Differentiation(P4)

1. Find Coordinates of Point \( P \)
   
   Substitute \( t = \frac{\pi}{4} \) into the parametric equations:
   \[ x = \sin^2 \left( \frac{\pi}{4} \right) = \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{1}{2} \]
   \[ y = 2\tan \left( \frac{\pi}{4} \right) = 2(1) = 2 \]
   So, \( P \) has coordinates \( \left( \frac{1}{2}, 2 \right) \).
   
2. Compute \(\frac{dy}{dx}\) (Gradient of Tangent)
   
   First, find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
   \[ \frac{dx}{dt} = 2\sin t \cos t = \sin 2t \]
   \[ \frac{dy}{dt} = 2\sec^2 t \]
   According to Chain Rule, the gradient of the tangent is:
   \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2\sec^2 t}{\sin 2t} \]
   Simplify using \( \sec t = \frac{1}{\cos t} \) and \(\sin 2t = 2\sin t \cos t\):
   \[ \frac{dy}{dx} = \frac{2/\cos^2 t}{2\sin t \cos t} = \frac{1}{\sin t \cos^3 t} \]
   At \( t = \frac{\pi}{4} \):
   \[ \frac{dy}{dx} = \frac{1}{\left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{2}}{2} \right)^3} = \frac{1}{\frac{\sqrt{2}}{2} \cdot \frac{2\sqrt{2}}{8}} = \frac{1}{\frac{4}{8}} = \frac{1}{\frac{1}{2}} = 2 \]
   
3. Find Gradient of Normal Line \( l \)
   
   The normal line is perpendicular to the tangent, so its gradient \( m \) satisfies:
   \[ m \times 2 = -1 \implies m = -\frac{1}{2} \]
   
4. Equation of Normal Line \( l \)
   
   Using point-slope form with \( P \left( \frac{1}{2}, 2 \right) \):
   \[ y – 2 = -\frac{1}{2} \left( x – \frac{1}{2} \right) \]
   Multiply through by 8 to eliminate fractions:
   \[ 8(y – 2) = -4 \left( x – \frac{1}{2} \right) \]
   \[ 8y – 16 = -4x + 2 \]
   Rearrange to standard form:
   \[ 8y + 4x = 18 \]
   \[ 4y + 2x = 9 \]
   \[ 8y + 2x = 17 \]
   

1. Evaluating Total Area S:
   The region S is bounded between the curve, line and x-axis. The shaded region may be divided into two parts. The first part is the area under the graph from \( x = 0 \) to \( x = \frac{1}{2} \). And the second part is enclosed between the line and x-axis which makes a triangle as shown below. The y-intercept of line is 17/2. And with the help of y-coordinate of point P the height of the triangle is 2.
2. Therefore,
   
   \[ Area = \int_{0}^{\frac{1}{2}} y \, dx + Area \, of \, Triangle \]
   \[ \int_{0}^{\frac{1}{2}} y \, dx = \int_{0}^{\frac{\pi}{4}} y \times \frac{dx}{dt} \times dt \]
   Note, the limit has changed as when \( x = \frac{1}{2} t \) is equals to \(\frac{\pi}{4}\).
   Where,
   \[ y = 2 \tan t \]
   \[ \frac{dx}{dt} = 2 \sin t \cos t \]
   So,
   \[ \int_{0}^{\frac{\pi}{4}} y \, dx = \int_{0}^{\frac{\pi}{4}} 2 \tan t \cdot 2 \sin t \cos t \, dt \]
   \[ \int_{0}^{\frac{\pi}{4}} y \, dx = 4 \int_{0}^{\frac{\pi}{4}} \frac{\sin t}{\cos t} \times \sin t \cos t \, dt \]
   \[ \int_{0}^{\frac{\pi}{4}} y \, dx = 4 \int_{0}^{\frac{\pi}{4}} \sin^2 t \, dt \]
   \[ \int_{0}^{\frac{\pi}{4}} y \, dx = 4 \int_{0}^{\frac{\pi}{4}} \frac{1}{2} (1 – \cos 2t) \, dt \]
   \[ \int_{0}^{\frac{\pi}{4}} y \, dx = 2 \int_{0}^{\frac{\pi}{4}} (1 – \cos 2t) \, dt \]
   \[ \int_{0}^{\frac{\pi}{4}} y \, dx = \int_{0}^{\frac{\pi}{4}} (2 – 2 \cos 2t) \, dt \]
   \[ \int_{0}^{\frac{\pi}{4}} y \, dx = [2t – \sin 2t]_{0}^{\frac{\pi}{4}} \]
   \[ \int_{0}^{\frac{1}{2}} y \, dx = \frac{\pi}{2} – \sin \frac{\pi}{2} – [0 – 0] \]
   \[ \int_{0}^{\frac{1}{2}} y \, dx = \frac{\pi}{2} – 1 \]
   

The exact area of S is: