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AP Unit 9 Solutions

Worked solutions to all thirty Unit 9 problems. Each solution gives the mathematical steps and the reason for using them. Compare your setup and justification as well as the final answer.

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Problem 1

For $x = t^{2} + 1$ , $y = t^{3} - 3 t$ , find the tangent line at $t = 2$ .

The point is $(5, 2)$ . Why: Substitute t=2 into both coordinate functions.
$x' = 2 t$ , $y' = 3 t^{2} - 3$ , so the slope is $9 / 4$ . Why: At t=2, divide y′=9 by x′=4≠0.
$y - 2 = (9 / 4) (x - 5)$ . Why: Use the calculated point and slope in point-slope form.

Final answer

y - 2 = (9 / 4) (x - 5)

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Problem 2

For $x = t^{2}$ , $y = t$ , classify the tangent at $t = 0$ and explain its equation.

The point is $(0, 0)$ . Why: Both coordinates are evaluated at the specified parameter.
$x' (0) = 0$ , $y' (0) = 1$ . Why: Differentiate each component; the vertical rate is nonzero.
The tangent is vertical: $x = 0$ . Why: Zero horizontal rate with nonzero vertical rate gives a vertical tangent.

Final answer

Vertical tangent

x = 0

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Problem 3

The graph is the ellipse $x = 2 \cos t$ , $y = 3 \sin t$ , traced counterclockwise from (2,0). At $t = π / 2$ , find the point, tangent and instantaneous direction of motion.

$r (π / 2) = 〈 0, 3 〉$ . Why: cos(π/2)=0 and sin(π/2)=1.
$r' (π / 2) = 〈 - 2, 0 〉$ . Why: Differentiate to 〈−2sin t,3cos t〉 and evaluate.
The tangent is $y = 3$ , and the point moves left. Why: The slope is 0/(−2)=0; the nonzero negative x-rate gives leftward motion.

Final answer

Point (0,3); tangent

y = 3

; moving left.

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Problem 4

For $x = t^{2}$ , $y = t^{3}$ , find $d^{2} y / d x^{2}$ at $t = 2$ .

$d y / d x = 3 t / 2$ for $t \neq 0$ . Why: Divide 3t² by 2t on a valid branch.
$d / d t (d y / d x) = 3 / 2$ . Why: Differentiate the slope with respect to t.
$d^{2} y / d x^{2} = 3 / (4 t) = 3 / 8$ at t=2. Why: Divide that derivative by dx/dt=2t again.

Final answer

3 / 8

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Problem 5

For $x = t^{3} - 3 t$ , $y = t^{2}$ , determine concavity at $t = 0$ .

$d y / d x = 2 t / (3 t^{2} - 3)$ . Why: The horizontal derivative is −3 at t=0, so the quotient is valid locally.
The t-derivative of the slope at 0 is $- 2 / 3$ . Why: The quotient rule gives (2(3t²−3)−12t²)/(3t²−3)².
$d^{2} y / d x^{2} = (- 2 / 3) / (- 3) = 2 / 9 > 0$ . Why: Divide by x′(0); a positive x-second derivative means concave up.

Final answer

2 / 9

; concave up.

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Problem 6

A curve has $x = t^{2}$ , $y = t^{3}$ on $t < 0$ . Is its y-versus-x graph concave up or down? State the domain restriction behind your calculation.

$x' = 2 t \neq 0$ on the negative-t branch. Why: The branch restriction allows division by x′.
$d y / d x = 3 t / 2$ , then $d^{2} y / d x^{2} = 3 / (4 t)$ . Why: Differentiate the slope with respect to t and divide by x′.
It is concave down throughout this branch. Why: The second derivative is negative when t is negative; the argument does not assign a value at t=0.

Final answer

Concave down for

t < 0

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Problem 7

Find the length of $x = 3 t$ , $y = 4 t$ for $0 \leq t \leq 2$ .

Coordinate derivatives are 3 and 4. Why: Differentiate before forming speed.
Speed is $\sqrt{9 + 16} = 5$ . Why: Use the Pythagorean magnitude of the derivative vector.
$L = {\int_{0}}^{2} 5 d t = 10$ . Why: Accumulate traveled length over the given interval.

