A curve given parametrically has $x$ and $y$ each written in terms of a third variable, the parameter $t$. You do not need to find the Cartesian equation to get the gradient — the chain rule does it directly.
A curve has parametric equations $x = t^2$, $y = 2t$. Find the equation of the tangent at the point where $t = 2$.
The ellipse $x = 3\cos t$, $y = 2\sin t$. Find the equation of the normal at the point where $t = \dfrac{\pi}{4}$.
| Condition | Meaning |
|---|---|
| $\dfrac{\mathrm{d}y}{\mathrm{d}t} = 0$ (with $\dfrac{\mathrm{d}x}{\mathrm{d}t} \neq 0$) | $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$: a stationary point (horizontal tangent) |
| $\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0$ (with $\dfrac{\mathrm{d}y}{\mathrm{d}t} \neq 0$) | Gradient undefined: a vertical tangent |
A curve has parametric equations $x = t^2 + 1$, $y = t^3 - 3t$. Find the stationary points and the point where the tangent is vertical.
Some curves, such as $x^2 + 3xy + y^2 = 11$, are given by an equation that mixes $x$ and $y$ and cannot easily be rearranged into $y = \ldots$. These are implicit equations. We differentiate every term with respect to $x$, treating $y$ as an (unknown) function of $x$.
- Differentiate every term on both sides with respect to $x$. Each time you differentiate something in $y$, multiply by $\dfrac{\mathrm{d}y}{\mathrm{d}x}$.
- Collect the $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ terms on one side and factorise.
- Divide to make $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ the subject. (Often it is quicker to substitute the point's coordinates before rearranging.)
Find the equation of the tangent to $x^2 + y^2 = 25$ at $(3, 4)$.
A curve has equation $x^2 + 3xy + y^2 = 11$. Find the equation of the normal at $(1, 2)$ in the form $ax + by + c = 0$.
Find the coordinates of the stationary points on the curve $x^2 - xy + y^2 = 12$.
Use implicit differentiation to show that if $y = a^x$ then $\dfrac{\mathrm{d}y}{\mathrm{d}x} = a^x\ln a$.
Parametric
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}$, in terms of $t$.
Find $t$ first
Work out the parameter value at the point before substituting.
Horizontal tangent
$\dfrac{\mathrm{d}y}{\mathrm{d}t} = 0$
Vertical tangent
$\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0$
$y^n$
$\to ny^{n-1}\dfrac{\mathrm{d}y}{\mathrm{d}x}$
$xy$
$\to x\dfrac{\mathrm{d}y}{\mathrm{d}x} + y$ (product rule)
Constants
Differentiate to $0$ — the right-hand side often vanishes.
Implicit stationary points
Numerator $= 0$, then substitute back into the curve.
Normal
Gradient $-1 \div$ tangent gradient.
Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $t$ for $x = 2t + 1$, $y = t^2$.
▶ Show solution
$\dfrac{\mathrm{d}x}{\mathrm{d}t} = 2$, $\dfrac{\mathrm{d}y}{\mathrm{d}t} = 2t$, so $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2t}{2} = t$.
Find the equation of the tangent to $x = t^3$, $y = t^2$ at the point where $t = 2$.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2t}{3t^2} = \dfrac{2}{3t}$; at $t = 2$ this is $\dfrac13$. The point is $(8, 4)$.
$y - 4 = \tfrac13(x - 8) \Rightarrow y = \tfrac13 x + \tfrac43$.
Find the equation of the tangent to $x^2 + y^2 = 13$ at $(2, -3)$, in the form $ax + by + c = 0$.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{x}{y} = -\dfrac{2}{-3} = \dfrac23$.
$y + 3 = \tfrac23(x - 2) \Rightarrow 3y + 9 = 2x - 4 \Rightarrow 2x - 3y - 13 = 0$.
A curve has equation $y^3 + x^2y = 10$. Find the gradient at $(1, 2)$.
▶ Show solution
Check: $8 + 2 = 10$ ✓
$3y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + \left(x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2xy\right) = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{2xy}{3y^2 + x^2}$.
At $(1, 2)$: $-\dfrac{4}{13}$.
A curve has parametric equations $x = 4\sin t$, $y = \cos 2t$. Show that $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\sin t$, and verify by finding the Cartesian equation.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-2\sin 2t}{4\cos t} = \dfrac{-4\sin t\cos t}{4\cos t} = -\sin t$.
