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Worksheet Assigned Problems for Section 13.3

The problems listed below are assigned to be included in your problem set portfolio. Note that a specific selection of these problems will also form the written homework assignments. I recommend working through all of them to ensure a solid grasp of the material. Reach out to Richard for help if you get stuck or have any questions.

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The solutions will be posted after the written homework due dates. If you have any questions about your work, talk to Richard and he is happy to discuss the process with you.

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13.3.5.

Compute the length of the curve traced by \(\v{r}(t) = \la t, 4t^\frac{3}{2}, 2t^\frac{3}{2} \ra\) over \(0 \leq t \leq 3\text{.}\)

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Solution.

First, we find the derivative of the vector-valued function:

\begin{equation*} \v{r}'(t) = \la 1, 6t^{1/2}, 3t^{1/2} \ra \end{equation*}

Next, we find the magnitude of the derivative (the speed):

\begin{align*} \|\v{r}'(t)\| \amp= \sqrt{1^2 + \lp 6t^{1/2} \rp^2 + \lp 3t^{1/2} \rp^2} \\ \amp= \sqrt{1 + 36t + 9t} \\ \amp= \sqrt{1 + 45t} \end{align*}

Now, we compute the arc length by integrating the speed over the interval \(0 \leq t \leq 3\text{:}\)

\begin{equation*} s = \int_0^3 \sqrt{1 + 45t} \, dt \end{equation*}

Using substitution with \(u = 1 + 45t\) and \(du = 45 \, dt\text{,}\) the limits change from \(u(0) = 1\) to \(u(3) = 136\text{:}\)

\begin{align*} s \amp= \frac{1}{45} \int_1^{136} u^{1/2} \, du \\ \amp= \frac{1}{45} \lp\frac{2}{3} u^{3/2} \rp \bigg|_1^{136} \\ \amp= \frac{2}{135} \lp 136^{3/2} - 1 \rp \\ \amp\approx 23.48176 \end{align*}

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13.3.9.

The curve shown in the figure is parametrized by \(\v{r}(t) = \la \cos(7t), \sin(7t), 2\cos(t) \ra\) for \(0 \leq t \leq 2\pi\text{.}\) Approximate its length.

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Figure 13.3.11. The curve traced by \(\v{r}(t) = \la \cos(7t), \sin(7t), 2\cos(t) \ra\) for \(0 \leq t \leq 2\pi\text{.}\)

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Solution.

First, find the derivative:

\begin{equation*} \v{r}'(t) = \la -7\sin(7t), 7\cos(7t), -2\sin(t) \ra \end{equation*}

Find its magnitude:

\begin{align*} \|\v{r}'(t)\| \amp= \sqrt{\lp -7\sin(7t) \rp^2 + \lp 7\cos(7t) \rp^2 + \lp -2\sin(t) \rp^2} \\ \amp= \sqrt{49\sin^2(7t) + 49\cos^2(7t) + 4\sin^2(t)} \\ \amp= \sqrt{49 + 4\sin^2(t)} \end{align*}

The exact length is given by the integral:

\begin{equation*} s = \int_0^{2\pi} \sqrt{49 + 4\sin^2(t)} \, dt \end{equation*}

This integral cannot be evaluated in terms of elementary functions. Using a CAS, we can approximate the length as

\begin{equation*} s \approx 44.8666 \end{equation*}

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13.3.11.

Compute the arc length function \(\displaystyle s(t) = \int_a^t \left\|\v{r}'(u)\right\| du\text{,}\) where

\begin{equation*} \v{r}(t) = \la t^2, 2t^2, t^3 \ra \end{equation*}

for the value \(a = 0\)

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Solution.

We begin by finding the derivative and its magnitude:

\begin{align*} \v{r}'(t) \amp= \la 2t, 4t, 3t^2 \ra \\ \|\v{r}'(t)\| \amp= \sqrt{4t^2 + 16t^2 + 9t^4} = \sqrt{20t^2 + 9t^4} = |t|\sqrt{20 + 9t^2} \end{align*}

Assuming \(t \geq 0\text{,}\) we have \(\|\v{r}'(t)\| = t\sqrt{20 + 9t^2}\text{.}\) Now, we compute the arc length function:

\begin{equation*} s(t) = \int_0^t u\sqrt{20 + 9u^2} \, du \end{equation*}

Use substitution with \(w = 20 + 9u^2\) and \(dw = 18u \, du\text{:}\)

\begin{align*} s(t) \amp= \frac{1}{18} \int_{20}^{20+9t^2} w^{1/2} \, dw \\ \amp= \frac{1}{18} \left[ \frac{2}{3}w^{3/2} \right]_{20}^{20+9t^2} \\ \amp= \frac{1}{27} \lp (20 + 9t^2)^{3/2} - 20^{3/2} \rp \end{align*}

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13.3.15.

