Question

If $r\left(t\right)=5t^{2}i+tj-t^{3}k,$ then
${\int }_1^2(r\times \frac{d^{2}r}{d{t}^2})dt=-14i+75j-15k$

• A. True
• B. False

Answer 1

The correct option is A True

$r\left(t\right)=5t^2\hat{i}+t\hat{j}-t^3\hat{k}$

$\frac{dr\left(t\right)}{dt}=10t\hat{i}+\hat{j}-3t^2\hat{k}$

$\frac{d^{2}r\left(t\right)}{d{t}^2}=10\hat{i}-6t\hat{k}$

$\overrightarrow{r}\times \frac{d^{2}r\left(t\right)}{d{t}^2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 5t^2& t& -t^3\\ 10& 0& -6t\end{array}\right|$

$=(-6t^2-0)\hat{i}-(-30t^3+10t^3)\hat{j}+(0-10t)\hat{k}$

$=-6t^2\hat{i}+20t^3\hat{j}-10t\hat{k}$

${\int }_1^2\overrightarrow{r}\times \frac{d^{2}r\left(t\right)}{d{t}^2}dt={\int }_1^2(-6t^2\hat{i}+20t^3\hat{j}-10t\hat{k})dt$

$={[\frac{-6t^3}{3}\hat{i}+\frac{20t^4}{4}\hat{j}-\frac{10t^2}{2}\hat{k}]}_1^2$

$={[-2t^3\hat{i}+5t^4\hat{j}-5t^2\hat{k}]}_1^2$

$=-2(2^3-1^3)\hat{i}+5(2^4-1^4)\hat{j}-5(2^2-1^2)\hat{k}$

$=-2(8-1)\hat{i}+5(16-1)\hat{j}-5(4-1)\hat{k}$

$=-2\left(7\right)\hat{i}+5\left(15\right)\hat{j}-5\left(3\right)\hat{k}$

$=-14\hat{i}+75\hat{j}-+15\hat{k}$

Hence the given statement is true.