Final answer

10

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Problem 8

Find the arc length of $x = \cos t$ , $y = \sin t$ on $0 \leq t \leq π$ . Describe which part of the curve is traced.

The speed is $\sqrt{\sin^{2} t + \cos^{2} t} = 1$ . Why: Differentiate the coordinates and use the identity.
$L = {\int_{0}}^{π} 1 d t = π$ . Why: The parameter interval has length π and unit speed.
The upper semicircle is traced once from (1,0) to (−1,0). Why: For 0≤t≤π, sin t≥0 and the point travels counterclockwise.

Final answer

π

; upper semicircle once.

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Problem 9

The unit circle $〈 \cos t, \sin t 〉$ is traced on $0 \leq t \leq 4 π$ . Find distance traveled and the length of the distinct geometric curve. Explain the difference.

Speed is 1. Why: The squared derivative components sum to 1.
Distance traveled is ${\int_{0}}^{4 π} 1 d t = 4 π$ . Why: Distance counts each traversal.
The distinct circle has circumference $2 π$ . Why: The parameter makes two complete laps; distinct geometric length counts the circle once.

Final answer

Distance

4 π

; distinct circumference

2 π

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Review the teaching: Topic 9.3 Topic 9.6

Problem 10

Given $r (t) = 〈 t^{2}, e^{t} 〉$ , find velocity, acceleration and speed at $t = 0$ .

$v (t) = 〈 2 t, e^{t} 〉$ , $a (t) = 〈 2, e^{t} 〉$ . Why: Differentiate each component once and twice.
$v (0) = 〈 0, 1 〉$ , $a (0) = 〈 2, 1 〉$ . Why: Use e⁰=1.
Speed is $\sqrt{0^{2} + 1^{2}} = 1$ . Why: Speed is the magnitude of velocity, not acceleration.

Final answer

Velocity

〈 0, 1 〉

; acceleration

〈 2, 1 〉

; speed

1

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Problem 11

For $r (t) = 〈 \sin (2 t), \cos (2 t) 〉$ , determine speed and acceleration magnitude at any t.

$v = 〈 2 \cos (2 t), - 2 \sin (2 t) 〉$ . Why: Apply the chain rule to both components.
$a = 〈 - 4 \sin (2 t), - 4 \cos (2 t) 〉$ . Why: Differentiate the velocity components with another factor 2.
$| v | = 2$ , $| a | = 4$ . Why: Use sin²(2t)+cos²(2t)=1 in each squared magnitude.

Final answer

Speed

2

; acceleration magnitude

4

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Problem 12

A derivative table gives instantaneous rates at time 2.

Exact coordinate derivative values at t=2
t	x′(t)	y′(t)
2	3	−4

Find the velocity, speed and tangent slope. Can this table determine the particle's position?

Velocity is $〈 3, - 4 〉$ . Why: The table gives the two exact instantaneous component rates.
Speed is $5$ and tangent slope is $- 4 / 3$ . Why: Take the magnitude for speed and divide y′ by nonzero x′ for slope.
Position cannot be determined from this table alone. Why: An instantaneous derivative does not supply coordinates or an initial position.

Final answer

v = 〈 3, - 4 〉

, speed

5

, slope

- 4 / 3

; position undetermined.

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Problem 13

Given $v (t) = 〈 2 t, 3 〉$ and $r (1) = 〈 4, 5 〉$ , find r(t) and r(3).

$r (t) = 〈 t^{2} + C_{x}, 3 t + C_{y} 〉$ . Why: Integrate components separately.
$C_{x} = 3$ , $C_{y} = 2$ . Why: Use r(1)=〈4,5〉 in both component formulas.
$r (t) = 〈 t^{2} + 3, 3 t + 2 〉$ , so $r (3) = 〈 12, 11 〉$ . Why: Evaluate the position formula at the new time.

Final answer

r (t) = 〈 t^{2} + 3, 3 t + 2 〉

;

r (3) = 〈 12, 11 〉

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Problem 14

A particle has $a (t) = 〈 6 t, - 2 〉$ , $v (0) = 〈 1, 4 〉$ , and $r (0) = 〈 2, - 1 〉$ . Find its position.

