Edexcel IAL June 2023 P4 (WMA14/01) Q2 Implicit Differentiation
Figure 2 shows a sketch of part of the curve C with parametric equations
\[ x = t + \frac{1}{t} \quad y = t – \frac{1}{t} \quad t > 0.7 \]
The curve C intersects the x-axis at the point Q.
(a) Find the x coordinate of Q.(1)
The line l is the normal to C at the point P as shown in Figure 2.
Given that \( t = 2 \) at P
(b) write down the coordinates of P
(1)
(c) Using calculus, show that an equation of l is
\[ 3x + 5y = 15 \]
(3)
Solution to Part a: Find the x-coordinate of Q
Key Concepts Used:
- Parametric axis intercepts: Set \( y = 0 \) to find x-axis intersection
- Algebraic solving: Solving \( t – \frac{1}{t} = 0 \)
- Domain consideration: Selecting valid t value
Step-by-Step Working:
-
Find when curve intersects x-axis
On the x-axis, \( y = 0 \):
\[ y = t – \frac{1}{t} = 0 \]
\[ t – \frac{1}{t} = 0 \]
\[ t = \frac{1}{t} \]
\[ t^2 = 1 \]
\[ t = 1 \quad (since \, t > 0.7) \]
-
Find x-coordinate at t = 1
\[ x = 1 + \frac{1}{1} = 2 \]
Final Answer
\[ 2 \]
Solution to Part b: Find the Coordinates of P
Key Concepts Used:
- Parametric point evaluation: Substituting parameter value into equations
- Fraction arithmetic: Adding and subtracting fractions
Step-by-Step Working:
-
Substitute t = 2 into parametric equations
\[ x = 2 + \frac{1}{2} = \frac{5}{2} \]
\[ y = 2 – \frac{1}{2} = \frac{3}{2} \]
Final Answer
\[ \left( \frac{5}{2}, \frac{3}{2} \right) \]
Solution to Part c: Equation of normal line l
Key Concepts Used:
- Parametric differentiation: \frac{dy}{dx} = \frac{dy/dt}{dx/dt}
- Normal gradient: Negative reciprocal of tangent gradient
- Line equation: Point-slope form and simplification to standard form
Step-by-Step Working:
-
Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
\[ x = t + t^{-1} \Rightarrow \frac{dx}{dt} = 1 – t^{-2} = 1 – \frac{1}{t^2} \]
\[ y = t – t^{-1} \Rightarrow \frac{dy}{dt} = 1 + t^{-2} = 1 + \frac{1}{t^2} \]
-
Find \(\frac{dy}{dx}\)
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1 + \frac{1}{t^2}}{1 – \frac{1}{t^2}} \]
Multiply numerator and denominator by \(t^2\):
\[ \frac{dy}{dx} = \frac{t^2 + 1}{t^2 – 1} \]
-
Evaluate \(\frac{dy}{dx}\) at \(t = 2\)
\[ \frac{dy}{dx} = \frac{4 + 1}{4 – 1} = \frac{5}{3} \]
So gradient of tangent at \(P\) is \(\frac{5}{3}\)
-
Find gradient of normal
The normal is perpendicular to the tangent, so:
\[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{3}{5} \]
-
Find equation of normal line
Using point-slope form at \(P\left(\frac{5}{2}, \frac{3}{2}\right)\):
\[ y – \frac{3}{2} = -\frac{3}{5} \left(x – \frac{5}{2}\right) \]
Multiply through by 10 to eliminate fractions:
\[ 10y – 15 = -6x + 15 \]
\[ 6x + 10y = 30 \]
Divide by 2:
\[ 3x + 5y = 15 \]
Final Answer
\[ 3x + 5y = 15 \]