Differentiation of Dot and Cross Products of Vector Functions; Tangential and Normal Components of Acceleration

Vector Calculus in Mechanics

Differentiating the Dot Product of Vector Functions

The dot product of two vector functions can be differentiated using a product rule analogous to that for scalar functions. This is essential in physics for analyzing quantities like work, power, and projections of vectors.

• Product Rule for Scalars: For scalar functions $f(t)$ and $g(t)$, the derivative is: $$ \frac{d}{dt}[f(t)g(t)] = \frac{df}{dt}g(t) + f(t)\frac{dg}{dt}$$

• Dot Product of Vector Functions: For $\mathbf{f}(t) = \langle f_1(t), f_2(t), f_3(t) \rangle$ and $\mathbf{g}(t) = \langle g_1(t), g_2(t), g_3(t) \rangle$, $$ \mathbf{f}(t) \cdot \mathbf{g}(t) = f_1(t)g_1(t) + f_2(t)g_2(t) + f_3(t)g_3(t)$$

• Product Rule for Dot Product: $$ \frac{d}{dt}[\mathbf{f}(t) \cdot \mathbf{g}(t)] = \frac{d\mathbf{f}}{dt} \cdot \mathbf{g}(t) + \mathbf{f}(t) \cdot \frac{d\mathbf{g}}{dt}$$

• Example: Given $\mathbf{f}(t) = \langle t-2, t^3+4, t-6 \rangle$, $\mathbf{g}(t) = \langle t^2-5, t+3, t^5+1 \rangle$, their derivatives, and applying the rule yields:

• $

Differentiating the Magnitude of a Vector Function

The magnitude of a vector function $\mathbf{f}(t)$ is $\|\mathbf{f}(t)\| = \sqrt{\mathbf{f}(t) \cdot \mathbf{f}(t)}$. Differentiating this requires the chain rule and the product rule for dot products.

• Key Formula: $$ \frac{d}{dt}\|\mathbf{f}(t)\| = \frac{1}{\|\mathbf{f}(t)\|} \left[ \mathbf{f}(t) \cdot \frac{d\mathbf{f}}{dt} \right]$$

• Example: For $\mathbf{f}(t) = \langle t^2, 1-t, \sqrt{t} \rangle$,

  • $\frac{d\mathbf{f}}{dt} = \langle 2t, -1, \frac{1}{2\sqrt{t}} \rangle$

  • $\|\mathbf{f}(t)\| = \sqrt{t^4 + (1-t)^2 + t}$

  • Thus, $$ \frac{d}{dt}\|\mathbf{f}(t)\| = \frac{4t^3 + 2t - 1}{2\sqrt{t^4 + t^2 - t + 1}}$$

Differentiating a Unit Vector Function

A unit vector function $\hat{\mathbf{f}}(t)$ is defined as $\hat{\mathbf{f}}(t) = \frac{\mathbf{f}(t)}{\|\mathbf{f}(t)\|}$ for $\mathbf{f}(t) \neq 0$. The derivative of a unit vector is always perpendicular to the unit vector itself.

• Orthogonality: $$ \hat{\mathbf{f}}(t) \cdot \frac{d\hat{\mathbf{f}}}{dt} = 0$$This means$\frac{d\hat{\mathbf{f}}}{dt}$is perpendicular to$\hat{\mathbf{f}}(t)$.

• Example: For $\mathbf{f}(t) = \langle \sin t, \cos t, 1 \rangle$, $\|\mathbf{f}(t)\| = \sqrt{2}$, so $\hat{\mathbf{f}}(t) = \langle \frac{\sin t}{\sqrt{2}}, \frac{\cos t}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle$ and $\frac{d\hat{\mathbf{f}}}{dt} = \langle \frac{\cos t}{\sqrt{2}}, -\frac{\sin t}{\sqrt{2}}, 0 \rangle$.

• Magnitude of Derivative: The derivative of a unit vector is not necessarily a unit vector. Normalizing it gives another unit vector perpendicular to the original.

