1. **State the problem:** We need to find the arc length of the curve given by the parametric equations $$x = e^t \sin(t), y = e^t \cos(t), z = 9$$ between $$t=0$$ and $$t=4$$. 2. **Recall the arc length formula for a space curve:** $$ L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt $$ 3. **Find the derivatives:** $$ \frac{dx}{dt} = \frac{d}{dt}\left(e^t \sin t\right) = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t) $$ $$ \frac{dy}{dt} = \frac{d}{dt}\left(e^t \cos t\right) = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t) $$ $$ \frac{dz}{dt} = 0 $$ since $$z$$ is constant. 4. **Square each derivative and add:** $$ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2 = \left(e^t (\sin t + \cos t)\right)^2 + \left(e^t (\cos t - \sin t)\right)^2 + 0 $$ $$ = e^{2t}(\sin t + \cos t)^2 + e^{2t}(\cos t - \sin t)^2 $$ 5. **Expand the squares:** $$ (\sin t + \cos t)^2 = \sin^2 t + 2 \sin t \cos t + \cos^2 t $$ $$ (\cos t - \sin t)^2 = \cos^2 t - 2 \sin t \cos t + \sin^2 t $$ 6. **Add the expanded terms:** $$ \sin^2 t + 2 \sin t \cos t + \cos^2 t + \cos^2 t - 2 \sin t \cos t + \sin^2 t = (\sin^2 t + \cos^2 t) + (\sin^2 t + \cos^2 t) + (2\sin t \cos t - 2\sin t \cos t) $$ $$ = 1 + 1 + 0 = 2 $$ using the Pythagorean identity. 7. **Simplify the expression under the square root:** $$ e^{2t} \times 2 = 2 e^{2t} $$ 8. **Take the square root:** $$ \sqrt{2 e^{2t}} = \sqrt{2} e^t $$ since $$e^t > 0$$. 9. **Compute the arc length integral:** $$ L = \int_0^4 \sqrt{2} e^t \, dt = \sqrt{2} \int_0^4 e^t \, dt $$ 10. **Integrate:** $$ \int_0^4 e^t \, dt = e^t \Big|_0^4 = e^4 - 1 $$ 11. **Final answer:** $$ L = \sqrt{2} (e^4 - 1) $$