Unit 8BC only

Parametric, Polar & Vectors

Derivatives, speed and arc length for parametric curves, area and slope in polar, and motion described by vectors.

Topics in this unit

Parametric Equations · Parametric Derivatives · Arc Length & Speed · Vector-Valued Functions · Vector Calculus · Polar Coordinates · Polar Slope · Polar Area

Parametric Derivatives

Question 1

BCno calculatormedium

At which points does the curve

x = t^{2} - 2 t, y = t^{3} - 12 t

shown have

horizontal

tangent lines?

Show the answer

The answer is B.

Horizontal:

\frac{d y}{d t} = 3 t^{2} - 12 = 0 \Rightarrow t = \pm 2

(and

\frac{d x}{d t} = 2 t - 2 \neq = 0

there). At

t = 2

(0, - 16)

; at

t = - 2

(8, 16)

Why not A:That is where $\frac{d x}{d t} = 2 t - 2 = 0$ (a $vertical$ tangent), not horizontal.
Why not C:Recompute $y$ : at $t = 2$ , $y = 8 - 24 = - 16$ ; at $t = - 2$ , $y = - 8 + 24 = 16$ .
Why not D:You set $y = 0$ . Horizontal tangents need $\frac{d y}{d t} = 0$ , not $y = 0$ .

Question 2

BCno calculatorhard

A curve is defined by

x = t^{2} + 1

and

y = t^{3} - 3 t

. What is

\frac{d ^{2} y}{d x ^{2}}

t = 2

Show the answer

The answer is C.

First

\frac{d y}{d x} = \frac{3 t ^{2} - 3}{2 t}

. Then

\frac{d ^{2} y}{d x ^{2}} = \frac{\frac{d}{d t} ( \frac{d y}{d x} )}{\frac{d x}{d t}}

. Since

\frac{d}{d t} (\frac{3 t ^{2} - 3}{2 t}) = \frac{3}{2} + \frac{3}{2 t ^{2}} = \frac{15}{8}

t = 2

, dividing by

\frac{d x}{d t} = 2 t = 4

gives

\frac{15}{32}

Why not A: $3 = \frac{d ^{2} y / d t ^{2}}{d x / d t} = \frac{12}{4}$ uses the wrong numerator; differentiate $\frac{d y}{d x}$ with respect to $t$ , then divide by $\frac{d x}{d t}$ .
Why not B: $\frac{15}{8}$ is $\frac{d}{d t} (\frac{d y}{d x})$ ; you must still divide by $\frac{d x}{d t} = 4$ to get $\frac{15}{32}$ .
Why not D: $6 = \frac{d ^{2} y / d t ^{2}}{d ^{2} x / d t ^{2}} = \frac{12}{2}$ ; the parametric second derivative is not the ratio of the two second derivatives.

Arc Length & Speed

Question 3

BCno calculatorhard

A curve is given by

x = \frac{1}{3} t^{3}

y = \frac{1}{2} t^{2}

for

0 \leq t \leq 2

. What is the length of the curve?

Show the answer

The answer is A.

\frac{d x}{d t} = t^{2}

and

\frac{d y}{d t} = t

, so the speed is

t^{4} + t^{2} = t t^{2} + 1

[0, 2]

. Then

L = \int_{0}^{2} t t^{2} + 1 d t = [\frac{1}{3} (t^{2} + 1)^{3/2}]_{0}^{2} = \frac{5 5 - 1}{3}

Why not B:With $u = t^{2} + 1$ , $d u = 2 t d t$ , so $\int t t^{2} + 1 d t = \frac{1}{3} (t^{2} + 1)^{3/2}$ ; forgetting the $\frac{1}{2}$ doubles the result.
Why not C: $t^{4} + t^{2} \neq = t^{2} + t$ ; factor first: $t^{2} (t^{2} + 1) = t t^{2} + 1$ .
Why not D:The lower limit contributes $\frac{1}{3} (0 + 1)^{3/2} = \frac{1}{3}$ ; subtracting it gives $\frac{5 5 - 1}{3}$ .

Question 4

BCno calculatormedium

The arc length of the polar curve

r = θ

from

θ = 0

θ = π

is given by which integral?

Show the answer

The answer is C.

Polar arc length is

\int r^{2} + (\frac{d r}{d θ})^{2} d θ

. With

r = θ

and

\frac{d r}{d θ} = 1

, this is

\int_{0}^{π} θ^{2} + 1 d θ

Why not A:This omits the radical entirely; arc length keeps $r^{2} + (r^{'})^{2}$ .
Why not B:You dropped the $(\frac{d r}{d θ})^{2} = 1$ term inside the radical.
Why not D:That would be $r^{2} + (r^{'})^{2}$ with $r = θ^{2}$ ; here $r = θ$ , so the radicand is $θ^{2} + 1$ .

