Find the point on the curve $r(t)=(5 \sin t) i+(5 \cos t) j+12 t k$ at a distance $52 \pi$ units along the curve from the point $(0,5,0)$ in the direction of increasing arc length. The point is (Type exact answers, using $\pi$ as needed.)
Solution
Step 1 ：First, we need to find the derivative of the function \(r(t) = (5 \sin t)i + (5 \cos t)j + 12tk\). The derivative \(r'(t)\) is \((5 \cos t)i - (5 \sin t)j + 12k\).
Step 2 ：Next, we find the magnitude of \(r'(t)\), which is \(\sqrt{(5 \cos t)^2 + (-5 \sin t)^2 + 12^2} = \sqrt{25 \sin^2 t + 25 \cos^2 t + 144}\).
Step 3 ：The arc length of the curve from the point \((0,5,0)\) to the point we are trying to find is given by the integral \(\int_0^t |r'(t)| dt\). We need to solve the equation \(\int_0^t |r'(t)| dt = 52\pi\) for \(t\).
Step 4 ：By solving the equation, we find that the value of \(t\) that satisfies the equation is \(4\pi\).
Step 5 ：Finally, we substitute this value into the original function \(r(t)\) to find the coordinates of the point on the curve that is \(52\pi\) units away from the point \((0,5,0)\). The point is \((5 \sin 4\pi, 5 \cos 4\pi, 12 \cdot 4\pi) = (0,5,48\pi)\).
Step 6 ：\(\boxed{(0,5,48\pi)}\) is the point on the curve \(r(t)=(5 \sin t) i+(5 \cos t) j+12 t k\) at a distance \(52 \pi\) units along the curve from the point \((0,5,0)\) in the direction of increasing arc length.
