Edexcel IAL January 2022 P4 (WMA14/01) Q5 Integration by Substitution, Parametric Equation
Figure 2 shows a sketch of the curve with parametric equations
\[x = \sqrt{9 – 4t}\]
\[y = \frac{t^3}{\sqrt{9 + 4t}}\]
\[0 \leq t \leq \frac{9}{4}\]
The curve touches the x-axis when t=0 and meets the y-axis when \( t = \frac{9}{4} \).
The region R, shown shaded in Figure 2, is bounded by the curve, the x-axis and the y-axis.
(a) Show that the area of R is given by
\[K \int_{0}^{\frac{9}{4}} \frac{t^3}{\sqrt{81 – 16t^2}} dt\]
where K is a constant to be found.
(b) Using the substitution \( u = 81 – 16t^2 \), or otherwise, find the exact area of R.
(Solutions relying on calculator technology are not acceptable.)
Solution to Part a: Establishing the Integral Form
Key Concepts Used:
- Parametric Differentiation: Using chain rule to express \(\frac{dx}{dt}\).
- Algebraic Manipulation: Combining square root terms using difference of squares formula.
- Limit Adjustment: Reversing integration limits and adjusting signs.
Step-by-Step Solution:
1. Compute \(\frac{dx}{dt}\)
Given parametric equation:
1. Compute \(\frac{dx}{dt}\)
Differentiate using chain rule:
\[\frac{dx}{dt} = \frac{1}{2}(9 – 4t)^{-1/2} \cdot (-4) = \frac{-2}{\sqrt{9 – 4t}}\]
2. Set Up Area Integral
Area formula for parametric curves:
\[Area = \int_{t_1}^{t_2} y \frac{dx}{dt} dt\]
Given:
\[y = \frac{t^3}{\sqrt{9 + 4t}}\]
Limits: \( t = 0 \) to \( t = \frac{9}{4} \)
Thus:
\[Area = \int_{\frac{9}{4}}^{0} \frac{t^3}{\sqrt{9 + 4t}} \cdot \frac{-2}{\sqrt{9 – 4t}} dt\]
3. Simplify the Integrand
Combine square roots:
\[\sqrt{(9 + 4t)(9 – 4t)} = \sqrt{81 – 16t^2}\]
Thus:
\[Area = 2 \int_{0}^{\frac{9}{4}} \frac{t^3}{\sqrt{81 – 16t^2}} dt\]
4. Identify Constant K
Comparing with given form:
\[K \int_{0}^{\frac{9}{4}} \frac{t^3}{\sqrt{81 – 16t^2}} dt\]
We see \( K = 2 \).
Final Answer:
The area expression is valid with \( K = 2 \)
Solution to Part b: Evaluating the Integral
Key Concepts Used:
- Substitution Method: Let \( u = 81 – 16t^2 \) to simplify the integral.
- Radical Simplification: Expressing \( t \) in terms of \( u \).
- Power Rule Integration: Applying standard integration techniques to polynomial terms.
Step-by-Step Solution:
1. Apply Substitution
Let:
\[u = 81 – 16t^2\]
\[\frac{du}{dt} = -32t\]
\[dt = \frac{-du}{32t}\]
2. Change Limits
When \( t = 0 \): \( u = 81 \)
When \( t = \frac{9}{4} \): \( u = 0 \)
3. Rewrite the Integral
Original integral:
\[2 \int_{0}^{\frac{9}{4}} \frac{t^3}{\sqrt{81 – 16t^2}} dt\]
Substitute:
\[= 2 \int_{81}^{0} \frac{t^3}{\sqrt{u}} \cdot \frac{-du}{32t}\]
\[= \frac{2}{32} \int_{0}^{81} \frac{t^2}{\sqrt{u}} du\]
Express \( t^2 \) in terms of \( u \):
\[t^2 = \frac{81 – u}{16}\]
Thus:
\[= \frac{1}{16} \int_{0}^{81} \frac{81 – u}{16\sqrt{u}} du\]
\[= \frac{1}{256} \int_{0}^{81} (81u^{-1/2} – u^{1/2}) du\]
4. Integrate Term by Term
\[\frac{1}{256} \int_{0}^{81} (81u^{-1/2} – u^{1/2}) du = \frac{1}{256} \left[ 162u^{1/2} – \frac{2}{3}u^{3/2} \right]_{0}^{81}\]
5. Evaluate at Limits
At \( u = 81 \):
\[162 \times 9 – \frac{2}{3} \times 729 = 1458 – 486 = 972\]
At \( u = 0 \):
\[0 – 0 = 0\]
Thus:
\[\frac{1}{256} \times 972 = \frac{243}{64}\]
Final Answer:
\[\frac{243}{64}\]