Cartesian: $y = \cos 2t = 1 - 2\sin^2 t = 1 - 2\left(\dfrac{x}{4}\right)^2 = 1 - \dfrac{x^2}{8}$.
Then $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{x}{4} = -\dfrac{4\sin t}{4} = -\sin t$ ✓
Find the equation of the normal to the curve $x = \ln t$, $y = t^2 - 2$ at the point where $t = 1$.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2t}{1/t} = 2t^2$; at $t = 1$ the gradient is $2$ and the point is $(0, -1)$.
Normal gradient $-\tfrac12$: $y = -\tfrac12 x - 1$.
Find the stationary points of the curve $x = t^2$, $y = t^3 - 12t$.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}t} = 3t^2 - 12 = 0 \Rightarrow t = \pm 2$ (and $\dfrac{\mathrm{d}x}{\mathrm{d}t} = 2t \neq 0$ there).
$t = 2$: $(4, -16)$. $t = -2$: $(4, 16)$.
Find the exact coordinates of the stationary points on the curve $x^2 + 2xy + 2y^2 = 8$.
▶ Show solution
$2x + 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y + 4y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{x + y}{x + 2y}$.
Stationary when $y = -x$. Substitute: $x^2 - 2x^2 + 2x^2 = 8 \Rightarrow x^2 = 8 \Rightarrow x = \pm 2\sqrt2$.
Points $\left(2\sqrt2, -2\sqrt2\right)$ and $\left(-2\sqrt2, 2\sqrt2\right)$.
By taking logarithms, show that if $y = x^x$ (for $x > 0$) then $\dfrac{\mathrm{d}y}{\mathrm{d}x} = x^x(1 + \ln x)$, and find the exact $x$-coordinate of the stationary point.
▶ Show solution
$\ln y = x\ln x$. Differentiate: $\dfrac1y\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ln x + x \cdot \dfrac1x = \ln x + 1$.
So $\dfrac{\mathrm{d}y}{\mathrm{d}x} = x^x(1 + \ln x)$.
$x^x > 0$, so stationary when $\ln x = -1$: $x = \mathrm{e}^{-1} = \dfrac{1}{\mathrm{e}}$.
A curve $C$ has parametric equations $x = 2\cos t$, $y = \sin 2t$, $\;0 \leqslant t < 2\pi$.
(a) Show that $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{\cos 2t}{\sin t}$.
(b) Find the coordinates of the four stationary points.
(c) Find the equation of the tangent at the point where $t = \dfrac{\pi}{6}$.
(d) Show that the Cartesian equation of $C$ is $y^2 = x^2\left(1 - \dfrac{x^2}{4}\right)$.
▶ Show solution
(a) $\dfrac{\mathrm{d}x}{\mathrm{d}t} = -2\sin t$, $\dfrac{\mathrm{d}y}{\mathrm{d}t} = 2\cos 2t$, so $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2\cos 2t}{-2\sin t} = -\dfrac{\cos 2t}{\sin t}$.
(b) $\cos 2t = 0$ with $0 \leqslant 2t < 4\pi$: $2t = \tfrac{\pi}{2}, \tfrac{3\pi}{2}, \tfrac{5\pi}{2}, \tfrac{7\pi}{2}$, so $t = \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4}$ ($\sin t \neq 0$ at each).
Points: $\left(\sqrt2, 1\right)$, $\left(-\sqrt2, -1\right)$, $\left(-\sqrt2, 1\right)$, $\left(\sqrt2, -1\right)$.
(c) At $t = \tfrac{\pi}{6}$: $x = \sqrt3$, $y = \sin\tfrac{\pi}{3} = \tfrac{\sqrt3}{2}$, gradient $-\dfrac{\cos(\pi/3)}{\sin(\pi/6)} = -\dfrac{1/2}{1/2} = -1$.
$y - \tfrac{\sqrt3}{2} = -(x - \sqrt3) \Rightarrow y = -x + \tfrac{3\sqrt3}{2}$.
(d) $y = 2\sin t\cos t = x\sin t$, so $y^2 = x^2\sin^2 t = x^2(1 - \cos^2 t) = x^2\left(1 - \dfrac{x^2}{4}\right)$ since $\cos t = \dfrac{x}{2}$.