Calculus III Advanced (Course) (12.4) (Homework)

Section 12.4 Homework

From Calculus 10e by Larson and Edwards, p. 848. Exercises 10, 19, 22, 29, 30, 31, 32

Exercise 12.4.10 Finding a Tangent Line

Find the unit tangent vector \(\textbf{T}(t)\) and find the parametric equations for the line tangent to the curve at point \(P\).

\( \textbf{r}(t) = \langle t, t, \sqrt{4-t^{2}} \rangle, \: P(1,1,\sqrt{3}) \)

Solution

$$ \textbf{r}^{\prime}(t) = \left( 1,1,-\frac{t}{ \sqrt{4-t^{2}} } \right) $$

$$ \| \textbf{r}^{\prime}(t) \| = \sqrt { (1)^{2} + (1)^{2} + \left(-\frac{t}{ \sqrt{4-t^{2}} } \right)^{2} } = \sqrt{ \frac{t^{2}-8}{t^{2}-4} }$$

$$ \textbf{T}(t) = \frac{ \left( 1,1,-\frac{t}{ \sqrt{4-t^{2}} } \right) }{ \sqrt{ \frac{t^{2}-8}{t^{2}-4} } } = \left( \frac{ 1 }{ \sqrt{ \frac{t^{2}-8}{t^{2}-4} }}, \frac{ 1 }{ \sqrt{ \frac{t^{2}-8}{t^{2}-4} }}, -\frac{ t }{ \sqrt{ \frac{t^{2}-8}{t^{2}-4} } \sqrt{4-t^{2}} } \right)$$

$$ \hphantom{ \textbf{T}(t) } = \frac{ 1 }{ \sqrt{ \frac{t^{2}-8}{t^{2}-4} }} \left( \textbf{i} + \textbf{j} -\frac{t}{ \sqrt{4-t^{2}} }\textbf{k} \right) \:\:\:\: \color{red}{\text{Unit tangent vector}} $$

At the point \(P(1,1,\sqrt{3}) \) \(t=1\) and the unit tangent vector is

$$ \textbf{T}(1) = \frac{ 1 }{ \sqrt{ \frac{t^{2}-8}{t^{2}-4} }} \left( \textbf{i} + \textbf{j} -\frac{1}{ \sqrt{3}} \textbf{k} \right) $$

The unit direction numbers are \(a=1\), \(b=1\), and \(c=-\frac{1}{ \sqrt{3}} \). At the point \((x,y,z)=(1,1,\sqrt{3})\) the parametric equations are

\(x\)	\(=x_{1}+as \)	\(=1+s\)
\(y\)	\(=y_{1}+bs \)	\(=1+s\)
\(z\)	\(=z_{1}+cs \)	\(=\sqrt{3}-\frac{1}{ \sqrt{3}}s\)

Exercise 12.4.19 Finding the Principal Unit Normal Vector

Find the principal unit normal vector to the curve at the specified parameter.

\( \textbf{r}(t) = 6 \cos t \textbf{i} + 6 \sin t\textbf{j} + \textbf{k}, \: t=3 \pi/4 \)

Solution Differentiating yields

\( \textbf{r}^{\prime}(t) = -6 \sin t \textbf{i} + 6 \cos t\textbf{j} \)

\( \| \textbf{r}^{\prime}(t) \| = \sqrt{-6\sin^{2} t \textbf{i} +6 \cos^{2} t\textbf{j} } =6\)

\( \textbf{T}(t) \)	$$=\frac{ \textbf{r}^{\prime}(t) }{ \|\textbf{r}^{\prime}(t) \|} = \frac{1}{6}(-6\sin t \textbf{i} + 6\cos t\textbf{j})= -\sin t \textbf{i} + \cos t\textbf{j} \:\:\:\: \text{Tangent unit vector}$$
\( \textbf{T}^{\prime}(t) \)	\(= -\cos t \textbf{i} - \sin t\textbf{j} \)
\( \| \textbf{T}^{\prime}(t) \| \)	\(= \sqrt{-\cos^{2} t \textbf{i} - \sin^{2} t\textbf{j}}\)
\( \textbf{N}(t) \)	$$= \frac{ -\cos t \textbf{i} - \sin t\textbf{j} }{ \sqrt{-\cos^{2} t \textbf{i} - \sin^{2} t\textbf{j}} } $$
At the point \( t=3 \pi/4 \) the unit normal tangent vector is
$$ \textbf{N} \left( \frac{3 \pi}{4} \right)$$	$$ = \frac{ -\cos \frac{3 \pi}{4} \textbf{i} - \sin \frac{3 \pi}{4}\textbf{j} }{ \sqrt{-\cos^{2} \frac{3 \pi}{4} \textbf{i} - \sin^{2} \frac{3 \pi}{4}\textbf{j}} } $$
	$$= \frac{ -(-\frac{ \sqrt{2}}{2}) \textbf{i} - (\frac{ \sqrt{2}}{2})\textbf{j} }{ \sqrt{-(-\frac{ \sqrt{2}}{2})^{2} \textbf{i} - (\frac{ \sqrt{2}}{2})^{2} \textbf{j}} } $$
	$$= \frac{ \frac{ \sqrt{2}}{2} \textbf{i} - \frac{ \sqrt{2}}{2}\textbf{j} }{ \sqrt{-(-\frac{ \sqrt{2}}{2})^{2} \textbf{i} - (\frac{ \sqrt{2}}{2})^{2} \textbf{j}} } $$
	$$= \frac{ \sqrt{2}}{2} (\textbf{i} -\textbf{j}) $$

Exercise 12.4.22 Find Tangential and Normal Acceleration Components

Find \(\textbf{T}(t)\), \(\textbf{N}(t)\), \(a_{\textbf{T}}\), and \(a_{\textbf{N}}\) at the given time \(t\) for the plane curve \(\textbf{r}(t)\).

