Mathematics guide, Unit 9, step 28 of 33Previous step Next step

Step 28: Parametric, Polar and Vector-Valued Functions

By the EconTutorials team. 13-minute read with 6 practice questions.

SyllabusAP BC 9.1–9.9 · 9709 P2/P3 parametric derivative

In 30 seconds

AP Calculus BC only. This step is AP Unit 9. AB students can skip it. Cambridge 9709 Papers 2 and 3 use only the first part: the first derivative $ \dfrac{dy}{dx} $ of a parametric curve. 9709 has no polar curves, and IB AA has no parametric or polar calculus.
For a curve $ x = x(t) $, $ y = y(t) $: $ \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} $, and the second derivative is $ \dfrac{d}{dt}\Big(\dfrac{dy}{dx}\Big) $ divided by $ \dfrac{dx}{dt} $. Arc length is $ \displaystyle\int_a^b \sqrt{x'(t)^2 + y'(t)^2}\,dt $.
For motion, $ \mathbf{v} = \mathbf{r}'(t) $, $ \mathbf{a} = \mathbf{v}'(t) $, speed $ = |\mathbf{v}| $, and distance travelled is the integral of speed. A polar point $ (r, \theta) $ is $ (r\cos\theta,\ r\sin\theta) $; polar area is $ \tfrac{1}{2}\displaystyle\int r^2\,d\theta $.
Exam point: $ \dfrac{dr}{d\theta} $ is not the slope of a polar curve. The slope is $ \dfrac{dy/d\theta}{dx/d\theta} $ with $ x = r\cos\theta $ and $ y = r\sin\theta $.

Video lesson coming soonA short animated lesson for this step will appear here.

Parametric curves and their gradients

AP Calculus BC only (Cambridge 9709 P2/P3 need the first derivative in this section, nothing more). Some curves are much easier to describe as a journey than as an equation. Instead of linking $ y $ to $ x $ directly, give both coordinates in terms of a third variable, the parameter $ t $, which you can think of as time: $ x = x(t) $, $ y = y(t) $. As $ t $ runs, the point $ (x, y) $ traces the curve. The curve can loop, cross itself or double back, none of which a function $ y = f(x) $ is allowed to do.

Key termParametric curve: the set of points $ (x(t),\ y(t)) $ as the parameter $ t $ runs through an interval. The same curve can have many parametrisations.

You do not have to eliminate $ t $ to find a gradient. Along the curve, $ y $ depends on $ x $, which depends on $ t $, so the chain rule of Step 13 gives $ \dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot\dfrac{dx}{dt} $. Divide through:

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \frac{dx}{dt} \neq 0. \]

Two special cases come straight out of the fraction. The tangent is horizontal where $ \dfrac{dy}{dt} = 0 $ and $ \dfrac{dx}{dt} \neq 0 $. It is vertical where $ \dfrac{dx}{dt} = 0 $ and $ \dfrac{dy}{dt} \neq 0 $. If both are zero at once, the formula says nothing and the point needs a closer look.

Worked example

A curve is given by $ x = t^2 + t $, $ y = t^3-12t $. Find the equation of the tangent at the point where $ t = 1 $, and find the points where the tangent is horizontal or vertical.

Step 1Differentiate each coordinate: $ \dfrac{dx}{dt} = 2t + 1 $ and $ \dfrac{dy}{dt} = 3t^2-12 $. So $ \dfrac{dy}{dx} = \dfrac{3t^2-12}{2t + 1} $.

Step 2At $ t = 1 $: the point is $ (1 + 1,\ 1-12) = (2, -11) $ and the gradient is $ \dfrac{3-12}{3} = -3 $.

Step 3Tangent: $ y + 11 = -3(x-2) $, that is $ y = -3x-5 $.

Step 4Horizontal: $ 3t^2-12 = 0 $ gives $ t = \pm 2 $, and $ 2t + 1 \neq 0 $ there. The points are $ (6, -16) $ at $ t = 2 $ and $ (2, 16) $ at $ t = -2 $.

