Question Statement

If $r(t)$ is the position vector of a particle, find its domain, range at the given point, and its first and second derivatives.

(i) $r(t)=(t+1)\hat{i}+\left(t^2-1\right)\hat{j}+5t\hat{k},t=2$

(ii) $r(t)=\frac{t}{t+1}\hat{i}+\frac{1}{t}\hat{j}+t^3\hat{k},t=-\frac{1}{2}$

(iii) $r(t)=e^t\hat{i}+\frac{2}{9}{e}^{2t}\hat{j}+5e^{-t}\hat{k},t=\ln 2$

(iv) $r(t)=(\cos 2t)\hat{i}+(3\sin 2t)\hat{j}+5t\hat{k},t=0$

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Background and Explanation

This problem involves vector-valued functions, where a particle's position in 3D space is described by parametric equations for each coordinate ($x,y,z$) in terms of time $t$. The first derivative $r^{\prime }(t)$ represents the velocity vector, and the second derivative $r^{\prime \prime }(t)$ represents the acceleration vector. The domain consists of all valid $t$-values where the function is defined, while the range is the set of all possible position vectors.

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Solution

Part (i)

Given the position vector: $r(t)=(t+1)\hat{i}+\left(t^2-1\right)\hat{j}+5t\hat{k}$

First Derivative: Differentiate each component with respect to $t$: \begin{align*} r'(t) &= \frac{d}{d t}\left\{(t+1) \hat{i}+\left(t^{2}-1\right) \hat{j}+5 t \hat{k}\right\} \\ &=(1+0) \hat{i}+(2 t-0) \hat{j}+5 \hat{k} \\ &=\hat{i}+2 t \hat{j}+5 \hat{k} \end{align*}

At $t=2$: $\begin{array}{rl}{r}^{\prime }(2)& =\hat{i}+2(2)\hat{j}+5\hat{k}\\ & =\hat{i}+4\hat{j}+5\hat{k}\end{array}$

Second Derivative: Differentiate $r^{\prime }(t)$ with respect to $t$: $\begin{array}{rl}{r}^{\prime \prime }(t)& =\frac{d}{dt}(\hat{i}+2t\hat{j}+5\hat{k})\\ & =0+2\hat{j}+0\\ & =2\hat{j}\end{array}$

At $t=2$: $r^{\prime \prime }(2)=2\hat{j}$

Part (ii)

Given the position vector: $r(t)=\frac{t}{t+1}\hat{i}+\frac{1}{t}\hat{j}+t^3\hat{k}$

First Derivative: Apply the quotient rule to the first two components and the power rule to the third: \begin{align*} r'(t) &=\frac{d}{d t}\left\{\frac{t}{1+t} \hat{i}+\frac{1}{t} \hat{j}+t^{3} \hat{k}\right\} \\ &=\frac{(1+t) \frac{d}{d t}(t)-t \frac{d}{d t}(1+t)}{(1+t)^{2}} \hat{i}+\frac{t \cdot 0-1}{t^{2}} \hat{j}+3 t^{2} \hat{k} \\ &=\frac{1+t-t}{(1+t)^{2}} \hat{i}-\frac{1}{t^{2}} \hat{j}+3 t^{2} \hat{k} \\ &=\frac{1}{(1+t)^{2}} \hat{i}-\frac{1}{t^{2}} \hat{j}+3 t^{2} \hat{k} \end{align*}

At $t=-\frac{1}{2}$: $\begin{array}{rl}{r}^{\prime }\left(-\frac{1}{2}\right)& =\frac{1}{{\left(1-\frac{1}{2}\right)}^2}\hat{i}-\frac{1}{{\left(\frac{-1}{2}\right)}^2}\hat{j}+3{\left(\frac{-1}{2}\right)}^2\hat{k}\\ & =\frac{1}{{\left(\frac{1}{2}\right)}^2}\hat{i}-\frac{1}{\frac{1}{4}}\hat{j}+3\left(\frac{1}{4}\right)\hat{k}\\ & =4\hat{i}-4\hat{j}+\frac{3}{4}\hat{k}\end{array}$