\( \textbf{r}(t) = t^{2} \textbf{i} + 2t\textbf{j}, \: t=1 \)

Solution

\( \textbf{v}(t) \)	\(= \textbf{r}^{\prime} = 2t \textbf{i} + 2\textbf{j} \)
\( \| \textbf{v}(t) \| \)	\(= \sqrt{ 4t^{2} + 4} \)
\( a(t) \)	\(= \textbf{r}^{\prime \prime} = 2\textbf{i} \)
\( a_{\textbf{T}} \)	$$= \frac{\textbf{v} \cdot \textbf{a}}{\| \textbf{v} \| } = \frac{4t+4}{ \sqrt{ 4t^{2} + 4} } = \frac{4t+4}{ 2t + 2 } $$	Tangential acceleration component
\( a_{\textbf{T}}(1) \)	$$ = \frac{4(1)+4}{ 2(1) + 2 } = 2 $$
\( a_{\textbf{N}} \)	$$=\frac{\textbf{v} \times \textbf{a}}{\| \textbf{v} \| } = \frac{4}{ \sqrt{ 4t^{2} + 4} }$$
\( a_{\textbf{N}}(1) \)	$$=\frac{4}{ \sqrt{ 4(1)^{2} + 4} } = \frac{4}{\sqrt{8}}$$
\(\textbf{T}(t)\)	$$= \frac{ 2t \textbf{i} + 2\textbf{j} }{ 2t + 2 } = \frac{1}{t+1} (t \textbf{i} + \textbf{j}) $$
\(\textbf{T}(1)\)	$$= \frac{1}{1+1} (1 \textbf{i} + \textbf{j})= \frac{1}{2} ( \textbf{i} + \textbf{j}) = \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$
\(\textbf{T}^{\prime}(t)\)	$$= \frac{-t+1}{(t^{2}+1)^{3/2}}$$
\(\textbf{N}(t)\)	$$=\frac{-t+1}{(t^{2}+1)^{3}} $$
\(\textbf{N}(1)\)	$$=\frac{-1+1}{(1^{2}+1)^{3}} = 0 $$

Circular Motion for 29-32

Consider an object moving according to the position vector

\( \textbf{r}(t) = a \cos \omega t \textbf{i} + a \sin \omega t\textbf{j} \)

where \( \omega \) is the angular velocity.

Exercise 12.4.29

Find \(\textbf{T}(t)\), \(\textbf{N}(t)\), \(a_{\textbf{T}}\), and \(a_{\textbf{N}}\).
Solution

\( \textbf{v}(t) = \textbf{r}^{\prime}(t) \)	\(= - a \omega \sin (\omega t) \textbf{i} + a \omega \cos (\omega t) \textbf{j} \)
\( \| \textbf{v}(t) \| \)	\(= \sqrt{ (- a \omega \sin (\omega t) + a\omega \cos (\omega t) )^{2}} = - a \omega \sin (\omega t) + a \omega \cos (\omega t) \)
\( \textbf{a}(t) = \textbf{r}^{\prime \prime}(t) \)	\(= - a \omega^{2} \cos (\omega t) \textbf{i} - a \omega^{2} \sin (\omega t) \textbf{j} \)
\(\textbf{T}(t)\)	$$= \frac{- a \omega \sin (\omega t) \textbf{i} + a \omega \cos (\omega t) \textbf{j}}{ - a \omega \sin (\omega t) + a \omega \cos (\omega t) }$$
	$$= \frac{(a \omega) (-\sin (\omega t) \textbf{i} + \cos (\omega t) \textbf{j})}{ (a \omega )(-\sin (\omega t) + \cos (\omega t)) }$$
	\(= -\sin (\omega t) \textbf{i} + \cos (\omega t) \textbf{j}\)
\(\textbf{N}(t)\)	\(= -\cos (\omega t) \textbf{i} - \sin (\omega t) \textbf{j}\)
\(a_{\textbf{T}}\)	\(=0\)
\(a_{\textbf{N}}\)	\(=a \omega^{2}\)

Exercise 12.4.30

Determine the directions for \( \textbf{T} \) and \( \textbf{N} \) relative to the position vector \( \textbf{r} \).
Solution Since \( \textbf{T} \) and \( \textbf{N} \) are both negative, they point in the opposite direction from the path.

Exercise 12.4.31

Determine the object's speed at any time \(t\) and explain its value relative to the value for \(a_{\textbf{T}}\).
Solution

\( \| \textbf{v}(t) \| = a \omega\)

The speed is constant because \(a_{\textbf{T}} = 0\).

Exercise 12.4.32

When the angular velocity \( \omega \) is halved, by what factor is \(a_{\textbf{N}}\) changed?.
Solution Since \(a_{\textbf{N}} =a \omega^{2}\) halving \( \omega \) would reduce \(a_{\textbf{N}}\) by half.

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Parent Article: Calculus III Advanced (Course)