Figure 4 shows a sketch of the closed curve with equation
The curve is used to model the shape of a cycle track with both \( x \) and \( y \) measured in km.
The points \( P \) and \( Q \) represent points that are furthest north and furthest south of the origin \( O \), as shown in Figure 4.
(5)
Using the result from part (a),
(b) find how far the point \( Q \) is south of \( O \), giving your answer to the nearest 100 m.
(4)
- Implicit Differentiation
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Differentiate Both Sides with Respect to \( x \)
Given:
\[ (x + y)^3 + 10y^2 – 108x = 0 \]
Differentiate term-by-term:First term \(((x + y)^3)\):
\[ \frac{d}{dx} [(x + y)^3] = 3(x + y)^2 \cdot \left( 1 + \frac{dy}{dx} \right) \]
Second term \((10y^2)\):
\[ \frac{d}{dx} [10y^2] = 20y \cdot \frac{dy}{dx} \]
Third term \((-108x)\):
\[ \frac{d}{dx} [-108x] = -108 \]
Combined Result:
\[ 3(x + y)^2 \left( 1 + \frac{dy}{dx} \right) + 20y \frac{dy}{dx} – 108 = 0 \]
-
Expand and collect \(\frac{dy}{dx}\) terms
\[ 3(x + y)^2 + 3(x + y)^2 \frac{dy}{dx} + 20y \frac{dy}{dx} = 108 \]
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Factor out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} [3(x+y)^2 + 20y] = 108 – 3(x+y)^2 \]
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Solve for \(\frac{dy}{dx}\)
\[ \frac{dy}{dx} = \frac{108 – 3(x+y)^2}{3(x+y)^2 + 20y} \]
- Solving Quadratic Equation
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Identify Turning Points
At turning points (farthest north/south), \(\frac{dy}{dx} = 0\). From part (a):
\[ 108 – 3(x+y)^2 = 0 \]
\[ (x+y)^2 = 36 \]
\[ x+y = \pm 6 \]
-
Case 1: \(x + y = 6\)
Substitute \(x = 6 – y\) into the original equation:
\[ (6)^3 + 10y^2 – 108(6 – y) = 0 \]
\[ 216 + 10y^2 – 648 + 108y = 0 \]
\[ 10y^2 + 108y – 432 = 0 \]
Divide by 2:
\[ 5y^2 + 54y – 216 = 0 \]
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Quadratic Formula:
\[ y = \frac{-54 \pm \sqrt{54^2 – 4 \cdot 5 \cdot (-216)}}{2 \cdot 5} \]
\[ y = \frac{-54 \pm \sqrt{2916 + 4320}}{10} \]
\[ y = \frac{-54 \pm \sqrt{7236}}{10} \]
Approximate \(\sqrt{7236} \approx 85.06\):
\[ y \approx \frac{-54 + 85.06}{10} = 3.106 \, (\text{North}) \]
\[ y \approx \frac{-54 – 85.06}{10} = -13.906 \, (\text{South}) \]
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Case 2: \( x + y = -6 \)
This leads to complex solutions, so we discard it. -
Determine Q (Furthest South)
From Case 1, the southernmost y-coordinate is \( y \approx -13.906 \, \text{km} \).
Distance from O:
Since the units are in km (as per the question), convert \( y \):
\[ -13.906 \, \text{km} = -13906 \, \text{m} \]
Rounding to Nearest 100 m:
\[ -13906 \, \text{m} \approx -13900 \, \text{m} \]