Unit 9 · 10 to 15% of the exam

Unit 9 practice test: Parametric Equations, Polar Coordinates, and Vector-Valued Functions

twelve questions, every option explained

12 exam-style questions from the unit, free, with an explanation for every option. Most have four answer choices, as on the real exam, and a few ask you to type a value or an expression; sets share one graph or table. Covers: Parametric Equations and Their Derivatives; Second Derivatives of Parametric Equations; Arc Length of a Parametric Curve; Vector-Valued Functions and Their Derivatives; Integrating Vector-Valued Functions; Motion in the Plane; Polar Coordinates and Polar Derivatives; Area Inside a Polar Curve; Area Between Two Polar Curves.

Start at question 1

Table

A curve known only from a table

A curve is traced by

(x(t), y(t))

, where

x

and

y

are differentiable functions of

t

. The table gives selected values.

Values of $x$ , $y$ and their derivatives with respect to $t$
$t$	$x(t)$	$x'(t)$	$y(t)$	$y'(t)$
$0$	$1$	$3$	$4$	$-2$
$1$	$3$	$1$	$2$	$0$
$2$	$4$	$0$	$3$	$2$
$3$	$2$	$-4$	$6$	$3$

1. What does the table tell you about the tangent lines at $t = 1$ and at $t = 2$ ?

Horizontal at $t = 1$ ; vertical at $t = 2$
Vertical at $t = 1$ ; horizontal at $t = 2$
Horizontal at both times
Slope 2 at $t = 1$ ; slope $\frac{3}{4}$ at $t = 2$

Show the answer

At $t = 1$ : $y'(1) = 0$ and $x'(1) = 1 \ne 0$ , so the slope is 0 (horizontal). At $t = 2$ : $x'(2) = 0$ and $y'(2) = 2 \ne 0$ , so the tangent is vertical.

Correct: Horizontal at $t = 1$ ; vertical at $t = 2$ : Right: a zero numerator ( $y'$ ) gives a horizontal tangent; a zero denominator ( $x'$ ) with a nonzero $y'$ gives a vertical one.
Vertical at $t = 1$ ; horizontal at $t = 2$ : This swaps the two cases. $\frac{dy}{dx} = \frac{y'}{x'}$ is 0 when $y' = 0$ : horizontal.
Horizontal at both times: At $t = 2$ it is $x'$ that is 0, not $y'$ : that tangent is vertical.
Slope 2 at $t = 1$ ; slope $\frac{3}{4}$ at $t = 2$ : These divide the coordinates $y(t)$ by $x(t)$ ; slopes come from the derivatives.

2. In which direction is the point moving at $t = 3$ ?

Right and up
Left and up
Left and down
Right and down

Show the answer

$x'(3) = -4 < 0$ , so $x$ is decreasing: the point moves left. $y'(3) = 3 > 0$ , so $y$ is increasing: it moves up.

Right and up: $x'(3) = -4$ is negative, so $x$ is decreasing: left, not right.
Correct: Left and up: Right: $x' < 0$ (left) and $y' > 0$ (up).
Left and down: $y'(3) = 3$ is positive, so the point moves up.
Right and down: Both signs are reversed here: $x'(3) < 0$ and $y'(3) > 0$ .

3. A curve is given by $x = t^3$ , $y = t^2 + 1$ for $t \ne 0$ . Which describes its concavity?

Concave down for $t < 0$ and for $t > 0$
Concave up for $t < 0$ and for $t > 0$
Concave up for $t > 0$ , concave down for $t < 0$
Concave down for $t > 0$ , concave up for $t < 0$

Show the answer

$\frac{dy}{dx} = \frac{2t}{3t^2} = \frac{2}{3t}$ , whose $t$ -derivative is $-\frac{2}{3t^2}$ . Dividing by $\frac{dx}{dt} = 3t^2$ gives $\frac{d^2y}{dx^2} = -\frac{2}{9t^4}$ , negative on both sides of $t = 0$ .

