Question

Vector A=2ti+tj-t^3k and B=sinti+costj evaluate

A..d/dt(A.B)

B..d/dt(A.A)

C..d/dt(A×B)

D..show that d/dt(A×A) is equal to zero.

Answer 1

A.

$$A\cdot B=2t\sin t+t\cos t$$

$$\dfrac{d}{dt}(A\cdot B)=2\sin t+2t\cos t+\cos t-t\sin t$$

B.

$$A\cdot A=4t^2+t^2=5t^2$$

$$\dfrac{d}{dt}(A\cdot A)=10t$$

C.

$$A\times B=\begin{vmatrix} i & j & k \\ 2t & t & -t^3 \\ \sin t & \cos t & 0 \\ \end{vmatrix}$$

$$=i(0+t^3\cos t)-j(0+t^3\sin t)+k(2t\cos t-t\sin t)$$

$$\dfrac{d}{dt}(A\times B)=(3t^2\cos t-t^3\sin t)i-(3t^2\sin t+t^3\cos t)j$$

$$+(2\cos t-2t\sin t-\sin t-t\cos t)k$$

D.

$$A\times A=\begin{vmatrix} i & j & k \\ 2t & t & -t^3 \\ 2t & t & -t^3 \\ \end{vmatrix}$$

$$=i(-t^4+t^4)-j(-2t^4+2t^4)+k(2t^2-2t^2)=0$$

$$\dfrac{d}{dt}(A\times A)=0$$