Differentiating the Cross Product of Vector Functions

Product Rule for Cross Product

The cross product of two vector functions can also be differentiated using a product rule:

• Key Formula: $$ \frac{d}{dt}[\mathbf{f}(t) \times \mathbf{g}(t)] = \frac{d\mathbf{f}}{dt} \times \mathbf{g}(t) + \mathbf{f}(t) \times \frac{d\mathbf{g}}{dt}$$

• Special Case: For $\mathbf{f}(t) \times \frac{d\mathbf{f}}{dt}$, the first term vanishes because the cross product of any vector with itself is zero: $$ \frac{d}{dt}[\mathbf{f}(t) \times \frac{d\mathbf{f}}{dt}] = \mathbf{f}(t) \times \frac{d^2\mathbf{f}}{dt^2}$$

• Example: For $\mathbf{f}(t) = \langle 2t, 3t^2, t^3 \rangle$, $\mathbf{g}(t) = \langle 4t^2, 2t, 1 \rangle$,

  • $\frac{d\mathbf{f}}{dt} = \langle 2, 6t, 3t^2 \rangle$

  • $\frac{d\mathbf{g}}{dt} = \langle 8t, 2, 0 \rangle$

  • Applying the rule yields: $$ \frac{d}{dt}[\mathbf{f}(t) \times \mathbf{g}(t)] = \langle 6t - 8t^3, 20t^4 - 2, 8t - 48t^3 \rangle$$

Tangential and Normal Components of Acceleration

Decomposition of Acceleration

For a particle moving along a path, its acceleration can be decomposed into tangential and normal components. This is crucial for understanding motion along curved paths.

• Unit Tangent Vector: $\hat{\mathbf{T}}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|}$

• Velocity Decomposition: $\mathbf{v}(t) = \|\mathbf{v}(t)\| \hat{\mathbf{T}}(t)$

• Acceleration Decomposition: $$ \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = [a_T(t)]\hat{\mathbf{T}}(t) + [a_N(t)]\hat{\mathbf{N}}(t)$$ where:

  • $a_T(t) = \frac{\mathbf{v}(t) \cdot \mathbf{a}(t)}{\|\mathbf{v}(t)\|}$ (tangential component)

  • $a_N(t) = \frac{\|\mathbf{v}(t) \times \mathbf{a}(t)\|}{\|\mathbf{v}(t)\|}$ (normal component)

  • $\hat{\mathbf{N}}(t) = \frac{d\hat{\mathbf{T}}/dt}{\|d\hat{\mathbf{T}}/dt\|}$ (unit normal vector)

• Physical Meaning:

  • Tangential acceleration changes the speed along the path.

  • Normal acceleration changes the direction of the velocity (curvature of the path).

Example: Parabolic Motion

• Position Vector: $\mathbf{r}(t) = \langle t, \frac{1}{2}t^2, 0 \rangle$

• Velocity: $\mathbf{v}(t) = \langle 1, t, 0 \rangle$

• Acceleration: $\mathbf{a}(t) = \langle 0, 1, 0 \rangle$

• Unit Tangent Vector: $\hat{\mathbf{T}}(t) = \frac{\langle 1, t, 0 \rangle}{\sqrt{1 + t^2}}$

• At $t=1$ s:

  • $\mathbf{a}(1) = \langle 0, 1, 0 \rangle = \frac{1}{\sqrt{2}}\hat{\mathbf{T}}(1) + \frac{1}{\sqrt{2}}\hat{\mathbf{N}}(1)$

  • $\hat{\mathbf{T}}(1) = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \rangle$

  • $\hat{\mathbf{N}}(1) = \langle -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \rangle$

• Path Equation: Eliminating $t$ gives $y = \frac{1}{2}x^2$, a parabola in the $xy$-plane.

Summary Table: Tangential and Normal Components

Component	Formula	Physical Meaning
Tangential ($a_T$)	$a_T = \frac{\mathbf{v} \cdot \mathbf{a}}{\|\mathbf{v}\|}$	Rate of change of speed along the path
Normal ($a_N$)	$a_N = \frac{\|\mathbf{v} \times \mathbf{a}\|}{\|\mathbf{v}\|}$	Rate of change of direction (curvature)

Additional info: The decomposition of acceleration into tangential and normal components is foundational for understanding circular and curved motion, and is widely used in kinematics and dynamics.