Correct: Concave down for $t < 0$ and for $t > 0$ : Right: $-\frac{2}{9t^4} < 0$ for $t \ne 0$ .
Concave up for $t < 0$ and for $t > 0$ : The sign of $-\frac{2}{9t^4}$ is negative; $t^4 > 0$ cannot make it positive.
Concave up for $t > 0$ , concave down for $t < 0$ : Even powers of $t$ have one sign on both sides of 0, so the concavity does not change.
Concave down for $t > 0$ , concave up for $t < 0$ : $-\frac{2}{9t^4}$ is negative for $t < 0$ as well: $t^4$ is positive there too.

4. Find the length of the path $x = 1 + 3t$ , $y = 2 - 4t$ for $0 \le t \le 2$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\int_0^2 \sqrt{3^2 + (-4)^2}\,dt$ $= \int_0^2 5\,dt$ $= 10$ . Check: the path is the segment from $(1, 2)$ to $(7, -6)$ , of length $\sqrt{6^2 + 8^2} = 10$ .

Answer: 10

Table

A robot's position from a table

A robot moves on a warehouse floor. Its position at time

t

seconds is

\mathbf{r}(t) = \left\langle x(t), y(t) \right\rangle

, in meters, where

x

and

y

are differentiable. Selected values are given in the table.

Position of the robot at selected times
$t$ (seconds)	$x(t)$ (meters)	$y(t)$ (meters)
$0$	$0$	$5$
$1$	$2$	$4$
$3$	$8$	$0$
$4$	$9$	$-1$
$6$	$10$	$2$

5. Which vector is the best estimate of the robot's velocity vector at $t = 2$ , in meters per second?

$\left\langle 6, -4 \right\rangle$
$\left\langle 5, 2 \right\rangle$
$\left\langle -2, 3 \right\rangle$
$\left\langle 3, -2 \right\rangle$

Show the answer

Estimate each component over $[1, 3]$ : $\frac{8 - 2}{2} = 3$ and $\frac{0 - 4}{2} = -2$ . The velocity vector is $\left\langle x'(2), y'(2) \right\rangle \approx \left\langle 3, -2 \right\rangle$ .

$\left\langle 6, -4 \right\rangle$ : These are the changes over 2 seconds; divide each by 2 to get rates.
$\left\langle 5, 2 \right\rangle$ : This averages the positions at $t = 1$ and $t = 3$ : an estimate of where the robot is, not how fast it moves.
$\left\langle -2, 3 \right\rangle$ : The components are swapped: the $x$ -rate comes first.
Correct: $\left\langle 3, -2 \right\rangle$ : Right: one difference quotient per component.

6. Based on the table, in which direction is the robot most likely moving at $t = 5$ ?

Right and up
Right and down
Left and up
Left and down

Show the answer

Over $[4, 6]$ : $x'(5)$ $\approx \frac{10 - 9}{2}$ $= \frac{1}{2} > 0$ and $y'(5)$ $\approx \frac{2 - (-1)}{2}$ $= \frac{3}{2} > 0$ . Both components of the velocity are positive: right and up.

Correct: Right and up: Right: both estimated components are positive.
Right and down: $y$ goes from $-1$ to $2$ over $[4, 6]$ , so $y'(5)$ is estimated as positive: up.
Left and up: $x$ goes from 9 to 10, so $x'(5) \approx \frac{1}{2} > 0$ : right.
Left and down: Both coordinates increase over $[4, 6]$ , so the estimated velocity points right and up.

7. Find $\int_0^1 \left\langle 3t^2, 4t^3 + 1 \right\rangle\,dt$ .

$\left\langle 6, 12 \right\rangle$
$\left\langle 1, 1 \right\rangle$
$\left\langle 3, 5 \right\rangle$
$\left\langle 1, 2 \right\rangle$

Show the answer

Integrate each component: $\int_0^1 3t^2\,dt$ $= 1$ and $\int_0^1 (4t^3 + 1)\,dt$ $= (t^4 + t)\Big|_0^1$ $= 2$ .

