Edexcel IAL - Pure Maths 4- 6.5 Area Under Parametric Curves- Study notes - New syllabus
Edexcel IAL – Pure Maths 4- 6.5 Area Under Parametric Curves -Study notes- New syllabus
Edexcel IAL – Pure Maths 4- 6.5 Area Under Parametric Curves -Study notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- 6.5 Area Under Parametric Curves
Area Under a Curve Using Parametric Equations
When a curve is given in parametric form, both \( x \) and \( y \) are written in terms of a third variable, usually \( t \).
\( x = g(t), \quad y = h(t) \)
The area under the curve can still be found using integration, but the formula must be adapted.
Formula for Area Using Parametric Equations
If a curve is defined by:
\( x = g(t), \quad y = h(t) \)
then the area \( A \) under the curve from \( t = a \) to \( t = b \) is:
\( A = \displaystyle\int_{a}^{b} y \dfrac{dx}{dt} \, dt \)
This is because \( dx = \dfrac{dx}{dt}dt \).
Important Notes
- You are not required to sketch the curve.
- The limits must be in terms of the parameter \( t \).
- Always calculate \( \dfrac{dx}{dt} \) before substituting.
- The area is positive when the curve moves left to right.
Steps for Finding the Area
| Step | What to Do |
| 1 | Write down \( y \) and find \( \dfrac{dx}{dt} \). |
| 2 | Substitute into \( A = \int y \dfrac{dx}{dt} dt \). |
| 3 | Change the limits to values of \( t \). |
| 4 | Integrate and simplify. |
Example
The curve is given by \( x = t \), \( y = t^2 \), where \( 0 \le t \le 2 \). Find the area under the curve.
▶️ Answer / Explanation
\( y = t^2 \), \( \dfrac{dx}{dt} = 1 \)
\( A = \displaystyle\int_{0}^{2} t^2 \cdot 1 \, dt \)
\( = \left[ \dfrac{t^3}{3} \right]_{0}^{2} = \dfrac{8}{3} \)
Example
The curve is given by \( x = t^2 \), \( y = 2t \), where \( 0 \le t \le 3 \). Find the area under the curve.
▶️ Answer / Explanation
\( y = 2t \), \( \dfrac{dx}{dt} = 2t \)
\( A = \displaystyle\int_{0}^{3} 2t \cdot 2t \, dt = \int_{0}^{3} 4t^2 \, dt \)
\( = \left[ \dfrac{4t^3}{3} \right]_{0}^{3} = 36 \)
Example
The curve is given by \( x = \cos t \), \( y = \sin t \), where \( 0 \le t \le \dfrac{\pi}{2} \). Find the area under the curve.
▶️ Answer / Explanation
\( y = \sin t \), \( \dfrac{dx}{dt} = -\sin t \)
\( A = \displaystyle\int_{0}^{\pi/2} \sin t (-\sin t)\,dt \)
\( = -\int_{0}^{\pi/2} \sin^2 t \, dt \)
\( = -\int_{0}^{\pi/2} \dfrac{1 – \cos 2t}{2}\,dt \)
\( = -\left[ \dfrac{t}{2} – \dfrac{\sin 2t}{4} \right]_{0}^{\pi/2} \)
\( = \dfrac{\pi}{4} \)