5. let $vec{r}_1(t):=langle t^2,sin(t),cos(3t)\rangle$ and $vec{r}_2(t):=langle e^{5t},e^{7t},e^{2t}\rangle$…

5. let $vec{r}_1(t):=langle t^2,sin(t),cos(3t)\rangle$ and $vec{r}_2(t):=langle e^{5t},e^{7t},e^{2t}\rangle$. compute each of the following in two different ways. first by performing the vector algebra and then taking the derivative, and second by using the appropriate product rule.\n(a) $\frac{d}{dt}(ln(1 + t^2)vec{r}_1(t))$.\n(b) $\frac{d}{dt}(vec{r}_1(t)cdotvec{r}_2(t))$.\n(c) $\frac{d}{dt}(vec{r}_1(t)\timesvec{r}_2(t))$.

5. let $vec{r}_1(t):=langle t^2,sin(t),cos(3t)\rangle$ and $vec{r}_2(t):=langle e^{5t},e^{7t},e^{2t}\rangle$. compute each of the following in two different ways. first by performing the vector algebra and then taking the derivative, and second by using the appropriate product rule.\n(a) $\frac{d}{dt}(ln(1 + t^2)vec{r}_1(t))$.\n(b) $\frac{d}{dt}(vec{r}_1(t)cdotvec{r}_2(t))$.\n(c) $\frac{d}{dt}(vec{r}_1(t)\timesvec{r}_2(t))$.

Answer

Explanation:

Step1: Recall product - rule for vector - valued functions

If $\vec{u}(t)=\langle u_1(t),u_2(t),u_3(t)\rangle$ and $\vec{v}(t)=\langle v_1(t),v_2(t),v_3(t)\rangle$, and $y = \vec{u}(t)\cdot\vec{v}(t)$ or $y=\vec{u}(t)\times\vec{v}(t)$, then $\frac{d}{dt}(\vec{u}(t)\cdot\vec{v}(t))=\vec{u}'(t)\cdot\vec{v}(t)+\vec{u}(t)\cdot\vec{v}'(t)$ and $\frac{d}{dt}(\vec{u}(t)\times\vec{v}(t))=\vec{u}'(t)\times\vec{v}(t)+\vec{u}(t)\times\vec{v}'(t)$. First, find $\vec{r}_1'(t)$ and $\vec{r}_2'(t)$. Given $\vec{r}_1(t)=\langle t^2,\sin(t),\cos(3t)\rangle$, then $\vec{r}_1'(t)=\langle 2t,\cos(t), - 3\sin(3t)\rangle$. Given $\vec{r}_2(t)=\langle e^{5t},e^{7t},e^{2t}\rangle$, then $\vec{r}_2'(t)=\langle 5e^{5t},7e^{7t},2e^{2t}\rangle$.

Step2: Solve part (a)

We want to find $\frac{d}{dt}(\ln(1 + t^2)\vec{r}_1(t))$. By the product - rule for vector - valued functions $\frac{d}{dt}(f(t)\vec{r}(t))=f'(t)\vec{r}(t)+f(t)\vec{r}'(t)$. Here $f(t)=\ln(1 + t^2)$ and $\vec{r}(t)=\vec{r}_1(t)$. Since $f'(t)=\frac{2t}{1 + t^2}$, we have: [ \begin{align*} \frac{d}{dt}(\ln(1 + t^2)\vec{r}_1(t))&=\frac{2t}{1 + t^2}\langle t^2,\sin(t),\cos(3t)\rangle+\ln(1 + t^2)\langle 2t,\cos(t), - 3\sin(3t)\rangle\ &=\left\langle\frac{2t^3}{1 + t^2}+2t\ln(1 + t^2),\frac{2t\sin(t)}{1 + t^2}+\ln(1 + t^2)\cos(t),\frac{2t\cos(3t)}{1 + t^2}-3\ln(1 + t^2)\sin(3t)\right\rangle \end{align*} ]

