Question

if A = 5t2 + tj-t3k and B = sin t1-costj evaluate d/dt (AXB)

Answer 1

Answer on Question #79476 – Math – Differential Geometry | Topology

Question

if $A = 5t^2 + tj - t^3k$ and $B = \sin t^1 - \cos t^j$ evaluate $d/dt$ (AxB)

Solution

the derivative of the cross product of two vector functions is

$$\frac{d}{dt} (\vec{A} \times \vec{B}) = \frac{d\vec{A}}{dt} \times \vec{B} + \vec{A} \times \frac{d\vec{B}}{dt}$$ $$\vec{A} = a_1(t) \vec{i} + a_2(t) \vec{j} + a_3(t) \vec{k} = 5t^2 \vec{i} + t \vec{j} - t^3 \vec{k}$$ $$\frac{d\vec{A}}{dt} = \frac{da_1(t)}{dt} \vec{i} + \frac{da_2(t)}{dt} \vec{j} + \frac{da_3(t)}{dt} \vec{k} = 10t \vec{i} + \vec{j} - 3t^2 \vec{k}$$ $$\vec{B} = b_1(t) \vec{i} + b_2(t) \vec{j} + b_3(t) \vec{k} = \sin(t) \vec{i} - \cos(t) \vec{j}$$ $$\frac{d\vec{B}}{dt} = \frac{db_1(t)}{dt} \vec{i} + \frac{db_2(t)}{dt} \vec{j} + \frac{db_3(t)}{dt} \vec{k} = \cos(t) \vec{i} + \sin(t) \vec{j}$$

cross product of two vectors $u \times v$:

$$\vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = (u_2v_3 - u_3v_2)\vec{i} + (u_3v_1 - u_1v_3)\vec{j} + (u_1v_2 - u_2v_1)\vec{k}$$ $$\begin{aligned} \frac{d\vec{A}}{dt} \times \vec{B} &= \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 10t & 1 & -3t^2 \\ \sin(t) & -\cos(t) & 0 \end{vmatrix} \\ &= [1 \cdot 0 - (-3t^2)(-\cos(t))]\vec{i} + [(-3t^2)\sin(t) - 10t \cdot 0]\vec{j} \\ &+ [10t(-\cos(t)) - 1 \cdot \sin(t)]\vec{k} \\ &= -3t^2\cos(t)\vec{i} - 3t^2\sin(t)\vec{j} - (10t \cdot \cos(t) + \sin(t))\vec{k} \end{aligned}$$ $$\begin{aligned} \vec{A} \times \frac{d\vec{B}}{dt} &= \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 5t^2 & t & -t^3 \\ \cos(t) & \sin(t) & 0 \end{vmatrix} \\ &= [t \cdot 0 - (-t^3)\sin(t)]\vec{i} + [(-t^3)\cos(t) - 5t^2 \cdot 0]\vec{j} + [5t^2\sin(t) - t \cdot \cos(t)]\vec{k} \\ &= t^3\sin(t)\vec{i} - t^3\cos(t)\vec{j} + (5t^2\sin(t) - t \cdot \cos(t))\vec{k} \end{aligned}$$

So,

$$\begin{array}{l} \frac{d}{dt} (\vec{A} \times \vec{B}) = \frac{d\vec{A}}{dt} \times \vec{B} + \vec{A} \times \frac{d\vec{B}}{dt} \\ \quad = [t^3 \sin(t) - 3t^2 \cos(t)]\vec{i} - [t^3 \cos(t) + 3t^2 \sin(t)]\vec{j} \\ \quad + [5t^2 \sin(t) - t \cos(t) - 10t \cdot \cos(t) - \sin(t)]\vec{k} \\ \quad = t^2 (t \cdot \sin(t) - 3 \cos(t)) \vec{i} - t^2 (t \cdot \cos(t) + 3 \sin(t)) \vec{j} \\ \quad + ((5t^2 - 1) \sin(t) - 11t \cdot \cos(t)) \vec{k} \end{array}$$

Answer: $\frac{d}{dt} (\vec{A} \times \vec{B}) = t^2 (t \cdot \sin(t) - 3\cos(t)) \vec{i} - t^2 (t \cdot \cos(t) + 3\sin(t)) \vec{j} + ((5t^2 - 1)\sin(t) - 11t \cdot \cos(t)) \vec{k}$

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