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MTH 254 Calculus IV Winter 2026

Richard Zhao

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Worksheet Assigned Problems for Section 13.4

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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13.4.3.

Calculate \(\v{r}'(t)\) and \(\v{T}(t)\) for \(\v{r}(t) = \la 3+4t, 3-5t, 9t \ra\text{,}\) and evaluate \(\v{T}(1)\text{.}\)
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Solution.
First, we compute the derivative \(\v{r}'(t)\text{:}\)
\begin{equation*} \v{r}'(t) = \la 4, -5, 9 \ra \end{equation*}
Next, we find the magnitude of the velocity vector:
\begin{equation*} \|\v{r}'(t)\| = \sqrt{4^2 + (-5)^2 + 9^2} = \sqrt{16 + 25 + 81} = \sqrt{122} \end{equation*}
The unit tangent vector is the velocity vector divided by its speed:
\begin{equation*} \v{T}(t) = \frac{\v{r}'(t)}{\|\v{r}'(t)\|} = \frac{1}{\sqrt{122}} \la 4, -5, 9 \ra \end{equation*}
Since \(\v{T}(t)\) is constant, evaluating it at \(t=1\) yields the exact same vector:
\begin{equation*} \v{T}(1) = \la \frac{4}{\sqrt{122}}, -\frac{5}{\sqrt{122}}, \frac{9}{\sqrt{122}} \ra \end{equation*}
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13.4.9.

Calculate the curvature function \(\kappa(t)\) of \(\v{r}(t) = \la 4t+1, 4t-3, 2t \ra\) using Theorem 13.4.6.
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Solution.
First, we find the first and second derivatives of \(\v{r}(t)\text{:}\)
\begin{align*} \v{r}'(t) \amp= \la 4, 4, 2 \ra \\ \v{r}''(t) \amp= \la 0, 0, 0 \ra \end{align*}
Using Theorem 13.4.6, the curvature is:
\begin{equation*} \kappa(t) = \frac{\|\v{r}'(t) \times \v{r}''(t)\|}{\|\v{r}'(t)\|^3} \end{equation*}
Since \(\v{r}''(t) = \v{0}\text{,}\) the cross product \(\v{r}'(t) \times \v{r}''(t)\) is the zero vector \(\v{0}\text{.}\) Its magnitude is \(0\text{.}\) Thus, the curvature is:
\begin{equation*} \kappa(t) = \frac{0}{\|\v{r}'(t)\|^3} = 0 \end{equation*}
(This makes geometric sense, as the parametrization describes a straight line, which has zero curvature everywhere).
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13.4.13.

Evaluate the curvature of the curve \(\v{r}(t) = \la \cos(t), \sin(t), t^2 \ra\) at the point where \(t=\frac{\pi}{2}\) using Theorem 13.4.6.
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Solution.
We first compute the velocity and acceleration vectors:
\begin{align*} \v{r}'(t) \amp= \la -\sin(t), \cos(t), 2t \ra \\ \v{r}''(t) \amp= \la -\cos(t), -\sin(t), 2 \ra \end{align*}
Evaluating these at \(t = \frac{\pi}{2}\text{:}\)
\begin{align*} \v{r}'\lp \frac{\pi}{2} \rp \amp= \la -1, 0, \pi \ra \\ \v{r}''\lp \frac{\pi}{2} \rp \amp= \la 0, -1, 2 \ra \end{align*}
Next, compute the cross product:
\begin{align*} \v{r}'\lp \frac{\pi}{2} \rp \times \v{r}''\lp \frac{\pi}{2} \rp \amp= \begin{vmatrix} \v{i} \amp \v{j} \amp \v{k} \\ -1 \amp 0 \amp \pi \\ 0 \amp -1 \amp 2 \end{vmatrix} \\ \amp= \la 0 - (-\pi), 0 - (-2), 1 - 0 \ra = \la \pi, 2, 1 \ra \end{align*}
Now, find the magnitudes:
\begin{align*} \left\|\v{r}'\lp \frac{\pi}{2} \rp \times \v{r}''\lp \frac{\pi}{2} \rp\right\| \amp= \sqrt{\pi^2 + 2^2 + 1^2} = \sqrt{\pi^2 + 5} \\ \left\|\v{r}'\lp \frac{\pi}{2} \rp\right\| \amp= \sqrt{(-1)^2 + 0^2 + \pi^2} = \sqrt{1 + \pi^2} \end{align*}
Using Theorem 13.4.6, the curvature is:
\begin{equation*} \kappa\lp \frac{\pi}{2} \rp = \frac{\sqrt{\pi^2 + 5}}{\lp \sqrt{1 + \pi^2} \rp^3} = \frac{\sqrt{\pi^2 + 5}}{\lp 1 + \pi^2 \rp^{3/2}} \end{equation*}
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13.4.17.