Step 5Vertical: $ 2t + 1 = 0 $ gives $ t = -\tfrac{1}{2} $, where $ \dfrac{dy}{dt} = -11.25 \neq 0 $. The point is $ \big(-\tfrac{1}{4},\ \tfrac{47}{8}\big) $.

Notice that the curve passes through $ x = 2 $ twice, at $ t = 1 $ and at $ t = -2 $. That is why you must always work from the value of $ t $, not the value of $ x $.

x = t² + t, y = t³ − 12t

curve, −3.5 ≤ t ≤ 3.2tangent at t = 1: y = −3x − 5horizontal tangents (t = ±2)vertical tangent (t = −½)

Figure 1: the curve passes through x = 2 twice, at t = 1 and t = −2. Horizontal tangents sit where dy/dt = 0 and the vertical tangent where dx/dt = 0.

For concavity, AP BC also asks for $ \dfrac{d^2y}{dx^2} $. The trap is to divide the two second derivatives. Instead, treat $ \dfrac{dy}{dx} $ as a new function of $ t $ and apply the same gradient rule to it:

\[ \frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{dx/dt}. \]

For the curve above, the quotient rule gives $ \dfrac{d}{dt}\Big(\dfrac{3t^2-12}{2t + 1}\Big) = \dfrac{6t^2 + 6t + 24}{(2t + 1)^2} $, so $ \dfrac{d^2y}{dx^2} = \dfrac{6t^2 + 6t + 24}{(2t + 1)^3} $. At $ t = 1 $ this is $ \tfrac{36}{27} = \tfrac{4}{3} > 0 $: the curve is concave up there, which matches the graph.

Worked example

The ellipse $ x = 3\cos t $, $ y = 2\sin t $ is traced for $ 0 \le t \le 2\pi $. Find $ \dfrac{dy}{dx} $ and $ \dfrac{d^2y}{dx^2} $ at $ t = \dfrac{\pi}{4} $.

Step 1$ \dfrac{dx}{dt} = -3\sin t $ and $ \dfrac{dy}{dt} = 2\cos t $, so $ \dfrac{dy}{dx} = \dfrac{2\cos t}{-3\sin t} = -\dfrac{2}{3}\cot t $.

Step 2At $ t = \dfrac{\pi}{4} $, $ \cot t = 1 $, so $ \dfrac{dy}{dx} = -\dfrac{2}{3} $.

Step 3Differentiate the gradient with respect to $ t $: $ \dfrac{d}{dt}\Big(-\tfrac{2}{3}\cot t\Big) = \tfrac{2}{3}\csc^2 t $.

Step 4Divide by $ \dfrac{dx}{dt} $: $ \dfrac{d^2y}{dx^2} = \dfrac{\tfrac{2}{3}\csc^2 t}{-3\sin t} = -\dfrac{2}{9}\csc^3 t $. At $ t = \dfrac{\pi}{4} $, $ \csc t = \sqrt{2} $, so $ \dfrac{d^2y}{dx^2} = -\dfrac{2}{9}\cdot 2\sqrt{2} = -\dfrac{4\sqrt{2}}{9} $.

The negative value says the ellipse is concave down at that point, as it should be on its upper half.

Common mistakeWriting $ \dfrac{d^2y}{dx^2} = \dfrac{d^2y/dt^2}{d^2x/dt^2} $. For the ellipse that would give $ \dfrac{-2\sin t}{-3\cos t} = \tfrac{2}{3}\tan t $, which is positive at $ t = \tfrac{\pi}{4} $ and so gets even the concavity wrong. Always differentiate $ \dfrac{dy}{dx} $ with respect to $ t $, then divide by $ \dfrac{dx}{dt} $.