Second Derivative: Differentiate $r^{\prime }(t)$ using the chain rule and power rule: $\begin{array}{rl}{r}^{\prime \prime }(t)& =\frac{d}{dt}\left\{\frac{1}{(1+t{)}^2}\hat{i}-\frac{1}{t^2}\hat{j}+3t^2\hat{k}\right\}\\ & =\frac{d}{dt}(1+t{)}^{-2}\hat{i}-\frac{d}{dt}\left(t^{-2}\right)\hat{j}+3\frac{d}{dt}\left(t^2\right)\hat{k}\\ & =-2(1+t{)}^{-3}\cdot 1\hat{i}-(-2)t^{-3}\cdot 1\hat{j}+6t\hat{k}\\ & =\frac{-2}{(1+t{)}^3}\hat{i}+\frac{2}{t^3}\hat{j}+6t\hat{k}\end{array}$

At $t=-\frac{1}{2}$: $\begin{array}{rl}{r}^{\prime \prime }\left(-\frac{1}{2}\right)& =\frac{-2}{{\left(1-\frac{1}{2}\right)}^3}\hat{i}+\frac{2}{{\left(\frac{-1}{2}\right)}^3}\hat{j}+6\left(\frac{-1}{2}\right)\hat{k}\\ & =\frac{-2}{{\left(\frac{1}{2}\right)}^3}\hat{i}+\frac{2}{\frac{-1}{8}}\hat{j}-3\hat{k}\\ & =\frac{-2}{1/8}\hat{i}+2(-8)\hat{j}-3\hat{k}\\ & =-16\hat{i}-16\hat{j}-3\hat{k}\end{array}$

Part (iii)

Given the position vector: $r(t)=e^t\hat{i}+\frac{2}{9}{e}^{2t}\hat{j}+5e^{-t}\hat{k}$

First Derivative: Using the chain rule for exponential functions: \begin{align*} r'(t) &=\frac{d}{d t}\left(e^{t} \hat{i}+\frac{2}{9} e^{2 t} \hat{j}+5 e^{-t} \hat{k}\right) \\ &=e^{t} \hat{i}+\frac{2}{9} \cdot 2 e^{2 t} \hat{j}+5(-1) e^{-t} \hat{k} \\ &=e^{t} \hat{i}+\frac{4}{9} e^{2 t} \hat{j}-5 e^{-t} \hat{k} \end{align*}

At $t=\ln 2$ (using $e^{\ln 2}=2$, $e^{2\ln 2}=e^{\ln 4}=4$, and $e^{-\ln 2}=\frac{1}{2}$): $\begin{array}{rl}{r}^{\prime }(\ln 2)& =e^{\ln 2}\hat{i}+\frac{4}{9}{e}^{2\ln 2}\hat{j}-5e^{-\ln 2}\hat{k}\\ & =2\hat{i}+\frac{4}{9}\cdot 4\hat{j}-5\cdot \frac{1}{2}\hat{k}\\ & =2\hat{i}+\frac{16}{9}\hat{j}-\frac{5}{2}\hat{k}\end{array}$

Second Derivative: Differentiate $r^{\prime }(t)$: $\begin{array}{rl}{r}^{\prime \prime }(t)& =\frac{d}{dt}\left\{e^t\hat{i}+\frac{4}{9}{e}^{2t}\hat{j}-5e^{-t}\hat{k}\right\}\\ & =e^t\hat{i}+\frac{4}{9}\cdot 2e^{2t}\hat{j}-5(-1)e^{-t}\hat{k}\\ & =e^t\hat{i}+\frac{8}{9}{e}^{2t}\hat{j}+5e^{-t}\hat{k}\end{array}$