Find the speed of \(\v{r}(t) = \la t, \ln(t), \lp \ln(t) \rp^2 \ra\) at \(t = 1\text{.}\)

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Solution.

The speed is the magnitude of the velocity vector. First, we find \(\v{r}'(t)\text{:}\)

\begin{equation*} \v{r}'(t) = \la 1, \frac{1}{t}, \frac{2\ln(t)}{t} \ra \end{equation*}

Evaluate the velocity vector at \(t = 1\text{:}\)

\begin{equation*} \v{r}'(1) = \la 1, 1, 0 \ra \end{equation*}

Finally, compute the speed by taking the magnitude:

\begin{equation*} v(1) = \|\v{r}'(1)\| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2} \end{equation*}

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13.3.27.

Let \(\v{r}(t) = \la 3t+1,4t-5,2t \ra\text{.}\)

Evaluate the arc length integral \(\displaystyle s = g(t) = \int_0^t \|\v{r}'(u)\|\, du\text{.}\)
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Find the inverse \(g^{-1}(s)\) of \(g(t)\text{.}\)
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Verify that \(\v{r}_1(s) = \v{r}\lp g^{-1}(s)\rp\) is an arc length parametrization.
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Solution.

First, find the derivative and its magnitude:

\begin{align*} \v{r}'(t) \amp= \la 3, 4, 2 \ra \\ \|\v{r}'(t)\| \amp= \sqrt{3^2 + 4^2 + 2^2} = \sqrt{9 + 16 + 4} = \sqrt{29} \end{align*}

Now evaluate the arc length integral:

\begin{equation*} s = g(t) = \int_0^t \sqrt{29} \, du = \sqrt{29}t \end{equation*}

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To find the inverse, solve \(s = \sqrt{29}t\) for \(t\text{:}\)

\begin{equation*} t = g^{-1}(s) = \frac{s}{\sqrt{29}} \end{equation*}

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Substitute \(t\) into the original parametrization:

\begin{align*} \v{r}_1(s) \amp= \v{r}\lp \frac{s}{\sqrt{29}} \rp \\ \amp= \la 3\lp\frac{s}{\sqrt{29}}\rp + 1, 4\lp\frac{s}{\sqrt{29}}\rp - 5, 2\lp\frac{s}{\sqrt{29}}\rp \ra \end{align*}

To verify it is an arc length parametrization, we check if \(\|\v{r}_1'(s)\| = 1\text{:}\)

\begin{align*} \v{r}_1'(s) \amp= \la \frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}}, \frac{2}{\sqrt{29}} \ra \\ \|\v{r}_1'(s)\| \amp= \sqrt{\lp \frac{3}{\sqrt{29}} \rp^2 + \lp \frac{4}{\sqrt{29}} \rp^2 + \lp \frac{2}{\sqrt{29}} \rp^2} \\ \amp= \sqrt{\frac{9+16+4}{29}} \\ \amp= \sqrt{\frac{29}{29}} \\ \amp= 1 \end{align*}

Since the speed is 1, it is indeed an arc length parametrization.

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13.3.31.

Find a path that traces the circle in the plane \(y = 10\) with radius \(4\) and center \((2,10,-3)\) with constant speed \(8\text{.}\)

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Solution.

A standard parametrization for a circle in the plane \(y = 10\) with radius \(4\) and center \((2, 10, -3)\) is:

\begin{equation*} \v{c}(t) = \la 2 + 4\cos(t), 10, -3 + 4\sin(t) \ra \end{equation*}

The speed of this standard parametrization is:

\begin{equation*} \|\v{c}'(t)\| = \sqrt{\lp -4\sin(t)\rp^2 + 0^2 + \lp 4\cos(t)\rp^2} = \sqrt{16(\sin^2(t) + \cos^2(t))} = 4 \end{equation*}

Since we want the path to have a constant speed of \(8\) (which is twice the current speed), we need to traverse the curve twice as fast. We can achieve this by substituting an inner function of \(t = 2u\text{:}\)

\begin{equation*} \v{r}(u) = \la 2 + 4\cos(2u), 10, -3 + 4\sin(2u) \ra \end{equation*}

You can quickly verify this new path has a constant speed of \(8\) by computing \(\|\v{r}'(u)\|\text{.}\)

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13.3.33.

