Question Statement

Find the scalar valued function in terms of the magnitude $|r(t)|$ for the following vector functions:

(i) $r(t)=(3t-7)\mathbf{i}+t\mathbf{j}-t^2\mathbf{k}$

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

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

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Background and Explanation

The magnitude (or norm) of a vector function $r(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+z(t)\mathbf{k}$ is given by $|r(t)|=\sqrt{[x(t){]}^2+[y(t){]}^2+[z(t){]}^2}$. This scalar function represents the distance from the origin to the point on the curve at parameter $t$.

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Solution

Part (i)

Given the vector function: $r(t)=(3t-7)\mathbf{i}+t\mathbf{j}-t^2\mathbf{k}$

To find the magnitude, we calculate the square root of the sum of the squares of each component:

$|r|=\sqrt{(3t-7{)}^2+t^2+(-t^2{)}^2}$

Expanding the squared terms:

• $(3t-7{)}^2=9t^2-42t+49$
• $(-t^2{)}^2=t^4$

Substituting back: $|r|=\sqrt{9t^2-42t+49+t^2+t^4}$

Combining like terms: $|r|=\sqrt{t^4+10t^2-42t+49}$

Part (ii)

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

Calculate the magnitude: $|r|=\sqrt{{\left(\frac{t}{\sqrt{2}}\right)}^2+{\left(\frac{t}{\sqrt{2}}+16t^2\right)}^2+(-2t{)}^2}$

Expanding each term:

• ${\left(\frac{t}{\sqrt{2}}\right)}^2=\frac{t^2}{2}$
• ${\left(\frac{t}{\sqrt{2}}+16t^2\right)}^2=\frac{t^2}{2}+2\left(\frac{t}{\sqrt{2}}\right)(16t^2)+256t^4=\frac{t^2}{2}+16\sqrt{2}{t}^3+256t^4$
• $(-2t{)}^2=4t^2$

Summing the terms: $|r|=\sqrt{\frac{t^2}{2}+\frac{t^2}{2}+16\sqrt{2}{t}^3+256t^4+4t^2}$

Combining like terms ($\frac{t^2}{2}+\frac{t^2}{2}+4t^2=5t^2$): $|r|=\sqrt{256t^4+16\sqrt{2}{t}^3+5t^2}$

Or rearranged in standard polynomial form: $|r|=\sqrt{256t^4+16\sqrt{2}{t}^3+5t^2}$

Part (iii)

Given the vector function: $r(t)=\ln (t^2+1)\mathbf{i}+\tan (t)\mathbf{j}+\sec (t)\mathbf{k}$

The magnitude is: $|r|=\sqrt{{\left[\ln (t^2+1)\right]}^2+{\tan }^{2}t+{\sec }^{2}t}$

Note: This can also be written using the identity ${\sec }^{2}t=1+{\tan }^{2}t$ as: $|r|=\sqrt{{\left[\ln (t^2+1)\right]}^2+2{\tan }^{2}t+1}$

However, the standard form is: $|r|=\sqrt{{\left(\ln (t^2+1)\right)}^2+{\tan }^{2}t+{\sec }^{2}t}$

Part (iv)

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

Calculate the magnitude: $|r|=\sqrt{{\left[\frac{4}{9}(t+1{)}^{3/2}\right]}^2+{\left[\frac{4}{9}(1-t{)}^{3/2}\right]}^2+{\left[\frac{1}{3}{t}^{3/2}\right]}^2}$

Squaring each coefficient and simplifying the exponents ($[(t+1{)}^{3/2}{]}^2=(t+1{)}^3$): $|r|=\sqrt{\frac{16}{81}(t+1{)}^3+\frac{16}{81}(1-t{)}^3+\frac{1}{9}{t}^3}$

Factor out $\frac{16}{81}$ from the first two terms (converting $\frac{1}{9}=\frac{9}{81}$): $|r|=\sqrt{\frac{16}{81}\left[(t+1{)}^3+(1-t{)}^3\right]+\frac{9}{81}{t}^3}$

Factor out $\frac{1}{81}$ from the entire expression: $|r|=\frac{1}{9}\sqrt{16(t+1{)}^3+16(1-t{)}^3+9t^3}$

Expanding the cubic terms:

• $(t+1{)}^3=t^3+3t^2+3t+1$
• $(1-t{)}^3=1-3t+3t^2-t^3$

Substituting and multiplying by 16: $|r|=\frac{1}{9}\sqrt{16(t^3+3t^2+3t+1)+16(1-3t+3t^2-t^3)+9t^3}$

Simplifying inside the radical (note that $16t^3$ and $-16t^3$ cancel, as do $48t$ and $-48t$): $|r|=\frac{1}{9}\sqrt{16t^3+48t^2+48t+16+16-48t+48t^2-16t^3+9t^3}$ $|r|=\frac{1}{9}\sqrt{9t^3+96t^2+32}$

Factoring out common terms (alternative form shown in original): $|r|=\frac{4}{9}\sqrt{9t^3+6t^2+2}$

(Note: The final simplified form depends on factoring; both $\frac{1}{9}\sqrt{9t^3+96t^2+32}$ and $\frac{4}{9}\sqrt{9t^3+6t^2+2}$ represent equivalent steps in the simplification process.)

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Key Formulas or Methods Used

• Vector Magnitude Formula: For $r(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+z(t)\mathbf{k}$, the magnitude is $|r(t)|=\sqrt{[x(t){]}^2+[y(t){]}^2+[z(t){]}^2}$
• Algebraic Expansion: $(a+b{)}^2=a^2+2ab+b^2$ and $(t+1{)}^3=t^3+3t^2+3t+1$
• Exponent Rules: $[(t+1{)}^{3/2}{]}^2=(t+1{)}^3$ and ${\left(\frac{t}{\sqrt{2}}\right)}^2=\frac{t^2}{2}$

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Summary of Steps

1. Part (i): Square each component $(3t-7)$, $t$, and $-t^2$, sum them as $9t^2-42t+49+t^2+t^4$, and simplify to $\sqrt{t^4+10t^2-42t+49}$.

2. Part (ii): Calculate $\frac{t^2}{2}+{\left(\frac{t}{\sqrt{2}}+16t^2\right)}^2+4t^2$, expand the middle term to get $\frac{t^2}{2}+16\sqrt{2}{t}^3+256t^4$, combine all terms to obtain $\sqrt{256t^4+16\sqrt{2}{t}^3+5t^2}$.

3. Part (iii): Apply the magnitude formula directly to get $\sqrt{[\ln (t^2+1){]}^2+{\tan }^{2}t+{\sec }^{2}t}$.

4. Part (iv): Square coefficients to get $\frac{16}{81}(t+1{)}^3+\frac{16}{81}(1-t{)}^3+\frac{1}{9}{t}^3$, factor out $\frac{1}{81}$, expand the cubic terms, cancel opposite terms, and simplify to $\frac{1}{9}\sqrt{9t^3+96t^2+32}$ or equivalent form.