Calculus with Vector Functions

Main idea

You differentiate and integrate a vector function one component at a time, so there's nothing new to learn mechanically. The payoff is the meaning: $\vec r\,'(t)$ points along the curve, the way you're heading, and for motion it's the velocity.

{\htmlData{sym=0}{\vec r\,'(t)}} = \langle f'(t),\; g'(t),\; h'(t)\rangle, \qquad \int\vec r(t)\,dt = \left\langle\int f,\; \int g,\; \int h\right\rangle + {\htmlData{sym=1}{\vec C}}

What the symbols mean

$\vec r\,'(t)$
Tangent vector
Points along the curve, in the direction of increasing t.
$\vec C$
Constant vector
One constant per component.

Explanation

Calculus with vector functions is simply the ordinary derivatives and integrals you already know, applied to each coordinate of a moving point. It answers the question "at this instant, which way is the object moving, and how fast?" The formula $\vec r\,'(t) = \langle f'(t), g'(t), h'(t)\rangle$ looks fancy, but it's three ordinary derivatives you already know how to take: $f'(t)$ (how fast you're moving east), $g'(t)$ (how fast north) and $h'(t)$ (how fast up). Put them together and you get one arrow pointing where you're headed. Integrals work the same way, one piece at a time.

Why does the derivative point along the curve? Go back to the definition: $\vec r\,'(t)$ is the limit of $\frac{\vec r(t + h) - \vec r(t)}{h}$ . The top is the arrow from where you are now to where you'll be a moment later, a short chord of the curve. As $h$ shrinks, that chord lines up with the curve itself, just like secant lines turning into tangent lines in Calculus I. Its length, divided by the time $h$ , becomes your speed.

Tangent lines. That gives the tangent line for free. At $t = a$ it passes through $\vec r(a)$ in the direction $\vec r\,'(a)$ , which is the point-plus-direction form of a line from earlier in this unit.

Product rules carry over to dot and cross products, with the same "differentiate one, leave the other, then swap" pattern:

(\vec u\cdot\vec v)' = \vec u\,'\cdot\vec v + \vec u\cdot\vec v\,', \qquad (\vec u\times\vec v)' = \vec u\,'\times\vec v + \vec u\times\vec v\,'.

For the cross product, keep the order of each pair as written, since swapping the two vectors flips the sign.

Constant length means perpendicular. Here's a nice use of the dot product rule. Suppose $\|\vec r(t)\|$ never changes, like a point moving on a sphere. Then $\vec r\cdot\vec r = c$ , and differentiating gives $2\,\vec r\cdot\vec r\,' = 0$ . So the velocity is always perpendicular to the position. It makes sense: to stay the same distance from the center, you can never move toward it or away from it, only sideways. We'll lean on this fact in the next lesson.

See it

A point climbing the helix r(t) = ⟨cos t, sin t, 0.3t⟩, with its velocity r′(t) = ⟨−sin t, cos t, 0.3⟩ drawn from the point. As you can see, the velocity always lies along the curve, pointing where the point is about to go. Its length never changes, √1.09 ≈ 1.04, because the helix is climbed at a steady speed.

Worked examples

Example 1

Find the tangent line to $\vec r(t) = \langle t^2, \sin t, e^{2t}\rangle$ at $t = 0$ .

1
The line passes through $\vec r(0) = \langle 0, 0, 1\rangle$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

For $\vec r(t) = \langle t^3, \cos t, e^t\rangle$ , find $\vec r\,'(0)$ .

Problem 2+5 XP

For $\vec r(t) = \langle 2t, t^2, \ln t\rangle$ , find $\vec r\,'(1)$ .

Problem 3+5 XP

Find $\displaystyle\int_0^1\langle 2t, 3t^2, 1\rangle\,dt$ .

Problem 4+5 XP

Find a direction vector for the tangent line to $\langle t, t^2, t^3\rangle$ at $t = 1$ .

Problem 5+5 XP

If $\|\vec r(t)\| = 5$ for all $t$ , what is $\vec r(t)\cdot\vec r\,'(t)$ ?

Problem 6+5 XP

A roller coaster car's position is $\vec r(t) = \left\langle 20\cos\frac t5, 20\sin\frac t5, 10 + 5\sin\frac t2\right\rangle$ (meters and seconds). How fast is it going at $t = 0$ ?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

The curves $\vec r_1(t) = \langle t, t^2, t^3\rangle$ and $\vec r_2(s) = \langle\sin s, \sin2s, s\rangle$ cross at the origin, at $t = s = 0$ . At what angle do they cross, in degrees?

Where it's used

Physics engines step forward in time by adding $\vec r\,'(t)\,\Delta t$ to each position; integration of acceleration gives velocity, and of velocity gives position, component by component.

Vector Functions Tangent, Normal and Binormal Vectors