Here is an economics reading of a parametric gradient. Over $ t $ months a firm’s sales are $ q = 100 + 10t $ units and its price is $ p = 60-4t + 0.2t^2 $ dollars. The path traced in the $ (q, p) $ plane has slope $ \dfrac{dp}{dq} = \dfrac{-4 + 0.4t}{10} $, which at $ t = 5 $ is $ -0.2 $: each extra unit sold has come with a price cut of about 20 cents. Revenue $ R = pq $ changes at $ \dfrac{dR}{dt} = p\,\dfrac{dq}{dt} + q\,\dfrac{dp}{dt} = 45 \times 10 + 150 \times (-2) = \$150 $ per month at that moment.

Arc length and motion along a path

AP Calculus BC only. In a short time $ \Delta t $ the point moves about $ \Delta x $ across and $ \Delta y $ up, so by Pythagoras it covers about $ \sqrt{(\Delta x)^2 + (\Delta y)^2} $. Dividing inside by $ \Delta t $ and adding up the pieces gives the length of the curve. The square root is the speed of the point, so the formula is just distance = speed × time, added up.

\[ L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt \]

This is valid when the curve is traced exactly once as $ t $ runs from $ a $ to $ b $. With $ x = t $ it turns back into the arc length formula of Step 27.

Worked example

Find the exact length of the curve $ x = t^2 $, $ y = \tfrac{2}{3}t^3 $ for $ 0 \le t \le \sqrt{3} $.

Step 1$ \dfrac{dx}{dt} = 2t $ and $ \dfrac{dy}{dt} = 2t^2 $, so the sum of squares is $ 4t^2 + 4t^4 = 4t^2(1 + t^2) $.

Step 2For $ t \ge 0 $ the square root is $ 2t\sqrt{1 + t^2} $.

Step 3By the reverse chain rule, $ L = \displaystyle\int_0^{\sqrt{3}} 2t(1 + t^2)^{1/2}\,dt = \Big[\tfrac{2}{3}(1 + t^2)^{3/2}\Big]_0^{\sqrt{3}} $.

Step 4$ = \tfrac{2}{3}\big(4^{3/2}-1\big) = \tfrac{2}{3}(8-1) = \tfrac{14}{3} $.

The length is $ \tfrac{14}{3} \approx 4.67 $ units. Most AP arc lengths need a calculator; exact ones like this are set so the square root simplifies.

Now let the point be a particle and write its position as a vector, $ \mathbf{r}(t) = \langle x(t),\ y(t) \rangle $. Calculus works one component at a time, exactly as for motion on a line in Step 15.

Quantity	Formula	Vector or number?
velocity	$ \mathbf{v}(t) = \mathbf{r}'(t) = \langle x'(t),\ y'(t) \rangle $	vector, tangent to the path
acceleration	$ \mathbf{a}(t) = \mathbf{v}'(t) = \langle x”(t),\ y”(t) \rangle $	vector
speed	$ \|\mathbf{v}(t)\| = \sqrt{x'(t)^2 + y'(t)^2} $	number
position from velocity	$ \mathbf{r}(b) = \mathbf{r}(a) + \displaystyle\int_a^b \mathbf{v}(t)\,dt $	vector
distance travelled	$ \displaystyle\int_a^b \|\mathbf{v}(t)\|\,dt $	number

Is the particle speeding up? Differentiate $ |\mathbf{v}|^2 = x’^2 + y’^2 $ to get $ 2(x’x” + y’y”) = 2\,\mathbf{v}\cdot\mathbf{a} $. So the speed is increasing when $ \mathbf{v}\cdot\mathbf{a} > 0 $ and decreasing when $ \mathbf{v}\cdot\mathbf{a} < 0 $.

Worked example

A particle has velocity $ \mathbf{v}(t) = \langle 3t^2-3,\ 6t \rangle $ for $ t \ge 0 $ and is at $ (2, -1) $ when $ t = 0 $. Find its position at time $ t $, its speed and whether it is speeding up at $ t = 1 $, and the distance it travels for $ 0 \le t \le 2 $.