Find the curvature of the plane curve \(y = t^4\) at the point where \(t=2\text{.}\)
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Solution.
We can treat this as a function \(f(t) = t^4\) and use the curvature formula for a graph from Theorem 13.4.13:
\begin{equation*} \kappa(t) = \frac{|f''(t)|}{\lp 1 + \lp f'(t) \rp^2 \rp^\frac{3}{2}} \end{equation*}
Compute the derivatives:
\begin{align*} f'(t) \amp= 4t^3 \implies f'(2) = 4(8) = 32 \\ f''(t) \amp= 12t^2 \implies f''(2) = 12(4) = 48 \end{align*}
Substitute these values into the formula:
\begin{equation*} \kappa(2) = \frac{|48|}{\lp 1 + (32)^2 \rp^\frac{3}{2}} = \frac{48}{\lp 1 + 1024 \rp^\frac{3}{2}} = \frac{48}{1025^\frac{3}{2}} \end{equation*}
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13.4.25.

Show that the curvature function of the parametrization \(\v{r}(t) = \la a\cos(t), b\sin(t) \ra \) of the ellipse \(\lp \dfrac{x}{a} \rp^2 + \lp \dfrac{y}{b} \rp^2 = 1\) is
\begin{equation*} \kappa(t) = \frac{ab}{\lp b^2\cos^2(t) + a^2\sin^2(t)\rp^\frac{3}{2}} \end{equation*}
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Solution.
We use the curvature formula for a plane curve \(\v{r}(t) = \la x(t), y(t) \ra\text{:}\)
\begin{equation*} \kappa(t) = \frac{|x'(t)y''(t) - y'(t)x''(t)|}{\lp \lp x'(t)\rp^2 + \lp y'(t)\rp^2 \rp^\frac{3}{2}} \end{equation*}
Compute the first and second derivatives of the components:
\begin{align*} x(t) \amp= a\cos(t) \amp y(t) \amp= b\sin(t) \\ x'(t) \amp= -a\sin(t) \amp y'(t) \amp= b\cos(t) \\ x''(t) \amp= -a\cos(t) \amp y''(t) \amp= -b\sin(t) \end{align*}
Now, evaluate the numerator:
\begin{align*} x'(t)y''(t) - y'(t)x''(t) \amp= (-a\sin(t))(-b\sin(t)) - (b\cos(t))(-a\cos(t)) \\ \amp= ab\sin^2(t) + ab\cos^2(t) \\ \amp= ab\lp \sin^2(t) + \cos^2(t) \rp = ab \end{align*}
Evaluate the term inside the denominator:
\begin{equation*} \lp x'(t)\rp^2 + \lp y'(t)\rp^2 = (-a\sin(t))^2 + (b\cos(t))^2 = a^2\sin^2(t) + b^2\cos^2(t) \end{equation*}
Substitute these into the curvature formula (and note that \(a,b \gt 0\) typically for an ellipse, so \(|ab| = ab\)):
\begin{equation*} \kappa(t) = \frac{ab}{\lp a^2\sin^2(t) + b^2\cos^2(t) \rp^\frac{3}{2}} \end{equation*}
Since addition is commutative, we have verified the given formula.
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13.4.31.

