9.2.1 Parametric Differentiation

Differentiation of Parametric Equations

Fact:

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\frac{dy}{dt} \times \frac{dt}{dx} = \frac{dy}{dx}

Parametric differentiation involves finding the derivative of parametric equations, where $x$ and $y$ are both functions of a parameter $t$ .

Step-by-step process:

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Differentiate $x = f(t)$ and $y = g(t)$ with respect to $t$ , giving $\frac{dx}{dt}$ and $\frac{dy}{dt}$ .
Find $\frac{dy}{dx}$ using the chain rule:

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

For second derivatives, use:

\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 $y = 5\sin\theta$ and $x = \cos\theta$ at the point where $\theta = \frac{\pi}{4}$ .

Find $\frac{dy}{dt}$ and $\frac{dx}{dt}$ :

\frac{dy}{dt} = 5\cos\theta

\frac{dx}{dt} = -\sin\theta

Calculate $\frac{dy}{dt} \times \frac{dt}{dx} = \frac{dy}{dx}$ at the given point:

\frac{dy}{dx} = 5\cos\theta \times \frac{-1}{\sin\theta}

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

Let $\theta = \frac{\pi}{4}$ :

\frac{dy}{dx} = 5\cos\left(\frac{\pi}{4}\right) \times \frac{-1}{\sin\left(\frac{\pi}{4}\right)} = -5

Calculate $x$ and $y$ values at the given point:

\text{When } \theta = \frac{\pi}{4} \quad y = 5\sin\left(\frac{\pi}{4}\right) = \frac{5\sqrt{2}}{2}

x = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}

\text{Point } \left(\frac{\sqrt{2}}{2}, \frac{5\sqrt{2}}{2}\right)

Equation of the tangent:

y - y_1 = m(x - x_1)

y - \frac{5\sqrt{2}}{2} = -5 \left(x - \frac{\sqrt{2}}{2}\right)

\Rightarrow y = -5x + 5\sqrt{2}

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📝Q6, (Jun 2011, Q8)

A curve has parametric equations:

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

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

Question

(i) Show that the value of $t$ at one of these points is - $2$ and find the value of $t$ at the other point.

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

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

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

Part (i)

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

t - 1 = 3\left(\frac{1}{t+1}\right)

\Rightarrow (t-1)(t + 1) = 3

\Rightarrow t^2 -1 = 3

\Rightarrow t^2 = 4 \Rightarrow t \pm 2

Part (ii)

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

x=(t-1)^{-1} \Rightarrow \frac {dx}{dt} \Rightarrow -(t+1)^{-2}=\frac {-1}{(t+1)^2}

y=t-1 \Rightarrow \frac {dy}{dt}=1 \Rightarrow \frac {dy}{dt}=1 \times -(t+1)^2 = -(t+1)^2

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

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

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

Part (iii)

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

Substitute $y = x - 2$ into the parametric equations:

t - 1 = \frac{1}{t+1} - 2

\Rightarrow (t-1)(t+1)=1-2(t+1)

\Rightarrow t^2-1=1-2t-2 \Rightarrow t^2+2t=0 \Rightarrow t(t+2)=0

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

Part (iv)

To find a Cartesian equation of the curve:

From the parametric equations:

t = y + 1 \quad \text{and} \quad x = \frac{1}{t+1}

Substitute $t = y + 1$ into the equation for $x$ :

y + 2 = \frac{1}{x}

So, the Cartesian equation of the curve is:

y = \frac{1}{x} - 2

Parametric Differentiation Simplified Revision Notes for A-Level AQA Maths Pure

9.2.1 Parametric Differentiation

Differentiation of Parametric Equations

Step-by-step process:

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

📝Q6, (Jun 2011, Q8)

500K+ Students Use These Powerful Tools to Master Parametric Differentiation For their A-Level Exams.

Other Revision Notes related to Parametric Differentiation you should explore

Parametric Differentiation Simplified Revision Notes for A-Level AQA Maths Pure

9.2.1 Parametric Differentiation

Differentiation of Parametric Equations

Step-by-step process:

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

📝Q6, (Jun 2011, Q8)

500K+ Students Use These Powerful Tools to Master Parametric Differentiation For their A-Level Exams.

Other Revision Notes related to Parametric Differentiation you should explore

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