m 408m - calculus quest solutions

brar (msb2938) Homework 1 lidman (56310) 1

This print-out should have 21 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.

001 10.0 points

Determine A so that the curve

y = 3x+ 11

can be written in parametric form as

x(t) = t 5 , y(t) = At 4 .

1. A = 3

2. A = 4

3. A = 3 correct

4. A = 2

5. A = 2

6. A = 4Explanation:

We have to eliminate t from the parametricequations for x and y. Now from the equationfor x it follows that t = x+ 5. Thus

y = 3x+ 11 = A(x+ 5) 4 .

Consequently

A = 3 .

002 10.0 points

Find a Cartesian equation for the curvegiven in parametric form by

x(t) = 2 cos2 4t , y(t) = 3 sin2 4t .

1.x

2 y

3=

1

6

2.x

3+

y

2=

1

6

3. 2x+ 3 y = 6

4. 3x 2 y = 6

5. 3x+ 2 y = 6 correct

6.x

3 y

2=

1

6Explanation:

We have to eliminate the parameter t fromthe equations for x and y. Now

cos2 + sin2 = 1 .

Thus x

2+

y

3= 1 .

But then after simplification, the curve hasCartesian form

3x+ 2y = 6 .

003 10.0 points

Determine a Cartesian equation for thecurve given in parametric form by

x(t) = 2 ln(9t) , y(t) =t .

1. y =1

2ex/6

2. y =1

3e4/x

3. y =1

2ex/3

4. y =1

3ex/2

5. y =1

2e6/x

6. y =1

3ex/4 correct

Explanation:

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

t =1

9ex/2 .


But then

y =(19ex/2

)1/2=

1

3ex/4 .

004 10.0 points

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

x(t) = 2t t3 , y(t) = 3 t2 ,

where the arrows indicate the direction ofincreasing t?

1.x

y

2.

x

y

3.

x

y

correct

4.

x

y

5.

x

y

6.

x

y

Explanation:

All the graphs are symmetric either aboutthe y-axis or the x-axis. Lets check which itis for the graph of

(x(t), y(t)) = (2t t3, 3 t2) .

Now

x(t) = 2(t) (t)3 = (2t t3) = x(t) ,and

y(t) = 3 (t)2 = 3 t2 = y(t) ,so

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


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

To decide which one of the remaining threeit is, we can check the path traced out as tranges from to + by looking at signcharts for x(t) and y(t) because this will tellus in which quadrant the graph lies:

1. x(t) = t(2 t2):

2

0+

0

0 +2

0

+

2. y(t) = 3 t2:

3

0 +3

0

+

Thus the graph starts in quadrant IV, crossesthe x-axis into quadrant I, then crosses the y-axis into quadrant II, crosses back into quad-rant I and so on. Consequently, the graphis

x

y

keywords: parametric curve, graph, direction,

005 10.0 points

Find parametric equations representing theline segment joining P (3, 1) to Q(1, 3) as

(x(t), y(t)) , 0 t 1 ,

where

P = (x(0), y(0)) , Q = (x(1), y(1))

and x(t), y(t) are linear functions of t.

1. (1 4t, 3 + 4t)

2. (3 + 4t2, 1 4t2)

3. (1 4t2, 3 + 4t2)

4.(1 4 sin2 pit

2, 1 4 cos2 pit

2

)

5.(3 + 4 sin2 pit

2, 3 + 4 cos2 pit

2

)

6. (3 + 4t, 1 4t) correct

Explanation:

All six possible answers represent the linesegment joining P (3, 1) toQ(1, 3) in para-metric form

(x(t), y(t)) , 0 t 1 .

But only in the cases

(3 + 4t, 1 4t)

and

(1 4t, 3 + 4t)are x(t), y(t) linear functions of t. This al-ready eliminates four of the possible answers.On the other hand, the first of these starts atP and ends at Q, while the second starts at Qand ends at P . Consequently, only

(3 + 4t, 1 4t)

represents the line segment in parametricform by linear functions x(t), y(t) so that

P = (x(0), y(0)) , Q = (x(1), y(1)) .

006 10.0 points

A ladder 10 feet in length slides down a wallas its bottom is pulled away from the wall asshown in


10

x

y

P

Using the angle as parameter, find theparametric equations for the path followed bythe point P located 2 feet from the top of theladder.

1. (8 cos , 2 sin )

2. (2 sin , 8 cos )

3. (2 sec , 8 tan )

4. (8 sin , 2 cos )

5. (2 tan , 8 sec )

6. (8 tan , 2 sec )

7. (2 cos , 8 sin ) correct

8. (8 sec , 2 tan )

Explanation:

By right triangle trigonometry, the coordi-nates (x, y) are given respectively by

x = 2 cos , y = (10 2) sin .Consequently, the curve traced out by P hasthe parametric form

(2 cos , 8 sin )

for 0 pi/2. Eliminating , we see thatP traces out the portion of the ellipse

x2

4+

y2

64= 1

in the first quadrant.

keywords: parametric form, circle, ellipse,

007 10.0 points

When a mortar shell is fired with an initialvelocity of v0 ft/sec at an angle above thehorizontal, then its position after t seconds isgiven by the parametric equations

x = (v0 cos)t , y = (v0 sin)t 16t2 .