Compute the curvature of \(\la t\cos(t), \sin(t) \ra\) at the point where \(t=\pi\) using [insert formula].
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Solution.
We use the curvature formula for a plane curve. First, find the derivatives:
\begin{align*} x(t) \amp= t\cos(t) \implies x'(t) = \cos(t) - t\sin(t) \\ x''(t) \amp= -\sin(t) - \lp \sin(t) + t\cos(t) \rp = -2\sin(t) - t\cos(t) \\ y(t) \amp= \sin(t) \implies y'(t) = \cos(t) \implies y''(t) = -\sin(t) \end{align*}
Evaluate these derivatives at \(t = \pi\text{:}\)
\begin{align*} x'(\pi) \amp= \cos(\pi) - \pi\sin(\pi) = -1 - 0 = -1 \\ x''(\pi) \amp= -2\sin(\pi) - \pi\cos(\pi) = 0 - \pi(-1) = \pi \\ y'(\pi) \amp= \cos(\pi) = -1 \\ y''(\pi) \amp= -\sin(\pi) = 0 \end{align*}
Substitute these into the curvature formula:
\begin{align*} \kappa(\pi) \amp= \frac{|x'(\pi)y''(\pi) - y'(\pi)x''(\pi)|}{\lp \lp x'(\pi)\rp^2 + \lp y'(\pi)\rp^2 \rp^\frac{3}{2}} \\ \amp= \frac{|(-1)(0) - (-1)(\pi)|}{\lp (-1)^2 + (-1)^2 \rp^\frac{3}{2}} \\ \amp= \frac{|\pi|}{\lp 2 \rp^\frac{3}{2}} = \frac{\pi}{2\sqrt{2}} \end{align*}
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13.4.37.

Find the normal vector \(\v{N}(t)\) to \(\v{r} = \la 4, \sin(2t), \cos(2t) \ra\text{.}\)
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Solution.
First, find the velocity vector and its magnitude:
\begin{align*} \v{r}'(t) \amp= \la 0, 2\cos(2t), -2\sin(2t) \ra \\ \|\v{r}'(t)\| \amp= \sqrt{0^2 + 4\cos^2(2t) + 4\sin^2(2t)} = \sqrt{4\lp \cos^2(2t) + \sin^2(2t) \rp} = \sqrt{4} = 2 \end{align*}
Next, find the unit tangent vector \(\v{T}(t)\text{:}\)
\begin{equation*} \v{T}(t) = \frac{\v{r}'(t)}{\|\v{r}'(t)\|} = \frac{1}{2} \la 0, 2\cos(2t), -2\sin(2t) \ra = \la 0, \cos(2t), -\sin(2t) \ra \end{equation*}
To find the normal vector \(\v{N}(t)\text{,}\) compute \(\v{T}'(t)\) and its magnitude:
\begin{align*} \v{T}'(t) \amp= \la 0, -2\sin(2t), -2\cos(2t) \ra \\ \|\v{T}'(t)\| \amp= \sqrt{0^2 + 4\sin^2(2t) + 4\cos^2(2t)} = \sqrt{4} = 2 \end{align*}
Finally, the normal vector is:
\begin{equation*} \v{N}(t) = \frac{\v{T}'(t)}{\|\v{T}'(t)\|} = \frac{1}{2} \la 0, -2\sin(2t), -2\cos(2t) \ra = \la 0, -\sin(2t), -\cos(2t) \ra \end{equation*}
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13.4.39.