$v (t) = 〈 3 t^{2} + 1, - 2 t + 4 〉$ . Why: Integrate acceleration and use both initial velocity components.
$r (t) = 〈 t^{3} + t + C_{x}, - t^{2} + 4 t + C_{y} 〉$ . Why: Integrate velocity componentwise.
$r (t) = 〈 t^{3} + t + 2, - t^{2} + 4 t - 1 〉$ . Why: The initial position fixes Cx=2 and Cy=−1.

Final answer

r (t) = 〈 t^{3} + t + 2, - t^{2} + 4 t - 1 〉

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Problem 15

A particle starts at $〈 1, 4 〉$ at t=0. Use the trapezoidal rule on each component to estimate its position at t=2.

Velocity samples; meters per second
t (s)	x′	y′
0	0	2
1	2	0
2	2	−2

Estimate x-displacement by $(0 + 2) / 2 + (2 + 2) / 2 = 3$ . Why: Each subinterval has width 1; average its endpoint rates.
Estimate y-displacement by $(2 + 0) / 2 + (0 - 2) / 2 = 0$ . Why: Apply the same trapezoidal rule to the y-component.
Estimated position is $〈 4, 4 〉$ meters. Why: Add the estimated displacement 〈3,0〉 to initial position 〈1,4〉; sampled data alone do not make this exact.

Final answer

Approximately

〈 4, 4 〉

meters.

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Review the teaching: Topic 9.5 Topic 9.6

Problem 16

For $r (t) = 〈 t^{2}, t^{3} 〉$ , find speed at t=1 and decide whether the particle is speeding up there.

$v (1) = 〈 2, 3 〉$ , $a (1) = 〈 2, 6 〉$ . Why: Differentiate position and evaluate both vectors.
Speed is $\sqrt{13}$ . Why: Use the magnitude of velocity.
$v (1) \cdot a (1) = 22 > 0$ , so speed is increasing. Why: For nonzero speed, its derivative equals v·a/|v|; here it is 22/√13>0.

Final answer

Speed

\sqrt{13}

; increasing.

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Problem 17

A particle follows $r (t) = 〈 \cos t, \sin t 〉$ from t=0 to t=2π. Find displacement and total distance, then explain why they differ.

Both endpoint positions are $〈 1, 0 〉$ , so displacement is $〈 0, 0 〉$ . Why: Displacement is the vector difference of endpoints.
Speed is 1, giving distance $2 π$ . Why: Integrate the magnitude of 〈−sin t,cos t〉 over the full interval.
The journey makes one complete lap. Why: Returning to the starting point cancels net position change but does not cancel distance.

Final answer

Displacement

〈 0, 0 〉

; distance

2 π

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Problem 18

A particle moves with constant velocity $〈 3, 4 〉$ m/s for 2 seconds. Find displacement, distance and average velocity.

Displacement is ${\int_{0}}^{2} 〈 3, 4 〉 d t = 〈 6, 8 〉$ meters. Why: Integrate both constant velocity components.
Distance is ${\int_{0}}^{2} 5 d t = 10$ meters. Why: The speed magnitude is 5 m/s.
Average velocity is $〈 3, 4 〉$ m/s. Why: Divide displacement by the 2-second elapsed time.

Final answer

Displacement

〈 6, 8 〉

m; distance

10

m; average velocity

〈 3, 4 〉

m/s.

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Problem 19

On $r = 2$ , find the Cartesian point and tangent slope at $θ = π / 3$ . Explain why dr/dθ=0 does not give the slope.

$x = 1$ , $y = \sqrt{3}$ . Why: Convert with x=2cos θ and y=2sin θ.
$x' = - \sqrt{3}$ , $y' = 1$ , so $d y / d x = - 1 / \sqrt{3}$ . Why: Differentiate the Cartesian coordinates with respect to θ and divide.
dr/dθ measures radial change, not y-versus-x change. Why: The circle has constant radius while its tangent direction varies.

