Calculus with Vector Functions — Question 3

Question 3

Let $$\mathbf u(t)=\left\langle t,t^2,1\right\rangle,\qquad \mathbf v(t)=\left\langle \cos t,\sin t,t\right\rangle.$$ Without expanding first, compute at $t=0$:

Tasks

1. $\dfrac{d}{dt}\bigl(\mathbf u\cdot\mathbf v\bigr)$,

2. $\dfrac{d}{dt}\bigl(\mathbf u\times\mathbf v\bigr)$,

3. $\dfrac{d}{dt}\bigl[(t^2+1)\mathbf v(t)\bigr]$.

State the differentiation rule used in each part.

Question 3 – Solution

Strategy. Apply the dot-product, cross-product, and scalar–vector product rules before inserting $t=0$.

At $t=0$, $$\mathbf u=\left\langle 0,0,1\right\rangle,\quad \mathbf u'=\left\langle 1,0,0\right\rangle,\quad \mathbf v=\left\langle 1,0,0\right\rangle,\quad \mathbf v'=\left\langle 0,1,1\right\rangle.$$

Step 1: Dot product rule $$\begin{align*} \frac d{dt}(\mathbf u\cdot\mathbf v) &=\mathbf u'\cdot\mathbf v+\mathbf u\cdot\mathbf v',\\ \left.\frac d{dt}(\mathbf u\cdot\mathbf v)\right|_{0} &=\left\langle 1,0,0\right\rangle\cdot\left\langle 1,0,0\right\rangle +\left\langle 0,0,1\right\rangle\cdot\left\langle 0,1,1\right\rangle=\boxed{2}. \end{align*}$$

Step 2: Cross product rule Order matters: $$\begin{align*} \frac d{dt}(\mathbf u\times\mathbf v) &=\mathbf u'\times\mathbf v+\mathbf u\times\mathbf v',\\ \left.\frac d{dt}(\mathbf u\times\mathbf v)\right|_0 &=\mathbf 0+\left\langle 0,0,1\right\rangle\times\left\langle 0,1,1\right\rangle =\boxed{\left\langle -1,0,0\right\rangle}. \end{align*}$$

Step 3: Scalar–vector rule With $f(t)=t^2+1$, $$(f\mathbf v)'=f'\mathbf v+f\mathbf v' \quad\Longrightarrow\quad (f\mathbf v)'(0)=0\mathbf v(0)+1\mathbf v'(0) =\boxed{\left\langle 0,1,1\right\rangle}.$$