Find the normal vectors to \(\v{r}(t) = \la t, \cos(t) \ra\) at \(t = \dfrac{\pi}{4}\) and \(t = \dfrac{3\pi}{4}\text{.}\)
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Solution.
The normal vector to \(\v{r}(t) = \la t, \cos(t) \ra\) is \(\v{T}'(t)\text{,}\) where \(\v{T}(t) = \frac{\v{r}'(t)}{\|\v{r}'(t)\|}\) is the unit tangent vector. We have
\begin{align*} \v{r}'(t) \amp= \la 1, -\sin(t) \ra \\ \|\v{r}'(t)\| \amp= \sqrt{1 + \sin^2(t)} \end{align*}
Hence,
\begin{equation*} \v{T}(t) = \frac{1}{\sqrt{1 + \sin^2(t)}}\la 1, -\sin(t) \ra \end{equation*}
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We compute the derivative of \(\v{T}(t)\) to find the normal vector. We use the Product Rule and the Chain Rule to obtain
\begin{align*} \v{T}'(t) \amp= \frac{1}{\sqrt{1 + \sin^2(t)}} \frac{d}{dt} \la 1, -\sin(t) \ra + \lp \frac{1}{\sqrt{1 + \sin^2(t)}} \rp' \la 1, -\sin(t) \ra \\ \amp= \frac{1}{\sqrt{1 + \sin^2(t)}} \la 0, -\cos(t) \ra - \frac{1}{1 + \sin^2(t)} \cdot \frac{2\sin(t)\cos(t)}{2\sqrt{1 + \sin^2(t)}} \la 1, -\sin(t) \ra \\ \amp= \frac{1}{\sqrt{1 + \sin^2(t)}} \la 0,-\cos(t) \ra - \frac{\sin(2t)}{2\lp 1 + \sin^2(t) \rp^{3/2}} \la 1, -\sin(t) \ra \end{align*}
At \(t = \dfrac{\pi}{4}\text{,}\) we obtain the normal vector
\begin{align*} \v{T}'\lp \frac{\pi}{4} \rp \amp= \frac{1}{\sqrt{1 + \frac{1}{2}}} \la 0, -\frac{1}{\sqrt{2}} \ra - \frac{1}{2 \lp 1 + \frac{1}{2} \rp^{3/2}} \la 1, -\frac{1}{\sqrt{2}} \ra \\ \amp= \la 0, -\frac{1}{\sqrt{3}} \ra - \la \frac{\sqrt{2}}{3\sqrt{3}}, \frac{-1}{3\sqrt{3}} \ra \\ \amp= \la \frac{-\sqrt{2}}{3\sqrt{3}}, \frac{-2}{3\sqrt{3}} \ra \end{align*}
At \(t = \dfrac{3\pi}{4}\text{,}\) we obtain
\begin{align*} \v{T}'\lp \frac{3\pi}{4} \rp \amp= \frac{1}{\sqrt{1 + \frac{1}{2}}} \la 0, \frac{1}{\sqrt{2}} \ra - \frac{-1}{2 \lp 1 + \frac{1}{2} \rp^{3/2}} \la 1, -\frac{1}{\sqrt{2}} \ra \\ \amp= \la 0, \frac{1}{\sqrt{3}} \ra + \la \frac{\sqrt{2}}{3\sqrt{3}}, \frac{-1}{3\sqrt{3}} \ra \\ \amp= \la \frac{\sqrt{2}}{3\sqrt{3}}, \frac{2}{3\sqrt{3}} \ra \end{align*}
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13.4.43.