If the mortar shell hits the ground 400 feetfrom the mortar when = 75, determine v0.

1. v0 = 170 ft/sec

2. v0 = 160 ft/sec correct

3. v0 = 150 ft/sec

4. v0 = 180 ft/sec

5. v0 = 140 ft/sec

Explanation:

The mortar shell hits the ground when

y(t) = v0(sin 75)t 16t2 = 0 ,

i.e., after t1 = (v0 sin 75)/16 seconds, and it

will then be a distance of

400 = x(t1)

=v20 sin 75

cos 75

16=

v20 sin 150

32

feet from the mortar since

sin cos =1

2sin 2 .

But sin 150 = 1/2. Consequently,

v0 =64 400 = 160 ft/sec .


keywords: trajectory, parabola, distance,range, initial velocity

008 10.0 points

Find dy/dx when

x3 + 2y3 9xy + 1 = 0 .

1.dy

dx=

x2 3y2y2 + 3x

2.dy

dx=

x2 + 3y

2y2 3x

3.dy

dx=

3y + x2

2y2 + 3x

4.dy

dx=

x2 3y2y2 3x

5.dy

dx=

3y x22y2 3x correct

Explanation:

We use implicit differentiation. For then

3x2 + 6y2dy

dx 9y 9xdy

dx= 0 ,

which after solving for dy/dx and taking outthe common factor 3 gives

3((x2 3y) + dy

dx(2y2 3x)

)= 0 .

Consequently,

dy

dx=

3y x22y2 3x .

keywords: implicit differentiation, Folium ofDescartes, derivative,

009 10.0 points

Find an equation for the tangent line to thecurve given parametrically by

x(t) = 2t sin(pit2

), y(t) = 3t cos

(pit2

)

at the point P (x(1), y(1)).

1. y = 6 32pi

x

2. y =3

2pix 6

3. y =3

2pix+ 6

4. y =3pi

4(x 2)

5. y +3pi

4(x+ 2) = 0

6. y =3pi

4(2 x) correct

Explanation:

The point slope formula with t = 1 can beused to find an equation for the tangent lineat P .We first need to find the slope at P . Now

by the Chain Rule and Product Rule,

x(t) = 2(sin

(pit2

)+

pit

2cos

(pit2

)),

y(t) = 3(cos

(pit2

) pit

2sin

(pit2

)).

Thus

dy

dx=

y(t)

x(t)

=3

2

(2 cos(pit/2) pit sin(pit/2)2 sin(pit/2) + pit cos(pit/2)

),

so the tangent line at t = 1 has

slope = 3pi4

.

On the other hand,

P (x(1), y(1)) = (2, 0) .

By the point slope formula, therefore, anequation for the tangent line is

y = 3pi4(x 2) ,


which after simplification becomes

y =3pi

4(2 x) .

keywords: parametric curve, tangent line,slope, trig function,

010 10.0 points

Finddy

dxfor the curve given parametrically

by

x(t) = 1 + 4t2 , y(t) = 2t2 + t3 .

1.dy

dx=

4

4 + 3t

2.dy

dx=

1

4+

3

4t

3.dy

dx=

4

1 + 6t

4.dy

dx=

1

2+

3

8t correct

5.dy

dx=

8

1 + 6t

6.dy

dx=

8

4 + 3t

7.dy

dx=

1

2+

3

4t

8.dy

dx=

1

4+

3

2t

Explanation:

Differentiating with respect to t we see that

x(t) = 8t , y(t) = 4t+ 3t2 .

Consequently,

dy

dx=

y(t)

x(t)=

4t+ 3t2

8t=

1

2+

3

8t .

011 10.0 points

Determine all values of t for which the curvegiven parametrically by

x = 3t3 t2 + 4 , y = t3 + 3t2 4thas a vertical tangent?

1. t = 0 , 29

2. t = 2

3. t = 0 , 2

4. t = 2

5. t = 29

6. t = 0 ,2

9correct

7. t =2

9

8. t = 0 , 2Explanation:

After differentiation with respect to t wesee that

y(t) = 3t2 + 6t 4 , x(t) = 9t2 2t .Now

dy

dx=

y(t)

x(t)=

3t2 + 6t 49t2 2t ,

so the tangent line to the curve will be verticalat the solutions of

x(t) = t(9t 2) = 0 ,hence at

t = 0 ,2

9.

012 10.0 points

Determine all points P at which the tangentline to the curve given parametrically by

x(t) = t3 6t , y = 3t2

is parallel to the line (t, 2t).

1. P = (5, 1) , (4, 1)


2. P = (5, 1) , (4, 1)

3. P = (5, 1) , (4, 1)

4. P = (5, 3) , (4, 12)

5. P = (5, 3) , (4, 12) correct

6. P = (5, 12) , (4, 3)Explanation:

Since

x(t) = 3t2 6 , y(t) = 6t ,the slope of the tangent line to the curve isgiven by

dy

dx=

y(t)

x(t)= 6t

3t2 6 .

