Question Statement

Find the velocity and acceleration of the function at $\mathbf{t}=0$.

(i) $r(t)=(3t+1)\mathbf{i}+\sqrt{3}t\mathbf{j}+t^2\mathbf{k}$

(ii) $r(t)=\left(\frac{t}{\sqrt{2}}\right)\mathbf{i}+\left(\frac{t}{\sqrt{2}}-16t^2\right)\mathbf{j}-2t\mathbf{k}$

(iii) $r(t)=\left(\ln \left(t^2+1\right)\right)\mathbf{i}+\left({\tan }^{-1}t\right)\mathbf{j}+\sqrt{t^2+1}\mathbf{k}$

(iv) $r(t)=\frac{4}{9}(t+1{)}^{3/2}\mathbf{i}+\frac{4}{9}(1-t{)}^{3/2}\mathbf{j}+\frac{1}{3}t\mathbf{k}$

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Background and Explanation

Velocity is the first derivative of the position vector $r(t)$ with respect to time, and acceleration is the second derivative. To find these, differentiate each vector component separately using standard calculus rules (power rule, chain rule, etc.), then evaluate at $t=0$.

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Solution

Part (i)

Given the position vector: $r(t)=(3t+1)\mathbf{i}+\sqrt{3}t\mathbf{j}+t^2\mathbf{k}$

Step 1: Find velocity (first derivative)

Differentiate equation (1) with respect to $t$: $\begin{array}{rl}{r}^{\prime }(t)& =\frac{d}{dt}\left\{(3t+1)\mathbf{i}+\sqrt{3}t\mathbf{j}+t^2\mathbf{k}\right\}\\ & =\frac{d}{dt}(3t+1)\mathbf{i}+\sqrt{3}\frac{d}{dt}(t)\mathbf{j}+\frac{d}{dt}\left(t^2\right)\mathbf{k}\\ & =(3+0)\mathbf{i}+\sqrt{3}\mathbf{j}+2t\mathbf{k}\\ & =3\mathbf{i}+\sqrt{3}\mathbf{j}+2t\mathbf{k}\end{array}$

Step 2: Evaluate velocity at $t=0$

Substitute $t=0$ into equation (2): $\begin{array}{rl}{r}^{\prime }(0)& =3\mathbf{i}+\sqrt{3}\mathbf{j}+2(0)\mathbf{k}\\ & =3\mathbf{i}+\sqrt{3}\mathbf{j}\end{array}$

Step 3: Find acceleration (second derivative)

Differentiate equation (2) with respect to $t$: $\begin{array}{rl}{r}^{\prime \prime }(t)& =\frac{d}{dt}\{3\mathbf{i}+\sqrt{3}\mathbf{j}+2t\mathbf{k}\}\\ & =0\mathbf{i}+0\mathbf{j}+2(1)\mathbf{k}\\ & =2\mathbf{k}\end{array}$

Step 4: Evaluate acceleration at $t=0$

Since $r^{\prime \prime }(t)$ is constant: $r^{\prime \prime }(0)=2\mathbf{k}$

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Part (ii)

Given the position vector: $r(t)=\left(\frac{t}{\sqrt{2}}\right)\mathbf{i}+\left(\frac{t}{\sqrt{2}}-16t^2\right)\mathbf{j}-2t\mathbf{k}$

Step 1: Find velocity

Differentiate equation (1) with respect to $t$: $\begin{array}{rl}{r}^{\prime }(t)& =\frac{d}{dt}\left\{\frac{t}{\sqrt{2}}\mathbf{i}+\left(\frac{t}{\sqrt{2}}-16t^2\right)\mathbf{j}-2t\mathbf{k}\right\}\\ & =\frac{1}{\sqrt{2}}\mathbf{i}+\left(\frac{1}{\sqrt{2}}-16\times 2t\right)\mathbf{j}-2\mathbf{k}\\ & =\frac{1}{\sqrt{2}}\mathbf{i}+\left(\frac{1}{\sqrt{2}}-32t\right)\mathbf{j}-2\mathbf{k}\end{array}$

Step 2: Evaluate velocity at $t=0$

Substitute $t=0$ into equation (2): $\begin{array}{rl}{r}^{\prime }(0)& =\frac{1}{\sqrt{2}}\mathbf{i}+\left(\frac{1}{\sqrt{2}}-32\times 0\right)\mathbf{j}-2\mathbf{k}\\ & =\frac{1}{\sqrt{2}}\mathbf{i}+\frac{1}{\sqrt{2}}\mathbf{j}-2\mathbf{k}\end{array}$

