This print-out should have 20 questions.

Multiple-choice questions may continue on the next column or page – find all choices before answering.

001

3. y =

1 4/x e 4

4. y =

1 x/4 e 4

5. y =

1 x/2 e 4

6. y =

1 x/8 e correct

2

10.0 points

Determine A so that the curve y = 4x + 2 can be written in parametric form as x(t) = t − 1 ,

y(t) = At − 2 .

1. A = −4

Explanation:

We have to eliminate the parameter t from the equations for x and y. Now from the equation for x it follows that t =

2. A = 2

4. A = 3

y =

5. A = −3

Explanation:

We have to eliminate t from the parametric equations for x and y. Now from the equation for x it follows that t = x + 1. Thus y = 4x + 2 = A(x + 1) − 2 .

Consequently

A = 4

.

10.0 points

Determine a Cartesian equation for the curve given in parametric form by

√

x(t) = 4 ln(4t) , y(t) = t .

2. y =

1 8/x e 2

1 x/4 e 4

003

6. A = 4 correct

1

1. y = ex/4

2

1 x/4 e .

4

But then

3. A = −2

002

1

1/2

=

1 x/8 e 2

.

10.0 points

Describe the motion of a particle with position P (x, y) when x = 2 sin t ,

y = 3 cos t

as t varies in the interval 0 ≤ t ≤ 2π.

1. Moves once counterclockwise along the ellipse (2x)2 + (3y)2 = 1 , starting and ending at (0, 3).

2. Moves once counterclockwise along the ellipse x2 y 2

+

= 1,

4

9 starting and ending at (0, 3).

3. Moves along the line x y

+

= 1,

2 3 starting at (2, 0) and ending at (0, 3).

sheriff (aas3563) – Homework 1 – spice – (54070)

4. Moves along the line

2

keywords: motion on curve, ellipse

x y

+

= 1,

2 3

004

10.0 points

starting at (0, 3) and ending at (2, 0).

5. Moves once clockwise along the ellipse

Which one of the following could be the graph of the curve given parametrically by

x2 y 2

+

= 1,

4

9

x(t) = 3 − t2 ,

starting and ending at (0, 3). correct

6. Moves once clockwise along the ellipse

y(t) = t3 − 2t ,

where the arrows indicate the direction of increasing t? y (2x)2 + (3y)2 = 1 ,

1.

starting and ending at (0, 3).

Explanation:

Since cos2 t + sin2 t = 1

x

for all t, the particle travels along the curve given in Cartesian form by x2 y 2

+

= 1;

4

9 this is an ellipse centered at the origin. At t = 0, the particle is at (2 sin 0, 3 cos 0), i.e., at the point (0, 3) on the ellipse. Now as t increases from t = 0 to t = π/2, x(t) increases from x = 0 to x = 2, while y(t) decreases from y = 3 to y = 0 ; in particular, the particle moves from a point on the positive y-axis to a point on the positive x-axis, so it is moving clockwise. In the same way, we see that as t increases from π/2 to π, the particle moves to a point on the negative y-axis, then to a point on the negative x-axis as t increases from π to 3π/2, until finally it returns to its starting point on the positive y-axis as t increases from 3π/2 to

2π.

Consequently, the particle moves clockwise once around the ellipse x2 y 2

+

= 1

4

9 starting and ending at (0, 3).

y

2.

x

y

3.

corx

, rect sheriff (aas3563) – Homework 1 – spice – (54070)

3

so

y

(x(−t), y(−t)) = (x(t), −y(t)) .

4. x Thus the graph is symmetric about the x-axis, eliminating three choices.

To decide which one of the remaining three it is, we can check the path traced out as t ranges from −∞ to +∞ by looking at sign charts for x(t) and y(t) because this will tell us in which quadrant the graph lies:

1. x(t) = 3 − t2 :

y

−

5. x −∞

−∞

6. x 0

√

3

−

+∞

2. y(t) = t(t2 − 2):

−

y

+

0

√

− 3

0 + 0 − 0

√

√

0

2

− 2

+

+∞

Thus the graph starts in quadrant III, crosses the y-axis into quadrant IV, then crosses the x-axis into quadrant I, crosses back into quadrant IV and so on. Consequently, the graph is y

x

Explanation:

All the graphs are symmetric either about the y-axis or the x-axis. Let’s check which it is for the graph of

(x(t), y(t)) = (3 − t2 , t3 − 2t) .

keywords: parametric curve, graph, direction,

005

10.0 points

Now x(−t) = 3 − (−t)2 = 3 − t2 = x(t) , and y(−t) = (−t)3 − 2(−t) = −(t3 − 2t) = −y(t) ,

Which one of