Question

Let \( x(t), y(t), 1 \leq t \leq \pi \), be the curve defined by \[ x(t) = \int_1^t \frac{\cos z}{z^2} \, dz \quad \text{and} \quad y(t) = \int_1^t \frac{\sin z}{z^2} \, dz. \] Let \( L \) be the length of the arc of this curve from the origin to the point \( P \) on the curve at which the tangent is perpendicular to the x-axis. Then \( L \) equals

• A. \( \sqrt{2} \)
• B. \( \frac{\pi}{\sqrt{2}} \)
• C. \( 1 - \frac{2}{\pi} \)
• D. \( \frac{\pi}{2} + \sqrt{2} \)

Hint:

To compute the arc length, use the formula for the length of a curve, and apply the conditions for perpendicular tangents.

Answer 1

The Correct Option is C

Step 1: Compute the arc length.
The length of a curve is given by the formula: \[ L = \int_1^t \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \] First, compute the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). The derivatives are: \[ \frac{dx}{dt} = \frac{\cos t}{t^2}, \quad \frac{dy}{dt} = \frac{\sin t}{t^2} \] Substitute these into the arc length formula: \[ L = \int_1^t \sqrt{ \left( \frac{\cos t}{t^2} \right)^2 + \left( \frac{\sin t}{t^2} \right)^2 } \, dt = \int_1^t \frac{1}{t^2} \sqrt{\cos^2 t + \sin^2 t} \, dt = \int_1^t \frac{1}{t^2} \, dt \] The integral of \( \frac{1}{t^2} \) is \( -\frac{1}{t} \), so the arc length becomes: \[ L = \left[ -\frac{1}{t} \right]_1^t = 1 - \frac{1}{t} \]
Step 2: Find when the tangent is perpendicular to the x-axis.
The tangent is perpendicular to the x-axis when \( \frac{dx}{dt} = 0 \), which happens when \( \cos t = 0 \). This occurs at \( t = \frac{\pi}{2} \), so the point \( P \) is at \( t = \frac{\pi}{2} \).
Step 3: Conclusion.
Substituting \( t = \frac{\pi}{2} \) into the arc length formula, we get: \[ L = 1 - \frac{1}{\frac{\pi}{2}} = 1 - \frac{2}{\pi} \] Thus, the correct answer is \( 1 - \frac{2}{\pi} \), which corresponds to option (C).