Edexcel IAL June 2021 P4 (WMA14/01) Q5 Implicit Differentiation
A curve has equation
\( y^2 = ye^{-2x} – 3x \)
(a) Show that
\( \frac{dy}{dx} = \frac{2ye^{-2x} + 3}{e^{-2x} – 2y} \)
(4)
The curve crosses the y-axis at the origin and at the point P.
The tangent to the curve at the origin and the tangent to the curve at P meet at the point R.
(b) Find the coordinates of R.
(5)
(5)
Solution to Part a: Implicit Differentiation for \( y^2 = ye^{-2x} – 3x \)
Key Concepts Used:
- Implicit differentiation: Differentiate both sides w.r.t. x.
- Product rule: For \( ye^{-2x} \).
Step-by-Step Working:
Prove that the derivative of \( y \) with respect to \( x \) is \( \frac{2ye^{-2x} + 3}{e^{-2x} – 2y} \).
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Differentiate both sides implicitly
Given equation:
\( y^2 = ye^{-2x} – 3x \)Differentiate each term with respect to \( x \):
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Now, let’s work on the left Side (\( y^2 \)):
\( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \)
(Chain rule: differentiate \( y^2 \) to get 2y, then multiply by \( \frac{dy}{dx} \)) -
First Term on Right Side (\( ye^{-2x} \)):
\( \frac{d}{dx}(ye^{-2x}) = \frac{dy}{dx}e^{-2x} + y(-2e^{-2x}) \)
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Second Term on Right Side (\( -3x \)):
\( \frac{d}{dx}(-3x) = -3 \)
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Combined Differentiation:
\( 2y\frac{dy}{dx} = e^{-2x}\frac{dy}{dx} – 2ye^{-2x} – 3 \)
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Collect Terms with \( \frac{dy}{dx} \)
Move all terms containing \( \frac{dy}{dx} \) to one side:
\( 2y\frac{dy}{dx} – e^{-2x}\frac{dy}{dx} = -2ye^{-2x} – 3 \)Factor out \( \frac{dy}{dx} \):
\( \frac{dy}{dx}(2y – e^{-2x}) = -2ye^{-2x} – 3 \) -
Solve for \( \frac{dy}{dx} \)
\( \frac{dy}{dx} = \frac{-2ye^{-2x} – 3}{2y – e^{-2x}} \)
Multiply numerator and denominator by \( -1 \) to match the given form:
\( \frac{dy}{dx} = \frac{2ye^{-2x} + 3}{e^{-2x} – 2y} \)
(Product rule: \( \frac{d}{dx}(uv) = u’v + uv’ \), where \( u = y, v = e^{-2x} \))
Final Answer:
\( \frac{dy}{dx} = \frac{2ye^{-2x} + 3}{e^{-2x} – 2y} \)
Solution to Part b: Coordinates of Point R (Intersection of Tangents)
Key Concepts Used:
- Tangent equations: At origin and \( P(0,1) \).
- Intersection: Solve simultaneous equations.
Step-by-Step Working:
Find where the tangents at the origin and point P intersect.
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Find Point P
The curve crosses the y-axis where \( x = 0 \). Substitute \( x = 0 \) into the original equation:
\( y^2 = ye^0 – 0 \Rightarrow y^2 = y \Rightarrow y(y – 1) = 0 \)
\( y = 0 \, (\text{Origin}) \)
\( y = 1 \, (\text{Point P}) \) -
Find Tangent at Origin ((0,0))
Compute \( \frac{dy}{dx} \) at (0,0):
\( \frac{dy}{dx} = \frac{2(0)e^0 + 3}{e^0 – 2(0)} = \frac{3}{1} = 3 \)
Equation of tangent (using point-slope form):
\( y – 0 = 3(x – 0) \Rightarrow y = 3x \) -
Find Tangent at \( P(0,1) \)
Compute \( \frac{dy}{dx} \) at (0,1):
\( \frac{dy}{dx} = \frac{2(1)e^0 + 3}{e^0 – 2(1)} = \frac{5}{-1} = -5 \)
Equation of tangent:
\( y – 1 = -5(x – 0) \Rightarrow y = -5x + 1 \) -
Find Intersection Point R
Set the two tangent equations equal:
\( 3x = -5x + 1 \)
\( 8x = 1 \Rightarrow x = \frac{1}{8} \)Substitute \( x = \frac{1}{8} \) into \( y = 3x \):
\( y = 3 \left( \frac{1}{8} \right) = \frac{3}{8} \)Hence, the coordinates of \( R \):
\( R \left( \frac{1}{8}, \frac{3}{8} \right) \)
Final Answer:
\( R \left( \frac{1}{8}, \frac{3}{8} \right) \)