Derivatives of functions defined parametrically: dy/dx = (dy/dt) / (dx/dt) by the chain rule, horizontal and vertical tangents, tangents and normals to parametric curves (new in the 2024 Extension 1 syllabus)

$\frac{dx}{dt} = 6t$ and $\frac{dy}{dt} = 6t^2$, so $\frac{dy}{dx} = \frac{6t^2}{6t} = t$ for $t \neq 0$.

At $t = 1$ the point is $(3, 2)$ and the gradient is $1$, so the tangent is $y - 2 = x - 3$, that is $y = x - 1$.

Markers expect the two $t$-derivatives, the quotient, the point from the parameter value, and the tangent in any correct form.

(a) $\frac{dx}{d\theta} = -2\sin\theta$ and $\frac{dy}{d\theta} = \cos\theta$, so $\frac{dy}{dx} = -\frac{\cos\theta}{2\sin\theta} = -\frac{1}{2}\cot\theta$.

(b) Horizontal tangents need $\cos\theta = 0$ with $\sin\theta \neq 0$: $\theta = \frac{\pi}{2}$ and $\frac{3\pi}{2}$, giving the points $(0, 1)$ and $(0, -1)$.

(c) At $\theta = \frac{\pi}{4}$, $\frac{dy}{dx} = -\frac{1}{2} \cot \frac{\pi}{4} = -\frac{1}{2}$.

Markers award one mark for (a), two for (b) (both parameter values and both points), and one for (c).

Differentiate each coordinate with respect to $t$. $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2$.

Divide.

Marker's note: one mark for both $t$-derivatives, one for $\frac{3t}{2}$. Cancelling $t$ is only valid because $t \neq 0$.

Find $\frac{dy}{dx}$. $\frac{dx}{dt} = 2$ and $\frac{dy}{dt} = 2t$, so $\frac{dy}{dx} = \frac{2t}{2} = t$.

Substitute $t = 3$. The gradient is $3$.

Marker's note: one mark for $\frac{dy}{dx} = t$, one for the value. Substituting $t = 3$ into $x$ or $y$ is not needed for a gradient.

Gradient

$\frac{dx}{dt} = -3\sin t$ and $\frac{dy}{dt} = 3\cos t$, so $\frac{dy}{dx} = \frac{3\cos t}{-3\sin t} = -\cot t$. At $t = \frac{\pi}{3}$, $\frac{dy}{dx} = -\frac{1}{\sqrt{3}}$.

Point

$x = 3 \cos \frac{\pi}{3} = \frac{3}{2}$ and $y = 3 \sin \frac{\pi}{3} = \frac{3\sqrt{3}}{2}$.

Tangent

$y - \frac{3\sqrt{3}}{2} = -\frac{1}{\sqrt{3}} \left( x - \frac{3}{2} \right)$. Multiplying by $\sqrt{3}$: $\sqrt{3}\,y - \frac{9}{2} = -x + \frac{3}{2}$, so

Marker's note: one mark for $-\cot t$, one for the point, one for the equation. As a check, the tangent to a circle is perpendicular to the radius: the radius has gradient $\sqrt{3}$ and $\sqrt{3} \times \left( -\frac{1}{\sqrt{3}} \right) = -1$.

Gradient

$\frac{dx}{dt} = 4$ and $\frac{dy}{dt} = 4t$, so $\frac{dy}{dx} = t$, which is $p$ at $t = p$.

Point

$(4p, 2p^2)$.

Tangent

$y - 2p^2 = p(x - 4p) = px - 4p^2$, so $y = px - 2p^2$.

Marker's note: one mark each for the gradient, the point and the simplified equation.

Derivatives. $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2 - 3 = 3(t - 1)(t + 1)$.

Horizontal tangents need $\frac{dy}{dt} = 0$ with $\frac{dx}{dt} \neq 0$: $t = \pm 1$. At $t = 1$ the point is $(-3, -2)$; at $t = -1$ it is $(-3, 2)$.

Vertical tangents need $\frac{dx}{dt} = 0$ with $\frac{dy}{dt} \neq 0$: $t = 0$, giving $(-4, 0)$.

Marker's note: one mark for the horizontal points, one for the vertical point, one for stating the conditions (in particular that the other derivative is non-zero).

(a) $\frac{dx}{dt} = 20$ and $\frac{dy}{dt} = 20\sqrt{3} - 10t$, so

(b) The gradient of the path is $\tan 30^\circ = \frac{1}{\sqrt{3}}$:

(c) At the maximum height the path is horizontal, so $\frac{dy}{dx} = 0$: $t = 2\sqrt{3}$. Then

Marker's note: one mark for (a); two for (b) (the $\tan 30^\circ$ condition, then $t$); two for (c) ($t = 2\sqrt{3}$, then $60$ m).

Tangent gradient

$\frac{dx}{dt} = 2a$ and $\frac{dy}{dt} = 2at$, so $\frac{dy}{dx} = t$, which is $p$ at the point $(2ap, ap^2)$.

Normal gradient

$-\frac{1}{p}$ (for $p \neq 0$).

Equation

$y - ap^2 = -\frac{1}{p}(x - 2ap)$. Multiply by $p$: $py - ap^3 = -x + 2ap$, so

Marker's note: one mark for $\frac{dy}{dx} = p$, one for the normal gradient, one for the point-gradient equation, one for rearranging to the required form. (At $p = 0$ the normal is the vertical line $x = 0$, which the formula also gives.)

Differentiate. $\frac{dx}{dt} = \cos t$ and $\frac{dy}{dt} = -2\sin 2t$, so

In terms of $x$. Since $x = \sin t$, $\frac{dy}{dx} = -4x$.

Check with the Cartesian equation. $y = \cos 2t = 1 - 2\sin^2 t = 1 - 2x^2$, and $\frac{d}{dx}(1 - 2x^2) = -4x$.

Marker's note: one mark for the quotient of derivatives, one for the double-angle simplification, one for $-4x$.