AP Calculus BC glossary

Vector magnitude

Also called: Magnitude of a vector, Length of a vector

The magnitude of a vector is its length, computed as the square root of the sum of the squares of its components. The magnitude of a velocity vector is the speed of the particle, so it is a single non-negative number, and integrating it over a time interval gives total distance traveled.

\left|\langle a,\, b \rangle\right| = \sqrt{a^{2} + b^{2}}, \qquad \left|\mathbf{v}(t)\right| = \sqrt{\left(x'(t)\right)^{2} + \left(y'(t)\right)^{2}}

Magnitude keeps size and discards direction, so the result is a scalar and never negative. A particle in the plane is at rest exactly when the magnitude of its velocity is zero, and a sum of squares vanishes only when both components vanish, so both $x'(t)$ and $y'(t)$ must be zero at the same instant.

Collapsing a vector to one number is what makes comparison possible. A question asking for the maximum speed on an interval wants the largest value of $\sqrt{\left(x'(t)\right)^{2} + \left(y'(t)\right)^{2}}$ , not the largest component, and a fast $x'(t)$ paired with a fast negative $y'(t)$ can beat both.

The mistake

Reading the magnitude of the acceleration vector as the rate at which speed changes. They are different numbers. For $x(t) = \cos t$ and $y(t) = \sin t$ the velocity is $\langle -\sin t,\, \cos t \rangle$ , so the speed is $1$ at every time and its rate of change is $0$ , while the acceleration $\langle -\cos t,\, -\sin t \rangle$ has magnitude $1$ .

Appears in: Unit 9: Parametric, Polar, and Vector (BC)