To find the arc length of the graph of the vector function $\mathbf{r}(t) = (7 \sin t + 2) \, \mathbf{i} + (7 \cos t + 2) \, \mathbf{j} + 7t \, \mathbf{k}$ over the interval $0 \leq t \leq \pi$, we can use the arc length formula for a vector function:

$L = \int_{a}^{b} \|\mathbf{r}'(t)\| \, dt$

where $\mathbf{r}'(t)$ is the derivative of $\mathbf{r}(t)$ with respect to $t$, and $\|\mathbf{r}'(t)\|$ is the magnitude of $\mathbf{r}'(t)$.

Step 1: Differentiate $\mathbf{r}(t)$

Given: $\mathbf{r}(t) = (7 \sin t + 2) \, \mathbf{i} + (7 \cos t + 2) \, \mathbf{j} + 7t \, \mathbf{k}$

1. The derivative with respect to $t$ for each component:
   • $x(t) = 7 \sin t + 2 \Rightarrow x'(t) = 7 \cos t$
   • $y(t) = 7 \cos t + 2 \Rightarrow y'(t) = -7 \sin t$
   • $z(t) = 7t \Rightarrow z'(t) = 7$

So, $\mathbf{r}'(t) = 7 \cos t \, \mathbf{i} - 7 \sin t \, \mathbf{j} + 7 \, \mathbf{k}$.

Step 2: Find $\|\mathbf{r}'(t)\|$

The magnitude $\|\mathbf{r}'(t)\|$ is given by: $\|\mathbf{r}'(t)\| = \sqrt{(7 \cos t)^2 + (-7 \sin t)^2 + (7)^2}$

Calculating each term:

• $(7 \cos t)^2 = 49 \cos^2 t$
• $(-7 \sin t)^2 = 49 \sin^2 t$
• $(7)^2 = 49$

Thus: $\|\mathbf{r}'(t)\| = \sqrt{49 \cos^2 t + 49 \sin^2 t + 49} = \sqrt{49(\cos^2 t + \sin^2 t) + 49} = \sqrt{49 \cdot 1 + 49} = \sqrt{98} = 7\sqrt{2}$

Step 3: Integrate $\|\mathbf{r}'(t)\|$ over the interval $[0, \pi]$

Since $\|\mathbf{r}'(t)\| = 7\sqrt{2}$ is constant, the arc length $L$ becomes: $L = \int_{0}^{\pi} 7\sqrt{2} \, dt = 7\sqrt{2} \cdot \left[ t \right]_{0}^{\pi} = 7\sqrt{2} \cdot \pi$

Final Answer

$L = 7\pi\sqrt{2}$

Would you like further details, or do you have any questions?

Here are five related questions you might find helpful:

1. How is arc length computed for functions in different coordinate systems?
2. What happens to the arc length formula if the function is not in terms of sine and cosine?
3. Can this formula be applied to find arc length for functions in polar coordinates?
4. What are some applications of arc length in physics or engineering?
5. How would this solution change if the interval for $t$ was different?

Tip: When finding the arc length for vector functions, always simplify the magnitude expression first to see if it has a constant value, as it can make integration easier.