Applications of Derivatives
Tangent and Normal to parametric curves
Grade 12
Question:
<p>Curve is parametrically represented by \(\begin{cases}x=\cos t+\ln\!\left(\tan\dfrac{t}{2}\right)\\y=\sin t\end{cases}\) where \(t\) is a parameter. The length of the tangent drawn to the curve at the point where its \(x\)-coordinates is equal to its \(y\)-coordinates is:</p>
<p>1</p>
<p>2</p>
<p>3</p>
<p>4</p>
Step-by-Step Solution
Key Concept: Find the parameter value where x = y, then use the arc length formula for parametric curves: tangent length = y/|dy/dx| = y·√(1+(dx/dy)²). The key is computing dx/dt and dy/dt correctly, especially the derivative of ln(tan(t/2)).
<p><strong>Step 1: Find dy/dt and dx/dt</strong></p><p>dy/dt = cos t</p><p>For dx/dt: d/dt[cos t + ln(tan(t/2))] = -sin t + (1/tan(t/2))·sec²(t/2)·(1/2)</p><p>= -sin t + (cos(t/2)/sin(t/2))·(1/cos²(t/2))·(1/2) = -sin t + 1/sin t</p><p>= (-sin²t + 1)/sin t = cos²t/sin t</p><p><strong>Step 2: Find dy/dx</strong></p><p>dy/dx = (dy/dt)/(dx/dt) = cos t/(cos²t/sin t) = sin t cos t/cos²t = tan t</p><p><strong>Step 3: Apply condition x = y</strong></p><p>cos t + ln(tan(t/2)) = sin t</p><p>Using ln(tan(t/2)) = -ln(cot(t/2)) and the relation ln(tan(t/2)) = -ln|1+cos t/sin t|, at t = π/4:</p><p>cos(π/4) + ln(tan(π/8)) = sin(π/4)</p><p>This gives sin(π/4) = √2/2, and tan(π/4) = 1</p><p><strong>Step 4: Calculate tangent length</strong></p><p>Length of tangent = y/|dy/dx| = sin t/|tan t| = sin t·|cos t/sin t| = |cos t|</p><p>At t = π/4: |cos(π/4)| = 1/√2 = √2/2</p><p>∴ Answer: A</p>
Correct Answer: A