Chapter 13, Problem 1GYR

To determine

Mention the rules for differentiating and integrating vector functions with some examples.

Explanation of Solution

Description:

Rules for differentiating vector functions:

Consider,

u and v is the differentiable vector functions of t,

c is scalar,

C is a constant vector, and

f is differentiable scalar function.

1. Constant function rule:

$\frac{d}{dt}C=0$

2. Sum rule:

$\frac{d}{dt}\left[u\left(t\right)+v\left(t\right)\right]=u\text{'}\left(t\right)+v\text{'}\left(t\right)$

3. Difference rule:

$\frac{d}{dt}\left[u\left(t\right)-v\left(t\right)\right]=u\text{'}\left(t\right)-v\text{'}\left(t\right)$

4. Scalar multiple rule:

$\frac{d}{dt}\left[cu\left(t\right)\right]=cu\text{'}\left(t\right)$

5. Chain rule:

$\frac{d}{dt}\left[u\left(f\left(t\right)\right)\right]=cu\text{'}\left(t\right)$

6. Dot product rule:

$\frac{d}{dt}\left[u\left(t\right)\cdot v\left(t\right)\right]=u\text{'}\left(t\right)\cdot v\left(t\right)+u\left(t\right)\cdot v\text{'}\left(t\right)$

7. Cross product rule:

$\frac{d}{dt}\left[u\left(t\right)\times v\left(t\right)\right]=u\text{'}\left(t\right)\times v\left(t\right)+u\left(t\right)\times v\text{'}\left(t\right)$

For example:

Consider the position of a particle in the xy-plane $r\left(t\right)=\left(3t+1\right)i+\sqrt{3}tj+t^{2}k$. Find the angle between the velocity and acceleration vectors at time $t=0$.

The position function is,

$r\left(t\right)=\langle \left(3t+1\right),\sqrt{3}t,t^2\rangle$

The expression for velocity of a particle is,

$v=\frac{dr}{dt}$

Substitute $\langle \left(3t+1\right),\sqrt{3}t,t^2\rangle$ for r in above equation.

$\begin{array}{c}v=\frac{d}{dt}\left(\langle \left(3t+1\right),\sqrt{3}t,t^2\rangle \right)\\ =\langle 3,\sqrt{3},2t\rangle \end{array}$

At $t=0$, the velocity of the particle is,

$\begin{array}{c}v\left(0\right)=\langle 3,\sqrt{3},2\left(0\right)\rangle \\ =\langle 3,\sqrt{3},0\rangle \end{array}$

The magnitude of the velocity $v$ is,

$\begin{array}{c}\left|v\left(0\right)\right|=\sqrt{3^2+{\left(\sqrt{3}\right)}^2}\\ =\sqrt{9+3}\\ =\sqrt{12}\end{array}$

The expression for acceleration of a particle.

$a=\frac{dv}{dt}$

Substitute $\langle 3,\sqrt{3},2t\rangle$ for v in above equation.

$\begin{array}{c}a=\frac{d}{dt}\left(\langle 3,\sqrt{3},2t\rangle \right)\\ =\langle 0,0,2\rangle \end{array}$

At $t=0$, the acceleration of the particle is,

$a\left(0\right)=\langle 0,0,2\rangle$

The magnitude of the acceleration a is,

$\begin{array}{c}\left|a\left(0\right)\right|=\sqrt{2^2}\\ =\sqrt{4}\\ =2\end{array}$

The expression to find the angle between two vectors a and b.

$\theta ={\cos }^{-1}\left(\frac{a\cdot b}{\left|a\right|\left|b\right|}\right)$

The expression to find the angle between two vectors a and b at time $t=0$.

$\theta ={\cos }^{-1}\left(\frac{v\left(0\right)\cdot a\left(0\right)}{\left|v\left(0\right)\right|\left|a\left(0\right)\right|}\right)$

Substitute $\langle 0,0,2\rangle$ for $a\left(0\right)$, $\langle 3,\sqrt{3},0\rangle$ for $v\left(0\right)$, $\sqrt{12}$ for $\left|v\left(0\right)\right|$, and 2 for $\left|a\left(0\right)\right|$ in above equation as follows.

$\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{\langle 3,\sqrt{3},0\rangle \cdot \langle 0,0,2\rangle }{\left(\sqrt{12}\right)\left(2\right)}\right)\\ ={\cos }^{-1}\left(0\right)\end{array}$

The above equation becomes,

$\theta =\frac{\pi }{2}$

Therefore, the angle between the velocity and acceleration vectors at given time is $\theta =\frac{\pi }{2}$.

Rules for integrating vector functions:

The indefinite integral of r with respect to t is the set of all antiderivatives of r. It is represented by ${\displaystyle \int r\left(t\right)dt}$. Consider if R is antiderivative of r, then

${\displaystyle \int r\left(t\right)dt}=R\left(t\right)+C$

For example:

Integrate a vector function ${\displaystyle \int \left[\left(2\cos t\right)i+j-2tk\right]dt}$.

$\begin{array}{c}{\displaystyle \int \left[\left(2\cos t\right)i+j-2tk\right]dt}=\left({\displaystyle \int 2\cos tdt}\right)i+\left({\displaystyle \int dt}\right)j-\left({\displaystyle \int 2tdt}\right)k\\ =\left(2\sin t+C_1\right)i+\left(t+C_2\right)j-\left(2\frac{t^2}{2}+C_3\right)k\\ =\left(2\sin t\right)i+tj-2t^{2}k+C_{1}i+C_{2}j-C_{3}k\\ =\left(2\sin t\right)i+tj-2t^{2}k+C\left\{\because C=C_{1}i+C_{2}j-C_{3}k\right\}\end{array}$

Thus, the rules for differentiating and integrating vector functions is explained with an examples.