Step 1Integrate each component and use the start point: $ x = t^3-3t + 2 $ and $ y = 3t^2-1 $. So $ \mathbf{r}(t) = \langle t^3-3t + 2,\ 3t^2-1 \rangle $.

Step 2At $ t = 1 $: $ \mathbf{v}(1) = \langle 0, 6 \rangle $, so the particle at $ (0, 2) $ is moving straight up with speed 6.

Step 3$ \mathbf{a}(t) = \langle 6t,\ 6 \rangle $, so $ \mathbf{a}(1) = \langle 6, 6 \rangle $ and $ \mathbf{v}\cdot\mathbf{a} = 0 + 36 = 36 > 0 $. The speed is increasing.

Step 4Speed: $ (3t^2-3)^2 + 36t^2 = 9t^4 + 18t^2 + 9 = 9(t^2 + 1)^2 $, so $ |\mathbf{v}| = 3t^2 + 3 $.

Step 5Distance $ = \displaystyle\int_0^2 (3t^2 + 3)\,dt = \big[t^3 + 3t\big]_0^2 = 14 $.

Compare the displacement $ \mathbf{r}(2)-\mathbf{r}(0) = \langle 4, 11 \rangle-\langle 2, -1 \rangle = \langle 2, 12 \rangle $, of size $ \sqrt{148} \approx 12.17 $. The path bends, so the distance travelled is longer.

Path of r(t) = ⟨t³ − 3t + 2, 3t² − 1⟩, 0 ≤ t ≤ 2

path of the particle⅓ v(1) = ⟨0, 2⟩⅓ a(1) = ⟨2, 2⟩

Figure 2: the particle starts at (2, −1), swings left to (0, 2) and then climbs to (4, 11). At t = 1 the velocity points straight up, along the path; the acceleration points into the bend. Both arrows are drawn at one third of their length, on equal axis scales.

Exam tipOn calculator questions, write the set-up in full before pressing buttons, for example $ x(4) = x(0) + \displaystyle\int_0^4 x'(t)\,dt $. AP mark schemes give a point for the integral with its limits and the starting value, even if the final number is wrong. Speed is a number: never write it as a vector.

Polar coordinates and polar graphs

AP Calculus BC only (not in Cambridge 9709). Cartesian coordinates say how far across and how far up. Polar coordinates say how far out and in which direction: $ r $ is the distance from the origin (called the pole) and $ \theta $ is the angle from the positive $ x $-axis, anticlockwise.

\[ x = r\cos\theta, \quad y = r\sin\theta \]

and, going the other way, $ r^2 = x^2 + y^2 $ and $ \tan\theta = \dfrac{y}{x} $ (choose the quadrant of $ \theta $ from a sketch).

For example, the polar point $ \big(6, -\tfrac{\pi}{3}\big) $ is $ \big(6\cos(-\tfrac{\pi}{3}),\ 6\sin(-\tfrac{\pi}{3})\big) = (3, -3\sqrt{3}) $. Polar coordinates are not unique: adding $ 2\pi $ to $ \theta $ gives the same point, and a negative $ r $ means “face direction $ \theta $, then walk backwards”. To convert a polar equation, multiply by $ r $ where it helps: $ r = 6\sin\theta $ becomes $ r^2 = 6r\sin\theta $, so $ x^2 + y^2 = 6y $, which is the circle $ x^2 + (y-3)^2 = 9 $.

To sketch $ r = f(\theta) $, work out $ r $ at multiples of $ \tfrac{\pi}{6} $ and plot each point as a distance along its ray. On a computer, plot it as the parametric curve $ x = f(\theta)\cos\theta $, $ y = f(\theta)\sin\theta $. Learn to recognise these families (with $ a, b > 0 $):

Equation	Shape
$ r = a $	circle, centre the pole, radius $ a $
$ r = 2a\cos\theta $ or $ r = 2a\sin\theta $	circle of radius $ a $ through the pole
$ r = a + a\cos\theta $ or $ r = a + a\sin\theta $	cardioid (heart shape), with a cusp at the pole
$ r = a + b\cos\theta $	limaçon; it has an inner loop when $ b > a $
$ r = a\sin n\theta $ or $ r = a\cos n\theta $	rose with $ n $ petals if $ n $ is odd, $ 2n $ if $ n $ is even

Slopes of polar curves

AP Calculus BC only. A polar curve is a parametric curve in disguise, with $ \theta $ as the parameter: $ x = f(\theta)\cos\theta $ and $ y = f(\theta)\sin\theta $. The product rule on each coordinate gives the slope.

