Parametric Differentiation (AQA A-Level Mathematics): Revision Notes

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9.2.1 Parametric Differentiation

Differentiation of Parametric Equations

Fact:

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dydt×dtdx=dydx\frac{dy}{dt} \times \frac{dt}{dx} = \frac{dy}{dx}

Parametric differentiation involves finding the derivative of parametric equations, where xx and yy are both functions of a parameter tt.

Step-by-step process:

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  1. Differentiate x=f(t)x = f(t) and y=g(t)y = g(t) with respect to tt, giving dxdt\frac{dx}{dt} and dydt \frac{dy}{dt}.
  2. Find dydx\frac{dy}{dx} using the chain rule:
dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
  1. For second derivatives, use:
d2ydx2=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}

This method helps calculate the slope, concavity, and other properties of curves defined parametrically.

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Example: Find the equation of the tangent to y=5sin⁡θy = 5\sin\theta and x=cos⁡θx = \cos\theta at the point where θ=π4\theta = \frac{\pi}{4}.

  1. Find dydt\frac{dy}{dt} and dxdt\frac{dx}{dt}:
dydt=5cos⁡θ\frac{dy}{dt} = 5\cos\thetadxdt=−sin⁡θ\frac{dx}{dt} = -\sin\theta
  1. Calculate dydt×dtdx=dydx\frac{dy}{dt} \times \frac{dt}{dx} = \frac{dy}{dx} at the given point:
dydx=5cos⁡θ×−1sin⁡θ\frac{dy}{dx} = 5\cos\theta \times \frac{-1}{\sin\theta}

(Reciprocal of dxdt\frac{dx}{dt})

Let θ=π4\theta = \frac{\pi}{4}:

dydx=5cos⁡(π4)×−1sin⁡(π4)=−5\frac{dy}{dx} = 5\cos\left(\frac{\pi}{4}\right) \times \frac{-1}{\sin\left(\frac{\pi}{4}\right)} = -5
  1. Calculate xx and yy values at the given point:
When θ=π4y=5sin⁡(π4)=522\text{When } \theta = \frac{\pi}{4} \quad y = 5\sin\left(\frac{\pi}{4}\right) = \frac{5\sqrt{2}}{2}x=cos⁡(π4)=22x = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}Point (22,522)\text{Point } \left(\frac{\sqrt{2}}{2}, \frac{5\sqrt{2}}{2}\right)
  1. Equation of the tangent:
y−y1=m(x−x1)y - y_1 = m(x - x_1)y−522=−5(x−22)y - \frac{5\sqrt{2}}{2} = -5 \left(x - \frac{\sqrt{2}}{2}\right)⇒y=−5x+52\Rightarrow y = -5x + 5\sqrt{2}
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📝Q6, (Jun 2011, Q8)

A curve has parametric equations:

x=1t+1,y=t−1.x = \frac{1}{t + 1}, \quad y = t - 1.

The line y=3xy = 3x intersects the curve at two points.

Question

(i) Show that the value of tt at one of these points is -22 and find the value of tt at the other point.

(ii) Find the equation of the normal to the curve at the point for which t=−2t = -2.

(iii) Find the value of tt at the point where this normal meets the curve again.

(iv) Find a cartesian equation of the curve, giving your answer in the form y=f(x)y = f(x).

Part (i)

Given the equation y=3xy = 3x and the parametric equations x=1t+1,y=t−1x = \frac{1}{t+1}, y = t - 1:

t−1=3(1t+1)t - 1 = 3\left(\frac{1}{t+1}\right) ⇒(t−1)(t+1)=3\Rightarrow (t-1)(t + 1) = 3 ⇒t2−1=3\Rightarrow t^2 -1 = 3 ⇒t2=4⇒t±2\Rightarrow t^2 = 4 \Rightarrow t \pm 2

Part (ii)

To find the equation of the normal to the curve at t=−2t = -2:

x=(t−1)−1⇒dxdt⇒−(t+1)−2=−1(t+1)2x=(t-1)^{-1} \Rightarrow \frac {dx}{dt} \Rightarrow -(t+1)^{-2}=\frac {-1}{(t+1)^2} y=t−1⇒dydt=1⇒dydt=1×−(t+1)2=−(t+1)2y=t-1 \Rightarrow \frac {dy}{dt}=1 \Rightarrow \frac {dy}{dt}=1 \times -(t+1)^2 = -(t+1)^2

Let t=−2⇒dydx=−(−2+1)2=−1⇒t = -2 \Rightarrow \frac {dy}{dx} = -(-2+1)^2 = -1 \Rightarrow grad of form =−1−1=1= -\frac {1}{-1}=1

At t=−2,x=1−2+1=−1,y=−2−1=−3t=-2, x = \frac {1}{-2+1} = -1, y=-2-1=-3

y−—3=1(x−—1)⇒y+3=x+1⇒y=x−2y-—3=1(x-—1) \Rightarrow y+3 = x+1 \Rightarrow y=x-2

Part (iii)

To find the value of tt at the point where this normal meets the curve again:

  1. Substitute y=x−2y = x - 2 into the parametric equations:
t−1=1t+1−2t - 1 = \frac{1}{t+1} - 2 ⇒(t−1)(t+1)=1−2(t+1)\Rightarrow (t-1)(t+1)=1-2(t+1) ⇒t2−1=1−2t−2⇒t2+2t=0⇒t(t+2)=0\Rightarrow t^2-1=1-2t-2 \Rightarrow t^2+2t=0 \Rightarrow t(t+2)=0

∴t=0, −2\therefore t=0,\ \xcancel{-2} Already given previously

Part (iv)

To find a Cartesian equation of the curve:

  1. From the parametric equations:
t=y+1andx=1t+1t = y + 1 \quad \text{and} \quad x = \frac{1}{t+1}
  1. Substitute t=y+1t = y + 1 into the equation for xx:
y+2=1xy + 2 = \frac{1}{x}

So, the Cartesian equation of the curve is:

y=1x−2y = \frac{1}{x} - 2

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