Find an arc length parametrization of the curve parametrized by

\begin{equation*} \v{r}(t) = \la \cos(t), \sin(t), \frac{2}{3}t^\frac{3}{2} \ra \end{equation*}

with the parameter \(s\) measuring from \((1,0,0)\text{.}\)

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Solution.

Step 1. Find the arc length function starting from \(t=0\) (which corresponds to the given point \((1,0,0)\)):

\begin{equation*} \v{r}'(t) = \la -\sin(t), \cos(t), t^{1/2} \ra \end{equation*}

\begin{align*} \|\v{r}'(t)\| \amp= \sqrt{(-\sin(t))^2 + (\cos(t))^2 + (t^{1/2})^2} \\ \amp= \sqrt{\sin^2(t) + \cos^2(t) + t} = \sqrt{1 + t} \end{align*}

\begin{align*} s = g(t) \amp= \int_0^t \sqrt{1+u} \, du \\ \amp= \left[ \frac{2}{3}(1+u)^{3/2} \right]_0^t = \frac{2}{3} \lp (1+t)^{3/2} - 1 \rp \end{align*}

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Step 2. Solve for \(t\) in terms of \(s\text{:}\)

\begin{align*} s \amp= \frac{2}{3} \lp (1+t)^{3/2} - 1 \rp \\ \frac{3}{2}s \amp= (1+t)^{3/2} - 1 \\ (1+t)^{3/2} \amp= \frac{3}{2}s + 1 \\ 1+t \amp= \lp \frac{3}{2}s + 1 \rp^{2/3} \\ t \amp= \lp \frac{3}{2}s + 1 \rp^{2/3} - 1 \end{align*}

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Step 3. Substitute \(t = g^{-1}(s)\) into the original parametrization:

\begin{equation*} \v{r}_1(s) = \left\langle \cos\lp \lp \frac{3}{2}s + 1 \rp^{2/3} - 1 \rp, \sin\lp \lp \frac{3}{2}s + 1 \rp^{2/3} - 1 \rp, \frac{2}{3}\lp \lp \frac{3}{2}s + 1 \rp^{2/3} - 1 \rp^{3/2} \right\rangle \end{equation*}

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13.3.35.

Find an arc length parametrization of the curve parametrized by \(\v{r}(t) = \la t^2, t^3 \ra\text{.}\)

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Solution.

Assuming the arc length is measured from \(t = 0\text{:}\)

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Step 1. Compute the derivative and its magnitude:

\begin{align*} \v{r}'(t) \amp= \la 2t, 3t^2 \ra \\ \|\v{r}'(t)\| \amp= \sqrt{4t^2 + 9t^4} = t\sqrt{4 + 9t^2} \amp\amp \text{(assuming } t \geq 0) \end{align*}

Find the arc length function:

\begin{equation*} s = g(t) = \int_0^t u\sqrt{4 + 9u^2} \, du \end{equation*}

Using substitution \(w = 4+9u^2\text{,}\) \(dw = 18u \, du\text{:}\)

\begin{align*} s \amp= \frac{1}{18} \int_4^{4+9t^2} w^{1/2} \, dw \\ \amp= \frac{1}{18} \left[ \frac{2}{3} w^{3/2} \right]_4^{4+9t^2} \\ \amp= \frac{1}{27} \lp (4+9t^2)^{3/2} - 8 \rp \end{align*}

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Step 2. Solve for \(t\) in terms of \(s\text{:}\)

\begin{align*} 27s \amp= (4+9t^2)^{3/2} - 8 \\ (4+9t^2)^{3/2} \amp= 27s + 8 \\ 4+9t^2 \amp= (27s + 8)^{2/3} \\ 9t^2 \amp= (27s + 8)^{2/3} - 4 \\ t \amp= \frac{1}{3}\sqrt{(27s + 8)^{2/3} - 4} \end{align*}

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Step 3. Substitute back into the original parametrization:

\begin{align*} \v{r}_1(s) \amp= \left\langle \lp \frac{1}{3}\sqrt{(27s + 8)^{2/3} - 4} \rp^2, \lp \frac{1}{3}\sqrt{(27s + 8)^{2/3} - 4} \rp^3 \right\rangle \\ \amp= \left\langle \frac{1}{9}\lp (27s + 8)^{2/3} - 4 \rp, \frac{1}{27}\lp (27s + 8)^{2/3} - 4 \rp^{3/2} \right\rangle \end{align*}

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Worksheet Assigned Problems for Section 13.3

13.3.5.

13.3.9.

13.3.11.

13.3.15.

13.3.27.

13.3.31.

13.3.33.

13.3.35.

Worksheet Assigned Problems for Section 13.3