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta-f(\theta)\sin\theta} \]

Keep three derivatives apart. $ \dfrac{dr}{d\theta} $ says how fast the distance from the pole is changing: when $ r > 0 $, a positive value means the point is moving away from the pole and a negative value means it is moving towards it. $ \dfrac{dx}{d\theta} $ and $ \dfrac{dy}{d\theta} $ say whether the point moves left or right, up or down. Only $ \dfrac{dy}{dx} $ is the slope of the tangent.

Worked example

For the cardioid $ r = 2 + 2\sin\theta $: (a) find the slope of the tangent at $ \theta = 0 $ and its equation; (b) say whether the point is moving towards or away from the pole there; (c) show that the tangent at $ \theta = \tfrac{\pi}{6} $ is vertical.

Step 1Here $ f(\theta) = 2 + 2\sin\theta $ and $ f'(\theta) = 2\cos\theta $. At $ \theta = 0 $: $ f = 2 $, $ f’ = 2 $, $ \sin\theta = 0 $, $ \cos\theta = 1 $.

Step 2$ \dfrac{dy}{d\theta} = 2(0) + 2(1) = 2 $ and $ \dfrac{dx}{d\theta} = 2(1)-2(0) = 2 $, so $ \dfrac{dy}{dx} = 1 $.

Step 3The point is $ (r\cos 0,\ r\sin 0) = (2, 0) $, so the tangent is $ y = x-2 $.

Step 4$ \dfrac{dr}{d\theta} = 2 > 0 $ with $ r = 2 > 0 $, so the point is moving away from the pole.

Step 5At $ \theta = \tfrac{\pi}{6} $: $ f = 3 $, $ f’ = \sqrt{3} $. Then $ \dfrac{dx}{d\theta} = \sqrt{3}\cdot\tfrac{\sqrt{3}}{2}-3\cdot\tfrac{1}{2} = 0 $ while $ \dfrac{dy}{d\theta} = \sqrt{3}\cdot\tfrac{1}{2} + 3\cdot\tfrac{\sqrt{3}}{2} = 2\sqrt{3} \neq 0 $. So the tangent is vertical, at $ x = \tfrac{3\sqrt{3}}{2} $.

At $ \theta = \tfrac{\pi}{6} $ the curve reaches its furthest point to the right, which is exactly where a vertical tangent should be.

The cardioid r = 2 + 2 sin θ

r = 2 + 2 sin θtangent at θ = 0: y = x − 2vertical tangent at θ = π/6

Figure 3: plotted as x = r cos θ, y = r sin θ with equal scales. The cusp is at the pole (θ = 3π/2). At θ = 0 the slope is 1; at θ = π/6 the curve is furthest right and its tangent is vertical.

Area in polar coordinates

AP Calculus BC only. The natural slices of a polar region are thin wedges fanning out from the pole. A wedge of radius $ r $ and tiny angle $ \Delta\theta $ is almost a circular sector, with area $ \tfrac{1}{2}r^2\,\Delta\theta $. Adding the wedges and letting $ \Delta\theta \to 0 $ turns the sum into an integral.

\[ A = \frac{1}{2}\int_\alpha^\beta r^2\,d\theta, \qquad A_{\text{between}} = \frac{1}{2}\int_\alpha^\beta \big(R^2-r^2\big)\,d\theta \]

In the second formula $ R $ is the outer curve and $ r $ the inner one. It is a big sector minus a small one, so it is $ R^2-r^2 $, never $ (R-r)^2 $.

