Calculate the arc length of the indicated portion of the curve r(t).
-$$\mathbf { r } ( t ) = ( 4 - 6 t ) \mathbf { i } + ( 9 + 9 t ) \mathbf { j } + ( 2 t - 2 ) \mathbf { k } , - 10 \leq t \leq 1$$
A) -1089
B) -99
C) 121
D) 1331

Question

Quizplus · Accepted Answer

The arc length $s$ of a curve given by a vector function $\mathbf{r}(t)$ from $t = a$ to $t = b$ is calculated using the formula $s = \int_{a}^{b} \|\mathbf{r}'(t)\| dt$, where $\|\mathbf{r}'(t)\|$ is the magnitude of the derivative of $\mathbf{r}(t)$. First, find the derivative of $\mathbf{r}(t)$:$$\mathbf{r}'(t) = \frac{d}{dt}[(4 - 6t)\mathbf{i} + (9 + 9t)\mathbf{j} + (2t - 2)\mathbf{k}] = (-6)\mathbf{i} + (9)\mathbf{j} + (2)\mathbf{k}.$$The magnitude of $\mathbf{r}'(t)$ is:$$\|\mathbf{r}'(t)\| = \sqrt{(-6)^2 + 9^2 + 2^2} = \sqrt{36 + 81 + 4} = \sqrt{121} = 11.$$Since the magnitude of the derivative is constant, the arc length from $t = -10$ to $t = 1$ is:$$s = \int_{-10}^{1} 11 dt = 11t \big|_{-10}^{1} = 11(1 - (-10)) = 11 \times 11 = 121.$$

Calculate the Arc Length of the Indicated Portion of the Curve