Implicit and parametric differentiation
Differentiation
- A2
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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Find $\dfrac{dy}{dx}$ for the circle $x^2 + y^2 = 25$.
3 marks
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A curve is given by $x = t^2$ and $y = 2t$. Find $\dfrac{dy}{dx}$ in terms of $t$.
3 marks
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Find $\dfrac{dy}{dx}$ for the curve $x^2 + xy + y^2 = 7$.
4 marks
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A curve is given by $x = \cos t$ and $y = \sin t$. Find the gradient at $t = \dfrac{\pi}{4}$, and the coordinates of the corresponding point.
4 marks
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Find the gradient of the curve $x^2 y = 4$ at the point $(2, 1)$.
5 marks
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A curve is given by $x = t^2$ and $y = 2t^3$. Find the equation of the tangent at the point where $t = 1$.
5 marks
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A curve is given by $x = t^2$ and $y = t^3 - 3t$. Find the coordinates of the stationary points.
6 marks
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Find the equation of the normal to the curve $x^2 + xy - y^2 = 11$ at the point $(3, 1)$.
7 marks
Mark scheme
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Answer. $-\dfrac{x}{y}$
- Differentiate to get $2x + 2y\dfrac{dy}{dx} = 0$.
- $\dfrac{dy}{dx} = -\dfrac{x}{y}$.
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Answer. $\dfrac{1}{t}$
- $\dfrac{dy}{dt} = 2$ and $\dfrac{dx}{dt} = 2t$.
- $\dfrac{dy}{dx} = \dfrac{2}{2t} = \dfrac{1}{t}$.
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Answer. $-\dfrac{2x + y}{x + 2y}$
- Differentiate: $2x + y + x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0$.
- Collect the terms in $\dfrac{dy}{dx}$.
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Answer. The gradient is $-1$ at $\left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)$.
- $\dfrac{dy}{dx} = \dfrac{\cos t}{-\sin t} = -\cot t$.
- At $t = \dfrac{\pi}{4}$ the gradient is $-1$ and the point is $\left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)$.
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Answer. $-1$
- Differentiate: $2xy + x^2\dfrac{dy}{dx} = 0$.
- $\dfrac{dy}{dx} = -\dfrac{2y}{x}$.
- At $(2, 1)$ the gradient is $-1$.
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Answer. $y = 3x - 1$
- The point is $(1, 2)$.
- $\dfrac{dy}{dx} = \dfrac{6t^2}{2t} = 3t$, so the gradient at $t = 1$ is $3$.
- $y - 2 = 3(x - 1)$.
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Answer. $(1, -2)$ and $(1, 2)$
- $\dfrac{dy}{dx} = \dfrac{3t^2 - 3}{2t} = 0$ when $t = \pm 1$, and $t \ne 0$.
- When $t = 1$ the point is $(1, -2)$.
- When $t = -1$ the point is $(1, 2)$.
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Answer. $y - 1 = \dfrac{1}{7}(x - 3)$
- Check that $(3, 1)$ lies on the curve: $9 + 3 - 1 = 11$.
- Differentiate: $2x + y + x\dfrac{dy}{dx} - 2y\dfrac{dy}{dx} = 0$.
- At $(3, 1)$, $7 + \dfrac{dy}{dx} = 0$, so the gradient is $-7$.
- The normal has gradient $\dfrac{1}{7}$.