Worked example

Find the area of one petal of the rose $ r = 2\sin 3\theta $, and hence the area enclosed by the whole rose.

Step 1A petal starts and ends where $ r = 0 $: $ \sin 3\theta = 0 $ at $ \theta = 0 $ and $ \theta = \tfrac{\pi}{3} $, with $ r > 0 $ in between.

Step 2$ A = \tfrac{1}{2}\displaystyle\int_0^{\pi/3} 4\sin^2 3\theta\,d\theta = \int_0^{\pi/3} (1-\cos 6\theta)\,d\theta $, using $ \sin^2 u = \tfrac{1}{2}(1-\cos 2u) $.

Step 3$ = \Big[\theta-\tfrac{1}{6}\sin 6\theta\Big]_0^{\pi/3} = \tfrac{\pi}{3} $.

Step 4The rose has 3 petals ($ n = 3 $ is odd), so it encloses $ 3 \times \tfrac{\pi}{3} = \pi $.

Integrating from 0 to $ 2\pi $ would give $ 2\pi $, because the three petals get traced twice. Always check how much of the curve your limits draw.

Worked example

Find the area of the region inside the circle $ r = 3 $ and outside the cardioid $ r = 2 + 2\cos\theta $.

Step 1The curves meet where $ 2 + 2\cos\theta = 3 $, so $ \cos\theta = \tfrac{1}{2} $: $ \theta = \tfrac{\pi}{3} $ and $ \theta = \tfrac{5\pi}{3} $. Between these angles (on the left) the circle is outside.

Step 2Expand: $ (2 + 2\cos\theta)^2 = 4 + 8\cos\theta + 4\cos^2\theta = 6 + 8\cos\theta + 2\cos 2\theta $.

Step 3So $ A = \tfrac{1}{2}\displaystyle\int_{\pi/3}^{5\pi/3} \big(3-8\cos\theta-2\cos 2\theta\big)\,d\theta = \tfrac{1}{2}\Big[3\theta-8\sin\theta-\sin 2\theta\Big]_{\pi/3}^{5\pi/3} $.

Step 4At $ \tfrac{5\pi}{3} $: $ 5\pi + 4\sqrt{3} + \tfrac{\sqrt{3}}{2} $. At $ \tfrac{\pi}{3} $: $ \pi-4\sqrt{3}-\tfrac{\sqrt{3}}{2} $. The difference is $ 4\pi + 9\sqrt{3} $.

Step 5$ A = 2\pi + \tfrac{9\sqrt{3}}{2} \approx 14.08 $.

Sense check: the circle has area $ 9\pi \approx 28.27 $, and the shaded part is about half of it, which matches the picture.

Inside r = 3, outside r = 2 + 2 cos θ

circle r = 3cardioid r = 2 + 2 cos θregion: area 2π + 9√3/2 ≈ 14.08

Figure 4: the curves meet where cos θ = ½. From θ = π/3 round to 5π/3 the circle is the outer curve, so each wedge of the shaded region is a sector of radius 3 minus a sector of the cardioid.

Exam tipFind the limits by setting the two $ r $ values equal, then sketch to see which curve is outside on which interval. Also check the pole: two curves can both pass through the origin at different values of $ \theta $, and that meeting point never shows up when you solve the equation.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. A curve has $ x = 3t-1 $, $ y = t^2 + 2 $. Find $ \dfrac{dy}{dx} $ at $ t = 2 $ and the point it refers to. Show worked answer

$ \dfrac{dx}{dt} = 3 $ and $ \dfrac{dy}{dt} = 2t $, so $ \dfrac{dy}{dx} = \dfrac{2t}{3} $.
At $ t = 2 $: $ \dfrac{dy}{dx} = \dfrac{4}{3} $.
The point is $ (3(2)-1,\ 4 + 2) = (5, 6) $.

