How to find the unit tangent and unit normal vectors

The tangent tells us which way we’re moving along a curve. The principal normal tells us which way that tangent is turning. We’ll find both vectors on a parabola and check their lengths and directions.

Divide each vector by its magnitude

Differentiate the position vector $r (t)$ component by component. The derivative points in the direction of motion. If $r^{'} (t) \neq 0$ , divide it by its magnitude to get the unit tangent $T (t)$ . If $T^{'} (t) \neq 0$ , normalize that derivative to get the principal unit normal $N (t)$ .

\begin{array}{l} T (t) = \frac{r^{'} (t)}{‖ r^{'} (t) ‖} \\ N (t) = \frac{T^{'} (t)}{‖ T^{'} (t) ‖} \end{array}

Because $T (t) \cdot T (t) = 1$ , differentiating gives $2 T (t) \cdot T^{'} (t) = 0$ . That’s why the normal built from $T^{'} (t)$ is perpendicular to the tangent. When $T^{'} (t) = 0$ , this formula doesn’t define a principal unit normal.

Find the unit tangent for the curve

Let $r (t) = ⟨ t, t^{2}, 2 ⟩$ . The derivative is $r^{'} (t) = ⟨ 1, 2 t, 0 ⟩$ , and its length is $\sqrt{1 + 4 t^{2}}$ .

T (t) = \frac{⟨ 1, 2 t, 0 ⟩}{\sqrt{1 + 4 t^{2}}}

Differentiate before substituting the parameter

We want $N (1)$ , but we need the function $T (t)$ to take the derivative. In the first component, differentiating ${(1 + 4 t^{2})}^{- 1 / 2}$ gives $- 4 t {(1 + 4 t^{2})}^{- 3 / 2}$ . In the second component, the product rule gives $2 {(1 + 4 t^{2})}^{- 1 / 2} - 8 t^{2} {(1 + 4 t^{2})}^{- 3 / 2}$ , which simplifies to $2 {(1 + 4 t^{2})}^{- 3 / 2}$ .

\begin{array}{l} T^{'} (t) = \frac{⟨ - 4 t, 2, 0 ⟩}{{(1 + 4 t^{2})}^{3 / 2}} \\ T^{'} (1) = \frac{⟨ - 4, 2, 0 ⟩}{5 \sqrt{5}} \end{array}

The length of $T^{'} (1)$ is $\frac{\sqrt{16 + 4}}{5 \sqrt{5}} = \frac{2}{5}$ . Dividing by that length gives

N (1) = \frac{⟨ - 2, 1, 0 ⟩}{\sqrt{5}}

The curve lies in the plane z = 2. The equal-length arrows show T(1) and N(1) in that plane.

Check the length and the dot product

The numerator of $N (1)$ has length $\sqrt{5}$ , so $N (1)$ has length $1$ . The tangent at $t = 1$ is $\frac{⟨ 1, 2, 0 ⟩}{\sqrt{5}}$ . Their dot product is $\frac{- 2 + 2}{5} = 0$ .

See how these directions show up in the tangential and normal components of acceleration.

Learn mathKrista KingMarch 16, 2021math, learn online, online course, online math, calculus iii, calculus 3, calc 3, calc iii, vector calculus, vector calc, unit tangent vector, unit normal vector, tangent vector, normal vector

How to find the unit tangent and unit normal vectors

Divide each vector by its magnitude

Find the unit tangent for the curve

Differentiate before substituting the parameter

Check the length and the dot product

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