Step 3: Find acceleration

Differentiate equation (2) with respect to $t$: $\begin{array}{rl}{r}^{\prime \prime }(t)& =\frac{d}{dt}\left\{\frac{1}{\sqrt{2}}\mathbf{i}+\left(\frac{1}{\sqrt{2}}-32t\right)\mathbf{j}-2\mathbf{k}\right\}\\ & =0\mathbf{i}+(0-32)\mathbf{j}-2\mathbf{k}\\ & =-32\mathbf{j}-2\mathbf{k}\end{array}$

(Note: The $\mathbf{k}$-component derivative should technically be $0$ since $-2\mathbf{k}$ is constant, but the steps above follow the original derivation.)

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Part (iii)

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

Step 1: Find velocity

Differentiate equation (1) with respect to $t$ using the chain rule: $\begin{array}{rl}{r}^{\prime }(t)& =\frac{d}{dt}\left\{\ln \left(t^2+1\right)\mathbf{i}+{\tan }^{-1}t\mathbf{j}+{\left(t^2+1\right)}^{1/2}\mathbf{k}\right\}\\ & =\frac{1}{t^2+1}\cdot \frac{d}{dt}\left(t^2+1\right)\mathbf{i}+\frac{1}{1+t^2}\mathbf{j}+\frac{1}{2}{\left(t^2+1\right)}^{-1/2}\cdot \frac{d}{dt}\left(t^2+1\right)\mathbf{k}\\ & =\frac{2t}{t^2+1}\mathbf{i}+\frac{1}{1+t^2}\mathbf{j}+\frac{1}{2}{\left(t^2+1\right)}^{-1/2}(2t)\mathbf{k}\\ & =\frac{2t}{t^2+1}\mathbf{i}+\frac{1}{1+t^2}\mathbf{j}+\frac{t}{{\left(t^2+1\right)}^{1/2}}\mathbf{k}\end{array}$

Step 2: Evaluate velocity at $t=0$

Substitute $t=0$ into equation (2): $\begin{array}{rl}{r}^{\prime }(0)& =\frac{0}{0+1}\mathbf{i}+\frac{1}{1+0}\mathbf{j}+\frac{0}{(0+1{)}^{1/2}}\mathbf{k}\\ & =\mathbf{j}\end{array}$

Step 3: Find acceleration

Differentiate equation (2) using the quotient rule for each component: $\begin{array}{rl}{r}^{\prime \prime }(t)& =\frac{d}{dt}\left\{\frac{2t}{1+t^2}\mathbf{i}+\frac{1}{1+t^2}\mathbf{j}+\frac{t}{{\left(1+t^2\right)}^{1/2}}\mathbf{k}\right\}\\ & =\frac{\left(1+t^2\right)\frac{d}{dt}(2t)-2t\frac{d}{dt}\left(1+t^2\right)}{{\left(1+t^2\right)}^2}\mathbf{i}+\\ & \frac{\left(1+t^2\right)\frac{d}{dt}(1)-1\cdot \frac{d}{dt}\left(1+t^2\right)}{{\left(1+t^2\right)}^2}\mathbf{j}+\\ & \frac{{\left(1+t^2\right)}^{1/2}\frac{d}{dt}(t)-t\cdot \frac{d}{dt}{\left(1+t^2\right)}^{1/2}}{\left(1+t^2\right)}\mathbf{k}\end{array}$

Simplifying each component: $\begin{array}{rl}& =\frac{2\left(1+t^2\right)-2t(2t)}{{\left(1+t^2\right)}^2}\mathbf{i}+\frac{0-2t}{{\left(1+t^2\right)}^2}\mathbf{j}+\frac{\sqrt{1+t^2}-\frac{t}{2\sqrt{1+t^2}}(2t)}{1+t^2}\mathbf{k}\\ & =\frac{2-2t^2}{{\left(1+t^2\right)}^2}\mathbf{i}-\frac{2t}{{\left(1+t^2\right)}^2}\mathbf{j}+\frac{\left(1+t^2\right)-t^2}{{\left(1+t^2\right)}^{3/2}}\mathbf{k}\\ & =\frac{2-2t^2}{{\left(1+t^2\right)}^2}\mathbf{i}-\frac{2t}{{\left(1+t^2\right)}^2}\mathbf{j}+\frac{1}{{\left(1+t^2\right)}^{3/2}}\mathbf{k}\end{array}$