Final answer

Point

(1, \sqrt{3})

; slope

- 1 / \sqrt{3}

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Problem 20

For $r = 1 + \cos θ$ , compute $d y / d x$ and $d^{2} y / d x^{2}$ at $θ = π / 2$ .

$r = 1$ , $r' = - 1$ , $r ″ = 0$ at π/2. Why: Differentiate the radial function.
$x' = - 1$ , $y' = - 1$ , $x ″ = 2$ , $y ″ = - 1$ . Why: Differentiate x=r cos θ and y=r sin θ using the product rule.
Slope is $1$ and second derivative is $\frac{(- 1) (- 1) - (- 1) (2)}{(- 1)^{3}} = - 3$ . Why: Use y′/x′ and (y″x′−y′x″)/(x′)³; x′ is nonzero.

Final answer

Slope

1

;

d^{2} y / d x^{2} = - 3

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Review the teaching: Topic 9.7 Topic 9.2

Problem 21

On $r = 2 \sin θ$ , classify the tangent at $θ = π / 4$ and give its line.

$r = \sqrt{2}$ , so the Cartesian point is $(1, 1)$ . Why: Multiply the radius by cos(π/4) and sin(π/4).
$r' = \sqrt{2}$ , $x' = 0$ , $y' = 2$ . Why: Apply the product-rule coordinate derivative formulas.
The tangent is vertical, $x = 1$ . Why: Zero x-rate with nonzero y-rate is the vertical-tangent condition.

Final answer

Vertical tangent

x = 1

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Problem 22

A polar graph marks radius −2 at angle 0 on the curve $r = - 2$ . Locate the point in Cartesian coordinates, give an equivalent polar pair with positive radius, and classify the tangent there.

The signed radius points opposite the chosen ray.

$x = (- 2) \cos 0 = - 2$ , $y = (- 2) \sin 0 = 0$ . Why: A signed radius is multiplied by the usual sine and cosine.
An equivalent positive-radius polar pair is $(2, π)$ . Why: Changing the sign of radius reverses the ray, adding π to the angle.
$x' (0) = 0$ , $y' (0) = - 2$ , so the tangent is vertical, $x = - 2$ . Why: Here r′=0; the nonzero y-rate and zero x-rate establish a vertical tangent.

Final answer

Cartesian

(- 2, 0)

; polar

(2, π)

; vertical tangent

x = - 2

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Problem 23

Find the area of the full disk bounded by $r = 3$ .

Use angles $0 \leq θ \leq 2 π$ . Why: One full turn sweeps this disk once.
$A = \frac{1}{2} {\int_{0}}^{2 π} 9 d θ$ . Why: Use the polar sector area formula.
$A = 9 π$ . Why: The constant integrand contributes 9·2π/2.

Final answer

9 π

square units.

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Problem 24

The curve $r = 2 \cos θ$ is a unit circle centered at (1,0). Set bounds that trace it once and calculate its enclosed area.

Choose $- π / 2 \leq θ \leq π / 2$ . Why: The radius is nonnegative and the circle is traced once between its pole crossings.
$A = \frac{1}{2} {\int_{- π / 2}}^{π / 2} 4 \cos^{2} θ d θ$ . Why: Accumulate squared radial sector areas.
$A = π$ . Why: cos²θ=(1+cos2θ)/2; its integral on this interval is π/2.

Final answer

Bounds [

- π / 2, π / 2

]; area

π

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Problem 25

Find the area of one petal of $r = 2 \sin (3 θ)$ swept from θ=0 to θ=π/3. Explain why these bounds select one petal.

r=0 at both bounds and r>0 between them. Why: Consecutive zeros and a positive radius trace one petal.
$A = \frac{1}{2} {\int_{0}}^{π / 3} 4 \sin^{2} (3 θ) d θ$ . Why: Use half the integral of squared radius.
$A = π / 3$ . Why: The identity sin²3θ=(1−cos6θ)/2 leaves a constant contribution π/3 and a zero sine endpoint contribution.

Final answer

π / 3

square units.

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Problem 26

Find the total enclosed area of the cardioid $r = 1 + \cos θ$ , traced once on $0 \leq θ \leq 2 π$ .