Vector-Valued Functions

Question 5

BCno calculatorhard

A particle moves with position vector

r (t) = ⟨ t^{2} - 2 t, t^{3} - 4 t ⟩

for

t \geq 0

. The particle is at the origin at

t = 0

and later passes through the origin again. What is the speed of the particle when it passes through the origin the second time?

Show the answer

The answer is B.

Both coordinates are

0

when

t^{2} - 2 t = 0

and

t^{3} - 4 t = 0

, which for

t > 0

happens only at

t = 2

. There

v (2) = ⟨ 2 t - 2, 3 t^{2} - 4 ⟩ = ⟨ 2, 8 ⟩

, so the speed is

4 + 64 = 217

Why not A:This is the speed at $t = 0$ , the first time at the origin, where $v = ⟨ - 2, - 4 ⟩$ .
Why not C:This adds the components of $v (2) = ⟨ 2, 8 ⟩$ . Speed is $2^{2} + 8^{2}$ .
Why not D:This is $2^{2} + 8^{2}$ without the square root.

Vector Calculus

Question 6

BCno calculatormedium

A particle moves with position

r (t) = ⟨ 2 cos t, 2 sin t ⟩

for

0 \leq t \leq π

. The total distance traveled and the magnitude of the displacement are:

Show the answer

The answer is D.

Speed

= (- 2 sin t)^{2} + (2 cos t)^{2} = 2

, so total distance

= \int_{0}^{π} 2 d t = 2 π

. Displacement

= r (π) - r (0) = ⟨ - 2, 0 ⟩ - ⟨ 2, 0 ⟩ = ⟨ - 4, 0 ⟩

, with magnitude

4

Why not A:The radius is $2$ , so speed is $2$ (not $4$ ); distance over $[0, π]$ is $2 π$ .
Why not B:Displacement is $0$ only over a full loop $[0, 2 π]$ . Over $[0, π]$ the particle ends at the opposite point, a distance $4$ away.
Why not C:Total distance is $\int_{0}^{π} (speed) d t$ with speed $2$ , giving $2 π$ , not $π$ .

Polar Coordinates

Question 7

BCno calculatormedium

The polar curve shown is the graph of which equation?

Show the answer

The answer is A.

A cardioid

r = a + a cos θ

bulges along the

+ x

-axis (max

r = 2

θ = 0

, cusp at

θ = π

), matching the figure.

Why not B:That cardioid bulges $upward$ (max $r$ at $θ = \frac{π}{2}$ ); this one bulges right, so it uses $cos θ$ .
Why not C: $r = 2 cos θ$ is a circle, not a heart-shaped cardioid.
Why not D:That is a four-petal rose, not a cardioid.

Question 8

BCno calculatormedium

Which of the following is the graph of the polar curve

r = 2 cos θ

Show the answer

The answer is C.

r = 2 cos θ

converts to

(x - 1)^{2} + y^{2} = 1

: a circle of radius

1

centered at

(1, 0)

, passing through the origin and tangent to the

y

-axis. Its rightmost point is

(2, 0)

θ = 0

Why not A:This is $r = 2 sin θ$ , a circle of radius $1$ centered at $(0, 1)$ on the $y$ -axis.
Why not B:This is $r = 2$ , a circle of radius $2$ centered at the origin.
Why not D:This is the cardioid $r = 1 + cos θ$ , a heart shape with a cusp at the origin, not a circle.

Polar Area

Question 9

BCno calculatorhard

What is the area of the region that lies inside the circle

r = 2 sin θ

and outside the circle

r = 1

Show the answer

The answer is B.

The circles meet where

2 sin θ = 1

, i.e.

θ = \frac{π}{6}, \frac{5 π}{6}

. There

r = 2 sin θ

is the outer curve, so

A = \frac{1}{2} \int_{π /6}^{5 π /6} [(2 sin θ)^{2} - 1^{2}] d θ = \frac{1}{2} \int_{π /6}^{5 π /6} (1 - 2 cos 2 θ) d θ = \frac{π}{3} + \frac{3}{2}

Why not A: $\frac{2 π}{3} + 3$ omits the factor $\frac{1}{2}$ in $A = \frac{1}{2} \int (r_{out}^{2} - r_{in}^{2}) d θ$ .
Why not C: $\frac{2 π}{3} + \frac{3}{2}$ is the area inside $r = 2 sin θ$ alone over this arc; you must subtract the part inside $r = 1$ .
Why not D:The $3$ term is positive: evaluating $\frac{1}{2} \int_{π /6}^{5 π /6} (1 - 2 cos 2 θ) d θ$ gives $+ \frac{3}{2}$ .

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