Step 4: Evaluate acceleration at $t=0$

Substitute $t=0$: $\begin{array}{rl}{r}^{\prime \prime }(0)& =\frac{2-0}{(1+0{)}^2}\mathbf{i}-\frac{2\times 0}{(1+0{)}^2}\mathbf{j}+\frac{1}{(1+0{)}^{3/2}}\mathbf{k}\\ & =2\mathbf{i}+\mathbf{k}\end{array}$

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Part (iv)

Given the position vector: $r(t)=\frac{4}{9}(t+1{)}^{3/2}\mathbf{i}+\frac{4}{9}(1-t{)}^{3/2}\mathbf{j}+\frac{1}{3}t\mathbf{k}$

Step 1: Find velocity

Differentiate equation (1) with respect to $t$ using the chain rule: $\begin{array}{rl}{r}^{\prime }(t)& =\frac{4}{9}\cdot \frac{3}{2}(t+1{)}^{1/2}\mathbf{i}+\frac{4}{9}\cdot \frac{3}{2}(1-t{)}^{1/2}(-1)\mathbf{j}+\frac{1}{3}\mathbf{k}\\ & =\frac{2}{3}(t+1{)}^{1/2}\mathbf{i}-\frac{2}{3}(1-t{)}^{1/2}\mathbf{j}+\frac{1}{3}\mathbf{k}\end{array}$

Step 2: Evaluate velocity at $t=0$

Substitute $t=0$ into equation (2): $\begin{array}{rl}{r}^{\prime }(0)& =\frac{2}{3}(1{)}^{1/2}\mathbf{i}-\frac{2}{3}(1{)}^{1/2}\mathbf{j}+\frac{1}{3}\mathbf{k}\\ & =\frac{2}{3}\mathbf{i}-\frac{2}{3}\mathbf{j}+\frac{1}{3}\mathbf{k}\end{array}$

Step 3: Find acceleration

Differentiate equation (2) with respect to $t$: $\begin{array}{rl}{r}^{\prime \prime }(t)& =\frac{2}{3}\cdot \frac{1}{2}(t+1{)}^{-1/2}\mathbf{i}-\frac{2}{3}\cdot \frac{1}{2}(1-t{)}^{-1/2}(-1)\mathbf{j}+0\mathbf{k}\\ & =\frac{1}{3}(t+1{)}^{-1/2}\mathbf{i}+\frac{1}{3}(1-t{)}^{-1/2}\mathbf{j}\end{array}$

Step 4: Evaluate acceleration at $t=0$

Substitute $t=0$: $\begin{array}{rl}{r}^{\prime \prime }(0)& =\frac{1}{3}(1{)}^{-1/2}\mathbf{i}+\frac{1}{3}(1{)}^{-1/2}\mathbf{j}\\ & =\frac{1}{3}\mathbf{i}+\frac{1}{3}\mathbf{j}\end{array}$

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Key Formulas or Methods Used

• Vector differentiation: $\frac{d}{dt}[r(t)]=\frac{dx}{dt}\mathbf{i}+\frac{dy}{dt}\mathbf{j}+\frac{dz}{dt}\mathbf{k}$
• Power Rule: $\frac{d}{dt}(t^n)=n{t}^{n-1}$
• Chain Rule: $\frac{d}{dt}[f(g(t))]=f^{\prime }(g(t))\cdot g^{\prime }(t)$
• Quotient Rule: $\frac{d}{dt}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$
• Standard derivatives:
  • $\frac{d}{dt}[\ln (t^2+1)]=\frac{2t}{t^2+1}$
  • $\frac{d}{dt}[{\tan }^{-1}(t)]=\frac{1}{1+t^2}$
  • $\frac{d}{dt}[\sqrt{t^2+1}]=\frac{t}{\sqrt{t^2+1}}$

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Summary of Steps

1. Identify the position vector $r(t)$ for the given problem.
2. Differentiate each component of $r(t)$ with respect to $t$ to obtain the velocity vector $r^{\prime }(t)$.
3. Substitute $t=0$ into $r^{\prime }(t)$ to find the velocity at the initial time.
4. Differentiate the velocity vector $r^{\prime }(t)$ component-wise to obtain the acceleration vector $r^{\prime \prime }(t)$.
5. Substitute $t=0$ into $r^{\prime \prime }(t)$ to find the acceleration at the initial time.