Find \(\v{T}\text{,}\) \(\v{N}\text{,}\) and \(\v{B}\) for the curve \(\v{r}(t) = \la t,t^2,\dfrac{2}{3}t^3 \ra\) at the point \(\lp 1,1,\dfrac{2}{3} \rp\text{.}\)
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Hint.
After finding \(\v{T}'\text{,}\) plug in the specific value for \(t\) before computing \(\v{N}\) and \(\v{B}\text{.}\)
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Solution.
The point \(\lp 1, 1, \frac{2}{3} \rp\) corresponds to \(t=1\text{.}\) Let’s find the derivatives and speed.
\begin{align*} \v{r}'(t) \amp= \la 1, 2t, 2t^2 \ra \\ v(t) \amp= \sqrt{1 + 4t^2 + 4t^4} = \sqrt{(1+2t^2)^2} = 1+2t^2 \end{align*}
Unit Tangent Vector \(\v{T}\text{:}\)
\begin{align*} \v{T}(t) \amp= \frac{\v{r}'(t)}{v(t)} = \la \frac{1}{1+2t^2}, \frac{2t}{1+2t^2}, \frac{2t^2}{1+2t^2} \ra \\ \v{T}(1) \amp= \la \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \ra \end{align*}
Normal Vector \(\v{N}\text{:}\) We differentiate each component of \(\v{T}(t)\) using the Quotient Rule:
\begin{align*} x_{\v{T}}'(t) \amp= \frac{0 - 1(4t)}{(1+2t^2)^2} = \frac{-4t}{(1+2t^2)^2} \\ y_{\v{T}}'(t) \amp= \frac{2(1+2t^2) - 2t(4t)}{(1+2t^2)^2} = \frac{2 - 4t^2}{(1+2t^2)^2} \\ z_{\v{T}}'(t) \amp= \frac{4t(1+2t^2) - 2t^2(4t)}{(1+2t^2)^2} = \frac{4t}{(1+2t^2)^2} \end{align*}
Evaluate at \(t=1\text{:}\)
\begin{align*} \v{T}'(1) \amp= \la \frac{-4}{9}, \frac{-2}{9}, \frac{4}{9} \ra \\ \|\v{T}'(1)\| \amp= \sqrt{\frac{16}{81} + \frac{4}{81} + \frac{16}{81}} = \sqrt{\frac{36}{81}} = \frac{6}{9} = \frac{2}{3} \end{align*}
Normalize to find \(\v{N}(1)\text{:}\)
\begin{equation*} \v{N}(1) = \frac{3}{2} \la -\frac{4}{9}, -\frac{2}{9}, \frac{4}{9} \ra = \la -\frac{2}{3}, -\frac{1}{3}, \frac{2}{3} \ra \end{equation*}
Binormal Vector \(\v{B}\text{:}\)
\begin{align*} \v{B}(1) \amp= \v{T}(1) \times \v{N}(1) = \begin{vmatrix} \v{i} \amp \v{j} \amp \v{k} \\ 1/3 \amp 2/3 \amp 2/3 \\ -2/3 \amp -1/3 \amp 2/3 \end{vmatrix} \\ \amp= \frac{1}{9} \begin{vmatrix} \v{i} \amp \v{j} \amp \v{k} \\ 1 \amp 2 \amp 2 \\ -2 \amp -1 \amp 2 \end{vmatrix} = \frac{1}{9} \la 4 - (-2), -(2 - (-4)), -1 - (-4) \ra \\ \amp= \frac{1}{9} \la 6, -6, 3 \ra = \la \frac{2}{3}, -\frac{2}{3}, \frac{1}{3} \ra \end{align*}
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13.4.49.

Find \(\v{N}\) for the curve \(\la \dfrac{t^2}{2}, \dfrac{t^3}{3}, t \ra\) at the point where \(t=1\) using the formula
\begin{equation*} \v{N}(t) = \frac{v(t)\v{r}''(t) - v'(t)\v{r}'(t)}{\|v(t)\v{r}''(t) - v'(t)\v{r}'(t)\|} \end{equation*}
where \(v(t) = \|\v{r}'(t) \|\text{.}\)
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Solution.
First compute the velocity and acceleration vectors, and the speed:
\begin{align*} \v{r}'(t) \amp= \la t, t^2, 1 \ra \implies \v{r}'(1) = \la 1, 1, 1 \ra \\ \v{r}''(t) \amp= \la 1, 2t, 0 \ra \implies \v{r}''(1) = \la 1, 2, 0 \ra \\ v(t) \amp= \|\v{r}'(t)\| = \sqrt{t^2 + t^4 + 1} \implies v(1) = \sqrt{3} \\ v'(t) \amp= \frac{2t + 4t^3}{2\sqrt{t^2+t^4+1}} = \frac{t + 2t^3}{\sqrt{t^2+t^4+1}} \implies v'(1) = \frac{3}{\sqrt{3}} = \sqrt{3} \end{align*}
Now compute the unnormalized vector \(\v{u} = v(1)\v{r}''(1) - v'(1)\v{r}'(1)\text{:}\)
\begin{align*} \v{u} \amp= \sqrt{3} \la 1, 2, 0 \ra - \sqrt{3} \la 1, 1, 1 \ra \\ \amp= \la \sqrt{3} - \sqrt{3}, 2\sqrt{3} - \sqrt{3}, 0 - \sqrt{3} \ra = \la 0, \sqrt{3}, -\sqrt{3} \ra = \sqrt{3}\la 0, 1, -1 \ra \end{align*}
Find the magnitude and normalize to get \(\v{N}\text{:}\)
\begin{align*} \|\v{u}\| \amp= \sqrt{3}\sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{3}\sqrt{2} = \sqrt{6} \\ \v{N}(1) \amp= \frac{\sqrt{3}\la 0, 1, -1 \ra}{\sqrt{6}} = \frac{1}{\sqrt{2}} \la 0, 1, -1 \ra = \la 0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \ra \end{align*}
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13.4.55.