EasyQ2. Convert the polar point $ \big(4, \tfrac{3\pi}{4}\big) $ to Cartesian coordinates, and convert $ r = 8\cos\theta $ to a Cartesian equation. Show worked answer

$ x = 4\cos\tfrac{3\pi}{4} = -2\sqrt{2} $ and $ y = 4\sin\tfrac{3\pi}{4} = 2\sqrt{2} $, so the point is $ (-2\sqrt{2},\ 2\sqrt{2}) $.
Multiply $ r = 8\cos\theta $ by $ r $: $ x^2 + y^2 = 8x $.
Complete the square: $ (x-4)^2 + y^2 = 16 $, a circle with centre $ (4, 0) $ and radius 4.

MediumQ3. A particle has position $ \mathbf{r}(t) = \langle e^{2t},\ t^2-4t \rangle $. Find its speed at $ t = 1 $ and decide whether the speed is increasing or decreasing then. Show worked answer

$ \mathbf{v}(t) = \langle 2e^{2t},\ 2t-4 \rangle $, so $ \mathbf{v}(1) = \langle 2e^2,\ -2 \rangle $.
Speed $ = \sqrt{4e^4 + 4} = 2\sqrt{e^4 + 1} \approx 14.9 $.
$ \mathbf{a}(t) = \langle 4e^{2t},\ 2 \rangle $, so $ \mathbf{a}(1) = \langle 4e^2,\ 2 \rangle $.
$ \mathbf{v}\cdot\mathbf{a} = 8e^4-4 > 0 $, so the speed is increasing.

MediumQ4. Find the area enclosed by the limaçon $ r = 3 + \cos\theta $. Show worked answer

Here $ r > 0 $ always, and the curve is traced once for $ 0 \le \theta \le 2\pi $.
$ A = \tfrac{1}{2}\displaystyle\int_0^{2\pi} (9 + 6\cos\theta + \cos^2\theta)\,d\theta $.
Over a full turn, $ \int\cos\theta\,d\theta = 0 $ and $ \int\cos^2\theta\,d\theta = \pi $.
So $ A = \tfrac{1}{2}(18\pi + 0 + \pi) = \dfrac{19\pi}{2} \approx 29.8 $.

MediumQ5. Find the exact length of the curve $ x = e^t\cos t $, $ y = e^t\sin t $ for $ 0 \le t \le \ln 2 $. Show worked answer

$ \dfrac{dx}{dt} = e^t(\cos t-\sin t) $ and $ \dfrac{dy}{dt} = e^t(\sin t + \cos t) $.
Sum of squares: $ e^{2t}\big[(\cos t-\sin t)^2 + (\sin t + \cos t)^2\big] = 2e^{2t} $.
So the speed is $ \sqrt{2}\,e^t $.
$ L = \displaystyle\int_0^{\ln 2} \sqrt{2}\,e^t\,dt = \sqrt{2}\,(e^{\ln 2}-1) = \sqrt{2} $.

ExamQ6. A particle moves in the $ xy $-plane with velocity $ \mathbf{v}(t) = \langle 2t-4,\ 3t^2 \rangle $ for $ t \ge 0 $. At $ t = 0 $ it is at $ (3, 1) $. (a) Find the speed of the particle at $ t = 1 $. [1] (b) Find $ \mathbf{a}(1) $ and decide whether the speed is increasing at $ t = 1 $. Give a reason. [2] (c) Find the position of the particle at $ t = 2 $. [3] (d) Describe the direction of the tangent to the path at $ t = 2 $, and give the interval of time on which the particle moves to the left. [2] Show worked answer

