How to find tangential and normal components of acceleration

Acceleration can change your speed or your direction of motion. The video finds those two components for a helix. We’ll use the same steps with a polynomial position vector.

Separate the change in speed from the change in direction

Let $v = r^{'} (t)$ be velocity and $a = r^{''} (t)$ be acceleration. When the speed $‖ v ‖$ is nonzero, the unit tangent $T$ points in the direction of motion. The tangential component $a_{T}$ tells us how quickly speed changes. It can be negative when the object slows down.

The tangential component lies along the path. The normal component points toward the bend.

The normal component $a_{N}$ is the magnitude of the acceleration perpendicular to $T$ . It measures the part that turns the velocity vector. When that component is nonzero, it points along the principal unit normal $N$ .

\begin{array}{l} a_{T} = \frac{v \cdot a}{‖ v ‖} \\ a_{N} = \frac{‖ v \times a ‖}{‖ v ‖} \end{array}

The dot product divided by speed projects $a$ onto the unit tangent. The cross product has magnitude equal to the speed times the magnitude of the perpendicular component. That’s why we divide both expressions by $‖ v ‖$ .

Differentiate the position vector twice

r (t) = ⟨ 2 t^{2}, 4 t, 3 t^{3} ⟩

Differentiate each coordinate once for velocity and a second time for acceleration.

\begin{array}{l} v = ⟨ 4 t, 4, 9 t^{2} ⟩ \\ a = ⟨ 4, 0, 18 t ⟩ \end{array}

The second coordinate of $v$ is always $4$ , so the speed is never zero. Compute the dot product and the speed.

\begin{array}{l} v \cdot a = 16 t + 162 t^{3} \\ ‖ v ‖ = \sqrt{81 t^{4} + 16 t^{2} + 16} \end{array}

Find the cross product and its magnitude

For the middle coordinate of $v \times a$ , subtract $(4 t) (18 t)$ from $(9 t^{2}) (4)$ . That gives $- 36 t^{2}$ . The first and third coordinates are $72 t$ and $- 16$ .

\begin{array}{l} v \times a = ⟨ 72 t, - 36 t^{2}, - 16 ⟩ \\ = 4 ⟨ 18 t, - 9 t^{2}, - 4 ⟩ \\ ‖ v \times a ‖ = 4 \sqrt{81 t^{4} + 324 t^{2} + 16} \end{array}

Substitute into the component formulas

\begin{array}{l} a_{T} = \frac{162 t^{3} + 16 t}{\sqrt{81 t^{4} + 16 t^{2} + 16}} \\ a_{N} = \frac{4 \sqrt{81 t^{4} + 324 t^{2} + 16}}{\sqrt{81 t^{4} + 16 t^{2} + 16}} \end{array}

At $t = 0$ , the dot product is zero and the normal component is $4$ . So at that instant the acceleration turns the velocity vector without changing its speed.

How to find tangential and normal components of acceleration

Separate the change in speed from the change in direction

Differentiate the position vector twice

Find the cross product and its magnitude

Substitute into the component formulas

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