$\left\langle 6, 12 \right\rangle$ : These are the derivatives at $t = 1$ ; integrate instead.
$\left\langle 1, 1 \right\rangle$ : The $+ 1$ in the second component integrates to $t$ , which adds 1.
$\left\langle 3, 5 \right\rangle$ : These are the integrand's values at $t = 1$ , not the integrals.
Correct: $\left\langle 1, 2 \right\rangle$ : Right: $\left\langle 1, 2 \right\rangle$ .

8. A particle has position $\mathbf{r}(t) = \left\langle t^2, t^3 \right\rangle$ . What is its speed at $t = 1$ ?

$\frac{3}{2}$
$\sqrt{13}$
$5$
$\sqrt{2}$

Show the answer

$\mathbf{v}(t) = \left\langle 2t, 3t^2 \right\rangle$ , so $\mathbf{v}(1) = \left\langle 2, 3 \right\rangle$ and the speed is $\sqrt{4 + 9} = \sqrt{13}$ .

$\frac{3}{2}$ : That is $\frac{dy}{dx} = \frac{3}{2}$ , the slope of the path. Speed is the magnitude of the velocity.
Correct: $\sqrt{13}$ : Right: $\sqrt{2^2 + 3^2}$ .
$5$ : That adds the components $2 + 3$ ; use $\sqrt{2^2 + 3^2}$ .
$\sqrt{2}$ : That is the length of the position $\left\langle 1, 1 \right\rangle$ ; speed uses the velocity.

9. The rose $r = 2\cos(3\theta)$ passes through the pole at $\theta = \frac{\pi}{6}$ . Find the slope of its tangent line there.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

At $\theta = \frac{\pi}{6}$ , $r = 0$ and $\frac{dr}{d\theta} = -6\sin\frac{\pi}{2} = -6$ . Then $\frac{dx}{d\theta}$ $= -6\cos\frac{\pi}{6} - 0$ $= -3\sqrt{3}$ and $\frac{dy}{d\theta}$ $= -6\sin\frac{\pi}{6} + 0$ $= -3$ , so $\frac{dy}{dx}$ $= \frac{-3}{-3\sqrt{3}}$ $= \frac{\sqrt{3}}{3}$ . At the pole the curve is tangent to the ray $\theta = \frac{\pi}{6}$ , whose slope is $\tan\frac{\pi}{6}$ .

Answer: $\frac{\sqrt{3}}{3}$ ≈ 0.577

10. Find the area of the sector of the circle $r = 2$ with $0 \le \theta \le \frac{\pi}{3}$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\frac{1}{2}\int_0^{\pi/3} 2^2\,d\theta$ $= 2 \cdot \frac{\pi}{3}$ $= \frac{2\pi}{3}$ , one sixth of the disk's area $4\pi$ .

Answer: $\frac{2\,\pi}{3}$ ≈ 2.094

11. Find the area of the region enclosed by the cardioid $r = 1 - \sin\theta$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\frac{1}{2}\int_0^{2\pi} (1 - \sin\theta)^2\,d\theta$ $= \frac{1}{2}\int_0^{2\pi} \left(1 - 2\sin\theta + \frac{1 - \cos(2\theta)}{2}\right)d\theta$ $= \frac{1}{2}\left(2\pi - 0 + \pi\right)$ $= \frac{3\pi}{2}$ .

Answer: $\frac{3\,\pi}{2}$ ≈ 4.712

12. Find the area of the region inside $r = 2\cos\theta$ and outside $r = 1$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

The curves meet where $2\cos\theta = 1$ : $\theta = \pm\frac{\pi}{3}$ . $\frac{1}{2}\int_{-\pi/3}^{\pi/3} \left(4\cos^2\theta - 1\right)d\theta$ $= \frac{1}{2}\int_{-\pi/3}^{\pi/3} \left(1 + 2\cos(2\theta)\right)d\theta$ $= \frac{1}{2}\Big[\theta + \sin(2\theta)\Big]_{-\pi/3}^{\pi/3}$ $= \frac{\pi}{3}$ $+ \frac{\sqrt{3}}{2}$ .

Answer: $\frac{\pi}{3} + \frac{\sqrt{3}}{2}$ ≈ 1.913