(a) $ \mathbf{v}(1) = \langle -2, 3 \rangle $, so the speed is $ \sqrt{4 + 9} = \sqrt{13} $. [B1]
(b) $ \mathbf{a}(t) = \langle 2,\ 6t \rangle $, so $ \mathbf{a}(1) = \langle 2, 6 \rangle $. [B1]
$ \mathbf{v}(1)\cdot\mathbf{a}(1) = -4 + 18 = 14 > 0 $, so the speed is increasing. [A1]
(c) $ x(2) = 3 + \displaystyle\int_0^2 (2t-4)\,dt = 3 + (4-8) = -1 $. [M1 A1]
$ y(2) = 1 + \displaystyle\int_0^2 3t^2\,dt = 1 + 8 = 9 $, so the particle is at $ (-1, 9) $. [A1]
(d) At $ t = 2 $, $ \dfrac{dx}{dt} = 0 $ and $ \dfrac{dy}{dt} = 12 \neq 0 $, so the tangent is vertical (the particle moves straight up). [B1]
It moves left when $ \dfrac{dx}{dt} = 2t-4 < 0 $, that is for $ 0 \le t < 2 $. [B1]
Total: 8 marks.

Check yourself

1. For the curve $ x = t^3 $, $ y = t^2 $ (with $ t \neq 0 $), what is $ \dfrac{d^2y}{dx^2} $?

$ \dfrac{dy}{dx} = \dfrac{2t}{3t^2} = \dfrac{2}{3t} $. Its $ t $-derivative is $ -\dfrac{2}{3t^2} $; divide by $ \dfrac{dx}{dt} = 3t^2 $ to get $ -\dfrac{2}{9t^4} $. The first option divides the two second derivatives, which is the classic mistake.

2. Which integral gives the area enclosed by $ r = f(\theta) $ for $ \alpha \le \theta \le \beta $?

Each thin wedge is a sector of area $ \tfrac{1}{2}r^2\,\Delta\theta $, so the area is $ \tfrac{1}{2}\int r^2\,d\theta $.

3. For $ r = 2 + \cos\theta $, at $ \theta = \tfrac{\pi}{2} $, the point on the curve is:

$ \dfrac{dr}{d\theta} = -\sin\theta = -1 < 0 $ while $ r = 2 > 0 $, so the distance from the pole is decreasing.

Your score: 0 out of 3

The short version

This step is AP Calculus BC only. Cambridge 9709 P2/P3 need just the first derivative $ \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} $ of a parametric curve.
Horizontal tangent where $ \dfrac{dy}{dt} = 0 $, vertical where $ \dfrac{dx}{dt} = 0 $. For concavity, differentiate $ \dfrac{dy}{dx} $ with respect to $ t $, then divide by $ \dfrac{dx}{dt} $.
Arc length and distance travelled are $ \displaystyle\int \sqrt{x'(t)^2 + y'(t)^2}\,dt $, the integral of speed. Velocity $ \mathbf{v} = \mathbf{r}’ $, acceleration $ \mathbf{a} = \mathbf{v}’ $; speed rises when $ \mathbf{v}\cdot\mathbf{a} > 0 $.
Polar: $ x = r\cos\theta $, $ y = r\sin\theta $. The slope is $ \dfrac{dy/d\theta}{dx/d\theta} $, not $ \dfrac{dr}{d\theta} $. Area is $ \tfrac{1}{2}\int r^2\,d\theta $, and between curves $ \tfrac{1}{2}\int (R^2-r^2)\,d\theta $.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

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Next: step 29 of 33Series and Convergence Tests

Equation	Shape
\( r = a \)	circle, centre the pole, radius \( a \)
\( r = 2a\cos\theta \) or \( r = 2a\sin\theta \)	circle of radius \( a \) through the pole
\( r = a + a\cos\theta \) or \( r = a + a\sin\theta \)	cardioid (heart shape), with a cusp at the pole
\( r = a + b\cos\theta \)	limaçon; it has an inner loop when \( b > a \)
\( r = a\sin n\theta \) or \( r = a\cos n\theta \)	rose with \( n \) petals if \( n \) is odd, \( 2n \) if \( n \) is even

Step 28: Parametric, Polar and Vector-Valued Functions

Parametric curves and their gradients

Arc length and motion along a path

Polar coordinates and polar graphs

Slopes of polar curves

Area in polar coordinates

Practise: 6 questions

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