Chapter 4, Problem 4.1P

To determine

Prove that product rule for differentiation applies to the dot product of two vectors.

Answer to Problem 4.1P

It is proved that product rule for differentiation applies to the dot product of two vectors.

Explanation of Solution

Write the expression for the vector $a$ and $b$.

    $\begin{array}{c}a=a_{x}i+a_{y}j+a_{z}k\\ b=b_{x}i+b_{y}j+b_{z}k\end{array}$        (I)

Here, $a$ and $b$ are the vectors, $a_i$ is the components of vector $a$ with $i=x,y,z$, $b_i$ is the components of vector $b$ with $i=x,y,z$.

Write the expression for the dot product of two vectors $a$ and $b$.

    $a\cdot b=a_x{b}_x+a_y{b}_y+a_z{b}_z$        (II)

Write the expression for the differentiation of the dot product of two vectors.

    $\begin{array}{c}\frac{d}{dt}\left(a\cdot b\right)=\frac{d}{dt}\left(a_x{b}_x+a_y{b}_y+a_z{b}_z\right)\\ =\frac{d}{dt}\left(a_x{b}_x\right)+\frac{d}{dt}\left(a_y{b}_y\right)+\frac{d}{dt}\left(a_z{b}_z\right)\\ =a_x\frac{d{b}_x}{dt}+\frac{d{a}_x}{dt}{b}_x+a_y\frac{d{b}_y}{dt}+\frac{d{a}_y}{dt}{b}_y+a_z\frac{d{b}_z}{dt}+\frac{d{a}_z}{dt}{b}_z\\ =\left[a_x\frac{d{b}_x}{dt}+a_y\frac{d{b}_y}{dt}+a_z\frac{d{b}_z}{dt}\right]+\left[\frac{d{a}_y}{dt}{b}_y++\frac{d{a}_z}{dt}{b}_z+\frac{d{a}_x}{dt}{b}_x\right]\\ =a\cdot \frac{db}{dt}+\frac{da}{dt}\cdot b\end{array}$        (III)

Rearrange equation (III) to prove the product rule for differentiation applies to the dot product of two vectors.

    $\frac{d}{dt}\left(a\cdot b\right)=\frac{da}{dt}\cdot b+a\cdot \frac{db}{dt}$        (IV)

Conclusion:

Therefore, it is proved that product rule for differentiation applies to the dot product of two vectors.