r}{h}\) as \(h\) shrinks to 0. It starts at the point and points along the curve.
Its horizontal leg is the limit of \(\frac{\Delta x}{h}\), which is \(x'(t)\), and its vertical leg is the limit of \(\frac{\Delta y}{h}\), which is \(y'(t)\).
Move the point. A leg is green when its rate is positive, and red, pointing left or down, when its rate is negative.
The derivative of a vector function is therefore the vector of the derivatives of its components, in the same order.
Notation
A vector function lists one component function for each coordinate.
\[\mathbf r(t) = \langle x(t),\ y(t)\rangle\]The derivative is the limit of the average rate of change of the position over a step of time \(h\).
\[\mathbf r'(t) = \lim_{h\to 0}\dfrac{\mathbf r(t+h) - \mathbf r(t)}{h}\]The limit of a vector is the vector of the limits of its parts, so you differentiate each component by itself and keep the order.
\[\mathbf r'(t) = \cy{\langle x'(t),\ y'(t)\rangle}\]Each component must be differentiable at \(t\), because each limit must exist.
When it fails
Take \(\mathbf r(t) = \langle t,\ |t|\rangle\) at \(t = 0\). For \(h > 0\), the average rate of change is \(\langle 1,\ 1\rangle\), so the orange arrow points up and to the right. For \(h < 0\), it is \(\langle 1,\ -1\rangle\), so the arrow points down and to the right.
The two sides differ, so the limit \(\mathbf r'(0)\) does not exist. The cause is the second component, because \(|t|\) has no derivative at 0.
Example 1: A polynomial and a cosine
Find \(\mathbf r'(t)\) for \(\mathbf r(t) = \langle 5t - t^3,\ \cos 3t\rangle\).
The first component is a polynomial, so differentiate it term by term.
\[\dfrac{d}{dt}\left(5t - t^3\right) = 5 - 3t^2\]For the second component, the chain rule gives a factor of 3, because the inside function is \(3t\).
\[\dfrac{d}{dt}\cos 3t = -3\sin 3t\]Write the results in the same order. At \(t = 1\), the first part is 2, so its leg is green. The second part is \(-3\sin 3\), which is negative because \(\sin 3 > 0\), so its leg is red.
\[\mathbf r'(t) = \cy{\langle 5 - 3t^2,\ -3\sin 3t\rangle}\]The figure draws the arrow at 0.4 of its length.
Example 2: A product in a component
Find \(\mathbf r'(t)\) for \(\mathbf r(t) = \langle te^{2t},\ \sin 2t\rangle\).
The first component is a product, so it needs the product rule.
\[x(t) = te^{2t}\]A tempting wrong move: Differentiate each factor and multiply.
\[x'(t) = 1\cdot 2e^{2t} = 2e^{2t}\]At \(t = 0\), this gives \(x'(0) = 2\), so it claims the arrow \(\langle 2,\ 2\rangle\), along the red line. The orange arrow for \(h = 0.01\) has a horizontal part of only \(e^{0.02} \approx 1.02\), so it is much steeper than the red line.
Use the product rule: differentiate one factor at a time and add the two terms.
\[\begin{aligned} x'(t) &= 1\cdot e^{2t} + t\cdot 2e^{2t} \\ &= (1 + 2t)e^{2t} \end{aligned}\]By the chain rule, the derivative of \(\sin 2t\) is \(2\cos 2t\). Write both results in order.
\[\mathbf r'(t) = \cy{\langle (1 + 2t)e^{2t},\ 2\cos 2t\rangle}\]At \(t = 0\), this is \(\langle 1,\ 2\rangle\), the orange arrow.
Practice
Find \(\mathbf r'(t)\) for \(\mathbf r(t) = \langle 5t^2e^{-t},\ 2\ln(2t+1)\rangle\), with \(t \ge 0\).
Differentiate each component by itself. On the figure, the point moves left after t = 2, so your first component should be negative there.
Solution
The first component is a product, so use the product rule. The second component needs the chain rule, because \(2t + 1\) is inside the logarithm.
\[\begin{aligned} x'(t) &= 5\left(2te^{-t} - t^2e^{-t}\right) \\ &= 5(2t - t^2)e^{-t} \\ y'(t) &= \dfrac{4}{2t+1} \end{aligned}\]Write the vector with the components in order. The first part is 0 at \(t = 2\) and negative after it, so at \(t = 3\) the horizontal leg is red and points left.
\[\mathbf r'(t) = \cy{\left\langle 5(2t - t^2)e^{-t},\ \dfrac{4}{2t+1}\right\rangle}\]
Questions
Questions 1 and 2 use the curve \(\mathbf r(t) = \langle 3 - t,\ t^2\rangle\). The orange arrow is \(\mathbf r'(t)\), drawn from the point \(\mathbf r(t)\).
The orange arrow is \(\mathbf r'(2)\). Which vector is it?
- \(\langle -1,\ 4\rangle\)
- \(\langle 1,\ 4\rangle\)
- \(\langle 4,\ -1\rangle\)
Answer
\(\langle -1,\ 4\rangle\)
At which time is the orange arrow horizontal?
- \(t = 0\)
- \(t = 2\)
- \(t = 1\)
Answer
\(t = 0\)
Find \(\mathbf r'(t)\) for \(\mathbf r(t) = \langle \sin 4t,\ t\ln t\rangle\), with \(t > 0\).
- \(\langle 4\cos 4t,\ \ln t + 1\rangle\)
- \(\langle \cos 4t,\ \ln t + 1\rangle\)
- \(\langle 4\cos 4t,\ \ln t\rangle\)
- \(\langle \ln t + 1,\ 4\cos 4t\rangle\)
Answer
\(\langle 4\cos 4t,\ \ln t + 1\rangle\)
Summary
You differentiate each component of the vector function by itself and write the results in the same order.
This works because the average rate of change of the position has one part for each component, so its limit does too.
You use it when a position is given as a vector function of time.
This is one of four skills in topic 9.4 of Which & Why: AP Calculus.