Step3: Solve part (b)

We want to find $\frac{d}{dt}(\vec{r}_1(t)\cdot\vec{r}_2(t))$. By the dot - product rule $\frac{d}{dt}(\vec{r}_1(t)\cdot\vec{r}_2(t))=\vec{r}_1'(t)\cdot\vec{r}_2(t)+\vec{r}_1(t)\cdot\vec{r}_2'(t)$. [ \begin{align*} \vec{r}_1'(t)\cdot\vec{r}_2(t)&=(2t)e^{5t}+\cos(t)e^{7t}-3\sin(3t)e^{2t}\ \vec{r}_1(t)\cdot\vec{r}_2'(t)&=t^2(5e^{5t})+\sin(t)(7e^{7t})+\cos(3t)(2e^{2t})\ \frac{d}{dt}(\vec{r}_1(t)\cdot\vec{r}_2(t))&=(2t + 5t^2)e^{5t}+(7\sin(t)+\cos(t))e^{7t}+(2\cos(3t)-3\sin(3t))e^{2t} \end{align*} ]

Step4: Solve part (c)

We want to find $\frac{d}{dt}(\vec{r}_1(t)\times\vec{r}_2(t))$. By the cross - product rule $\frac{d}{dt}(\vec{r}_1(t)\times\vec{r}_2(t))=\vec{r}_1'(t)\times\vec{r}_2(t)+\vec{r}_1(t)\times\vec{r}_2'(t)$. The cross - product of two vectors $\vec{a}=\langle a_1,a_2,a_3\rangle$ and $\vec{b}=\langle b_1,b_2,b_3\rangle$ is $\vec{a}\times\vec{b}=\langle a_2b_3 - a_3b_2,a_3b_1 - a_1b_3,a_1b_2 - a_2b_1\rangle$. [ \begin{align*} \vec{r}_1'(t)\times\vec{r}_2(t)&=\left|\begin{array}{ccc} \vec{i}&\vec{j}&\vec{k}\ 2t&\cos(t)&- 3\sin(3t)\ e^{5t}&e^{7t}&e^{2t} \end{array}\right|\ &=\vec{i}(\cos(t)e^{2t}+3\sin(3t)e^{7t})-\vec{j}(2te^{2t}+3\sin(3t)e^{5t})+\vec{k}(2te^{7t}-\cos(t)e^{5t}) \end{align*} ] [ \begin{align*} \vec{r}_1(t)\times\vec{r}_2'(t)&=\left|\begin{array}{ccc} \vec{i}&\vec{j}&\vec{k}\ t^2&\sin(t)&\cos(3t)\ 5e^{5t}&7e^{7t}&2e^{2t} \end{array}\right|\ &=\vec{i}(2e^{2t}\sin(t)-7e^{7t}\cos(3t))-\vec{j}(2t^2e^{2t}-5e^{5t}\cos(3t))+\vec{k}(7t^2e^{7t}-5e^{5t}\sin(t)) \end{align*} ] Then $\frac{d}{dt}(\vec{r}_1(t)\times\vec{r}_2(t))$ is the sum of the above two cross - products.

Answer:

(a) $\left\langle\frac{2t^3}{1 + t^2}+2t\ln(1 + t^2),\frac{2t\sin(t)}{1 + t^2}+\ln(1 + t^2)\cos(t),\frac{2t\cos(3t)}{1 + t^2}-3\ln(1 + t^2)\sin(3t)\right\rangle$ (b) $(2t + 5t^2)e^{5t}+(7\sin(t)+\cos(t))e^{7t}+(2\cos(3t)-3\sin(3t))e^{2t}$ (c) $\vec{r}_1'(t)\times\vec{r}_2(t)+\vec{r}_1(t)\times\vec{r}_2'(t)$ (with $\vec{r}_1'(t)\times\vec{r}_2(t)$ and $\vec{r}_1(t)\times\vec{r}_2'(t)$ calculated as above)