Let \(\v{r}(t) = \la t,1-t,t^2 \ra\text{.}\)
  1. Find the general formulas for \(\v{T}\) and \(\v{N}\) as functions of \(t\text{.}\)
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  2. Find the general formula for \(\v{B}\) as a function of \(t\text{.}\)
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  3. What can you conclude about the osculating planes of the curve based on your answer to (b)?
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Solution.
  1. Differentiating gives \(\v{r}'(t) = \la 1, -1, 2t \ra\text{,}\) so that
    \begin{align*} \v{T}(t) \amp= \frac{\v{r}'(t)}{\|\v{r}'(t)\|} \\ \amp= \frac{1}{\sqrt{1^2 + (-1)^2 + (2t)^2}} \la 1, -1, 2t \ra \\ \amp= \left\langle \frac{1}{\sqrt{4t^2+2}}, -\frac{1}{\sqrt{4t^2+2}}, \frac{2t}{\sqrt{4t^2+2}} \right\rangle \end{align*}
    The derivatives of the components of \(\v{T}(t)\) are
    \begin{align*} \frac{d}{dt} \lp \frac{1}{\sqrt{4t^2+2}} \rp \amp= -\frac{4t}{(4t^2+2)^{3/2}} \\ \frac{d}{dt} \lp -\frac{1}{\sqrt{4t^2+2}} \rp \amp= \frac{4t}{(4t^2+2)^{3/2}} \\ \frac{d}{dt} \lp \frac{2t}{\sqrt{4t^2+2}} \rp \amp= \frac{(4t^2+2)^{1/2}(2) - 2t \cdot \frac{1}{2}(4t^2+2)^{-1/2} \cdot 8t}{4t^2+2} \\ \amp= \frac{2(4t^2+2) - 8t^2}{(4t^2+2)^{3/2}} = \frac{4}{(4t^2+2)^{3/2}} \end{align*}
    Thus,
    \begin{align*} \v{T}'(t) \amp= \left\langle -\frac{4t}{(4t^2+2)^{3/2}}, \frac{4t}{(4t^2+2)^{3/2}}, \frac{4}{(4t^2+2)^{3/2}} \right\rangle \\ \amp= \frac{1}{(4t^2+2)^{3/2}} \la -4t, 4t, 4 \ra \end{align*}
    To find \(\v{N}(t)\) we must find \(\|\v{T}'(t)\|\text{:}\)
    \begin{align*} \|\v{T}'(t)\| \amp= \frac{1}{(4t^2+2)^{3/2}} \sqrt{(-4t)^2 + (4t)^2 + 4^2} \\ \amp= \frac{1}{(4t^2+2)^{3/2}} \sqrt{32t^2+16} \\ \amp= \frac{\sqrt{8}\sqrt{4t^2+2}}{(4t^2+2)^{3/2}} = \frac{\sqrt{8}}{4t^2+2} \end{align*}
    So
    \begin{align*} \v{N}(t) \amp= \frac{\v{T}'(t)}{\|\v{T}'(t)\|} = \frac{\frac{1}{(4t^2+2)^{3/2}} \la -4t, 4t, 4 \ra}{\frac{\sqrt{8}}{4t^2+2}} \\ \amp= \frac{4t^2+2}{(4t^2+2)^{3/2}\sqrt{8}} \la -4t, 4t, 4 \ra \\ \amp= \frac{1}{\sqrt{4t^2+2}\sqrt{8}} \la -4t, 4t, 4 \ra \\ \amp= \frac{1}{\sqrt{32t^2+16}} \la -4t, 4t, 4 \ra \\ \amp= \frac{1}{\sqrt{16(2t^2+1)}} 4 \la -t, t, 1 \ra \\ \amp= \frac{1}{4\sqrt{2t^2+1}} 4 \la -t, t, 1 \ra = \frac{1}{\sqrt{2t^2+1}} \la -t, t, 1 \ra \end{align*}
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  2. Using the cross product formula \(\v{B}(t) = \v{T}(t) \times \v{N}(t)\text{:}\)
    \begin{align*} \v{B}(t) \amp= \lp \frac{1}{\sqrt{4t^2+2}} \rp \lp \frac{1}{\sqrt{2t^2+1}} \rp \begin{vmatrix} \v{i} \amp \v{j} \amp \v{k} \\ 1 \amp -1 \amp 2t \\ -t \amp t \amp 1 \end{vmatrix} \\ \amp= \frac{1}{\sqrt{2}\sqrt{2t^2+1}\sqrt{2t^2+1}} \la -1-2t^2, -(1-(-2t^2)), t-t \ra \\ \amp= \frac{1}{\sqrt{2}(2t^2+1)} \la -(1+2t^2), -(1+2t^2), 0 \ra \\ \amp= \frac{-(1+2t^2)}{\sqrt{2}(1+2t^2)} \la 1, 1, 0 \ra \\ \amp= \left\langle -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0 \right\rangle \end{align*}
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  3. Since the binormal vector \(\v{B}(t) = \la -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0 \ra\) is constant, the normal vector to the osculating plane never changes. Therefore, the osculating plane is the same for all \(t\text{.}\) All osculating planes are the fixed plane \(x + y = 1\text{.}\)
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13.4.67.

