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

Step 1: Identify when the tangent is perpendicular to the x-axis.
A tangent is perpendicular to the x-axis when dx/dt = 0.

Given:
x(t) = ∫₁ᵗ (cos z / z²) dz
Differentiate:
dx/dt = cos t / t²

Set dx/dt = 0 ⇒ cos t = 0.
In the interval 1 ≤ t ≤ π, the only solution is t = π/2.
So, point P corresponds to t = π/2.

Step 2: Arc length formula for parametric curve.
L = ∫ √[(dx/dt)² + (dy/dt)²] dt from t = 1 to t = π/2.

Compute derivatives:
dx/dt = cos t / t²
dy/dt = sin t / t²

(dx/dt)² + (dy/dt)² = (cos²t + sin²t) / t⁴ = 1 / t⁴.
So the integrand becomes √(1 / t⁴) = 1 / t².

Thus,
L = ∫₁^(π/2) (1 / t²) dt.

Step 3: Evaluate the integral.
∫ 1/t² dt = ∫ t⁻² dt = -t⁻¹.

L = [-1/t]₁^(π/2)
= (-2/π) - (-1)
= 1 - 2/π.

Final Answer:
1 - 2/π