At $t=\ln 2$: $\begin{array}{rl}{r}^{\prime \prime }(\ln 2)& =e^{\ln 2}\hat{i}+\frac{8}{9}{e}^{2\ln 2}\hat{j}+5e^{-\ln 2}\hat{k}\\ & =2\hat{i}+\frac{8}{9}\cdot 4\hat{j}+5\cdot \frac{1}{2}\hat{k}\\ & =2\hat{i}+\frac{32}{9}\hat{j}+\frac{5}{2}\hat{k}\end{array}$

Part (iv)

Given the position vector: $r(t)=(\cos 2t)\hat{i}+(3\sin 2t)\hat{j}+5t\hat{k}$

First Derivative: Using the chain rule for trigonometric functions: \begin{align*} r'(t) &=\frac{d}{d t}(\cos 2 t) \hat{i}+3 \frac{d}{d t}(\sin 2 t) \hat{j}+5 \frac{d}{d t}(t) \hat{k} \\ &=-\sin 2 t \cdot \frac{d}{d t}(2 t) \hat{i}+3 \cos 2 t \cdot \frac{d}{d t}(2 t) \hat{j}+5 \hat{k} \\ &=-2 \sin 2 t \hat{i}+6 \cos 2 t \hat{j}+5 \hat{k} \end{align*}

At $t=0$ (using $\sin 0=0$ and $\cos 0=1$): $\begin{array}{rl}{r}^{\prime }(0)& =-2\sin 0\hat{i}+6\cos 0\hat{j}+5\hat{k}\\ & =-2(0)\hat{i}+6(1)\hat{j}+5\hat{k}\\ & =6\hat{j}+5\hat{k}\end{array}$

Second Derivative: Differentiate $r^{\prime }(t)$: $\begin{array}{rl}{r}^{\prime \prime }(t)& =-2\frac{d}{dt}(\sin 2t)\hat{i}+6\frac{d}{dt}(\cos 2t)\hat{j}+\frac{d}{dt}(5)\hat{k}\\ & =-2\cos 2t\cdot 2\hat{i}+6(-\sin 2t\cdot 2)\hat{j}+0\hat{k}\\ & =-4\cos 2t\hat{i}-12\sin 2t\hat{j}\end{array}$

At $t=0$: $\begin{array}{rl}{r}^{\prime \prime }(0)& =-4\cos 0\hat{i}-12\sin 0\hat{j}\\ & =-4(1)\hat{i}-12(0)\hat{j}\\ & =-4\hat{i}\end{array}$

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Key Formulas or Methods Used

• Component-wise differentiation: $\frac{d}{dt}[f(t)\hat{i}+g(t)\hat{j}+h(t)\hat{k}]=f^{\prime }(t)\hat{i}+g^{\prime }(t)\hat{j}+h^{\prime }(t)\hat{k}$
• Quotient rule: $\frac{d}{dt}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$
• Chain rule: $\frac{d}{dt}f(g(t))=f^{\prime }(g(t))\cdot g^{\prime }(t)$
• Exponential derivatives: $\frac{d}{dt}{e}^{kt}=k{e}^{kt}$ and $\frac{d}{dt}{e}^{-t}=-e^{-t}$
• Trigonometric derivatives with chain rule: $\frac{d}{dt}\sin (kt)=k\cos (kt)$ and $\frac{d}{dt}\cos (kt)=-k\sin (kt)$
• Logarithm properties: $e^{\ln x}=x$, $e^{k\ln x}=x^k$, and $e^{-\ln x}=\frac{1}{x}$

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Summary of Steps

1. Identify components: Break down $r(t)$ into its $\hat{i}$, $\hat{j}$, and $\hat{k}$ scalar components
2. Differentiate each component: Apply appropriate differentiation rules (power, quotient, chain, exponential, or trigonometric) to each scalar component separately
3. Evaluate first derivative: Substitute the given $t$-value into $r^{\prime }(t)$ to find the velocity vector at that instant
4. Differentiate again: Differentiate $r^{\prime }(t)$ component-wise to obtain $r^{\prime \prime }(t)$
5. Evaluate second derivative: Substitute the given $t$-value into $r^{\prime \prime }(t)$ to find the acceleration vector at that instant