$A = \frac{1}{2} {\int_{0}}^{2 π} (1 + \cos θ)^{2} d θ$ . Why: The nonnegative radius traces the cardioid once on the stated interval.
Expand to $1 + 2 \cos θ + \cos^{2} θ$ . Why: Separate terms with familiar integrals.
$A = \frac{1}{2} (2 π + 0 + π) = 3 π / 2$ . Why: The cosine integrates to zero and cos² has integral π over a full turn.

Final answer

3 π / 2

square units.

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Problem 27

Find the area inside $r = 2$ and outside $r = 1 + \cos θ$ . State why one difference integral works on the full turn.

$0 \leq 1 + \cos θ \leq 2$ for every θ. Why: The cardioid remains inside the radius-2 disk, with no outer-boundary switch.
$A = \frac{1}{2} {\int_{0}}^{2 π} (4 - (1 + \cos θ)^{2}) d θ$ . Why: Subtract inner squared radius from outer squared radius.
$A = 4 π - 3 π / 2 = 5 π / 2$ . Why: The two full areas are the radius-2 disk and the cardioid.

Final answer

5 π / 2

square units.

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Problem 28

Find the area inside $r = 2 \sin θ$ but outside $r = 1$ in the upper half-plane.

Solve $2 \sin θ = 1$ : bounds $π / 6$ and $5 π / 6$ . Why: On the interval between them the nonnegative outer radius is at least 1.
$A = \frac{1}{2} {\int_{π / 6}}^{5 π / 6} (4 \sin^{2} θ - 1) d θ$ . Why: Subtract the unit-radius sectors on the common rays.
$A = [θ / 2 - \sin (2 θ) / 2 {]_{π / 6}}^{5 π / 6} = π / 3 + \sqrt{3} / 2$ . Why: Use sin²θ=(1−cos2θ)/2 and evaluate both endpoints.

Final answer

π / 3 + \sqrt{3} / 2

square units.

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Problem 29

Find the overlap area inside both $r = 1$ and $r = 2 \cos θ$ . Your solution must explain a change of boundary.

On $- π / 2 \leq θ \leq π / 2$ , the radius-2cosθ disk has nonnegative radius. Intersections are $θ = \pm π / 3$ . Why: Solve 1=2cosθ and restrict to the once-traced right disk.
The overlap upper radius is 1 when |θ|≤π/3 and 2cosθ nearer ±π/2. Why: A point in both disks must lie below the smaller radial upper bound.
$A = {\int_{0}}^{π / 3} 1 d θ + {\int_{π / 3}}^{π / 2} 4 \cos^{2} θ d θ = 2 π / 3 - \sqrt{3} / 2$ . Why: Symmetry cancels the 1/2 sector factor, then the piecewise squared bounds integrate separately.

Final answer

2 π / 3 - \sqrt{3} / 2

square units.

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Problem 30

Use numerical integration to find, to three decimals, the area in $0 \leq θ \leq π$ inside $r = 2 + \sin θ$ and outside $r = 1$ . Give the defining integral and justify the order of the boundaries. Use radians.

$2 + \sin θ \geq 2 > 1$ on [0,π]. Why: sinθ≥0 there, so the first curve is always the outer boundary.
$A = \frac{1}{2} {\int_{0}}^{π} ((2 + \sin θ)^{2} - 1) d θ$ . Why: Use the difference of squared radii for the stated angular sector.
Numerical integration gives $A \approx 9.498$ . An exact check is $7 π / 4 + 4$ . Why: Accumulate the sector difference in radians and round only the final value; expanding integrates to 3π/2+4+π/4.

Final answer

A \approx 9.498

square units.

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Problem 1

Final answer

Problem 2

Final answer

Problem 3

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Problem 4

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Problem 5

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Problem 6

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Problem 7

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Problem 8

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Problem 9

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Problem 10

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Problem 11

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Problem 12

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Problem 13

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Problem 14

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Problem 15

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Problem 16

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Problem 17

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Problem 18

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Problem 19

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Problem 20

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Problem 21

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Problem 22

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Problem 23

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Problem 24

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Problem 25

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Problem 26

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Problem 27

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Problem 28

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Problem 29

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Problem 30

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