Find the curvature for r(t) = 17i + 2t j + 3t² k at t = 0. C) 0 D) 1

Question

Quizplus · Accepted Answer

We can use the formula for curvature: 
$$
\kappa = \frac{\|r'(t) 	imes r''(t)\|}{\|r'(t)\|^3}
$$
where $	imes$ denotes the cross product.

First we compute $r'(t)$ and $r''(t)$:
\begin{align*}
r'(t) &= 2\mathbf{j} + 3S_1U_1P_{12}S_1S_1P_0\mathbf{k} \
r''(t) &= 3U_1P_{12}S_1S_1P_0\mathbf{k}.
\end{align*}
Note that since $S_1^2 = U_1^2 = P_0^2 = 1$ and $S_1P_0=-P_0S_1$, we have $S_1U_1P_{12}S_1S_1P_0 = -P_{12}$, which simplifies our computations.

Plugging in at $t=0$, we have:
\begin{align*}
r'(0) &= 2\mathbf{j} + 3P_{12}\mathbf{k} \
r''(0) &= 3P_{12}\mathbf{k}.
\end{align*}

Then,
\begin{align*}
r'(0) 	imes r''(0) &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \
0 & 2 & 3P_{12} \
0 & 0 & 3P_{12} \end{vmatrix} \
&= -3P_{12}\mathbf{i}.
\end{align*}

Finally,
\begin{align*}
\kappa &= \frac{\|r'(0) 	imes r''(0)\|}{\|r'(0)\|^3} \
&= \frac{3|P_{12}|}{(2^2 + 3^2|P_{12}|^2)^{3/2}}.
\end{align*}

Since $P_{12}=\frac{\sqrt{3}}{2}$, we have:
\begin{align*}
\kappa &= \frac{3|\frac{\sqrt{3}}{2}|}{(2^2 + 3^2\cdot \frac{3}{4})^{3/2}} \
&= \frac{3\sqrt{3}}{80}.
\end{align*}

Converting to the desired format, we have:
$$
	ext{Answer: } \mathbf{(E)}\;11ead0bc\_8dfd\_2da7\_99a0\_615a0505a957\_TB6988\_11
$$

Find the Curvature for R(t) = 17i + 2t