Find an equation of the osculating circle to the curve \(\v{r}(t) = \la 1-\sin(t), 1-2\cos(t) \ra\) at the point where \(t=\pi\) or indicate that none exists.
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Solution.
Step 1: Compute derivatives at \(t=\pi\)
\begin{align*} \v{r}(\pi) \amp= \la 1-\sin(\pi), 1-2\cos(\pi) \ra = \la 1, 3 \ra \\ \v{r}'(t) \amp= \la -\cos(t), 2\sin(t) \ra \implies \v{r}'(\pi) = \la 1, 0 \ra \\ \v{r}''(t) \amp= \la \sin(t), 2\cos(t) \ra \implies \v{r}''((\pi) = \la 0, -2 \ra \end{align*}
Step 2: Find the radius of curvature
\begin{align*} \kappa(\pi) \amp= \frac{|x'y'' - y'x''|}{\lp (x')^2 + (y')^2 \rp^{3/2}} = \frac{|(1)(-2) - (0)(0)|}{\lp 1^2 + 0^2 \rp^{3/2}} = 2 \\ R \amp= \frac{1}{\kappa(\pi)} = \frac{1}{2} \end{align*}
Step 3: Find the Normal Vector Since the velocity vector is \(\v{r}'(\pi) = \la 1, 0 \ra\text{,}\) the unit tangent vector is \(\v{T}(\pi) = \la 1, 0 \ra\text{.}\) In 2D, the principal normal vector \(\v{N}\) is orthogonal to \(\v{T}\) and points in the direction the curve is bending (concave side). Since the acceleration is \(\v{r}''(\pi) = \la 0, -2 \ra\text{,}\) the curve is bending downwards. Thus:
\begin{equation*} \v{N}(\pi) = \la 0, -1 \ra \end{equation*}
Step 4: Find the Center and Equation The center \(Q\) is found by translating from the point \(P(1,3)\) along the normal vector by the distance \(R\text{:}\)
\begin{equation*} \overrightarrow{OQ} = \v{r}(\pi) + R\v{N}(\pi) = \la 1, 3 \ra + \frac{1}{2}\la 0, -1 \ra = \la 1, \frac{5}{2} \ra \end{equation*}
With center \(\lp 1, \frac{5}{2} \rp\) and radius \(R = \frac{1}{2}\text{,}\) the equation of the osculating circle is:
\begin{equation*} (x - 1)^2 + \lp y - \frac{5}{2} \rp^2 = \frac{1}{4} \end{equation*}
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