But if this tangent line is to be parallel to theline defined parametrically by (t, 2t), then

6t3t2 6 = 2 .

After cross-multiplication and simplification,this becomes the equation

6t2 + 6t 12 = 6(t 1)(t+ 2) = 0 .Consequently, P is the point corresponding tothe values t = 1 and t = 2, in which case

P = (5, 3) , (4, 12) .

013 10.0 points

The curve traced out by a point P on thecircumference of a circle as the circle rollsalong a straight line shown in

P

x

y

is called a Cycloid and the shaded region isthe region, A, below an arch.If the circle has radius R, the cycloid is

given parametrically by

x(t) = R(t sin t) , y(t) = R(1 cos t) .Find the area of A when R = 3.

1. area(A) = 27pi correct

2. area(A) = 23pi

3. area(A) = 26pi

4. area(A) = 24pi

5. area(A) = 25piExplanation:

When R = 3, the base of the arch extendsfrom the point 0 = x(0) to the point 6pi =x(2pi) on the x-axis. In this case

area(A) = 6pi0

y dx =

x(2pi)x(0)

3(1 cos t) dx ,

where

x = x(t) = 3(t sin t) , dx = 3(1 cos t) dt .Thus after a change of variable from x to t wesee that

area(A) = 9 2pi0

(1 cos t)2 dt

= 9

2pi0

(1 2 cos t+ cos2 t) dt

= 9

2pi0

(32 2 cos t+ 1

2cos 2t

)dt .

Consequently,

area(A) = 27pi .

keywords: parametric curve, area, cycloid

014 10.0 points


Determine the arc length of the astroidshown in

x

y

and given parametrically by

x(t) = 4 cos3 t , y(t) = 4 sin3 t .

1. arc length = 12pi

2. arc length = 24pi

3. arc length = 4

4. arc length = 4pi

5. arc length = 24 correct

6. arc length = 12

Explanation:

The arc length of the parametric curve

(x(t), y(t)) , a t b

is given by the integral

I =

ba

(x(t))2 + (y(t))2 dt .

But when

x(t) = 4 cos3 t , y(t) = 4 sin3 t

we see that

x(t) = 12 sin t cos2 t ,y(t) = 12 cos t sin2 t ,

in which case

(x(t))2 + (y(t))2

= (12 cos t sin t)2(cos2 t+ sin2 t)

= (12 cos t sin t)2 .

Thus

I = 12

2pi0

| cos t sin t| dt

= 48

pi/20

cos t sin t dt = 24[sin2 t

]pi/20

.

Consequently, the astroid has

arc length = 24 .

keywords: arc length, parametric curve, as-troid trig functions, definite integral

015 10.0 points

Which, if any, of

A. (4, pi/6) ,

B. (4, 7pi/3) ,

C. (4, 4pi/3) ,are polar coordinates for the point given inCartesian coordinates by P (2, 2

3)?

1. B only

2. A only

3. all of them

4. none of them

5. B and C only correct

6. A and B only

7. C only

8. A and C only


Explanation:

To convert from Cartesian coordinates topolar coordinates we use the relations:

x = r cos , y = r sin ,

so that

r2 = x2 + y2 , tan =y

x.

For the point P (2, 23) in Cartesian co-

ordinates, therefore, one choice of r and isr = 4 and = pi/3, but there are equivalentsolutions for r < 0 as well as values of dif-fering by integer multiples of pi. For the givenchoices we thus see that

A. FALSE: incorrect.

B. TRUE: differs from pi/3 by 2pi.

C. TRUE:

4 cos(4pi/3) = 2 , 4 sin(4pi/3) = 23 .

016 10.0 points

Find the Cartesian coordinates, (a, b),of the point given in polar coordinates byP (4, pi/3).

1. (a, b) = (23, 2)

2. (a, b) = (4, 43)

3. (a, b) = (2, 23)

4. (a, b) = (4, 23)

5. (a, b) = (23, 2)

6. (a, b) = (43, 4)

7. (a, b) = (2, 4)

8. (a, b) = (2, 23) correct

Explanation:

Since the relationship between Cartesiancoordinates and polar coordinates is

x = r cos , y = r sin ,

the point P (4, pi/3) is given in Cartesian co-ordinates by

P (4, pi/3) =(4 cos

pi

3, 4 sin

pi

3

)= (2, 2

3) .

keywords: polar coordinates, Cartesian coor-dinates

017 10.0 points

Which one of the following shaded-regionsin the plane consists of all points whose polarcoordinates satisfy the inequalities

0 r < 3 , 16pi 11

12pi ?

1.

2 4

2.

2 4

3.

2 4

4.

2 4


5.

2 4

correct

6.

2 4

Explanation:

Using the definition of polar coordinates(r, ), we see that the region defined by theinequalities

0 r < 3 , 16pi 11

12pi

is

2 4

keywords: polar coordinates, inequalities, po-lar graph,

018 10.0 points

Find the slope of the tangent line to thegraph of

r = e 2at = pi/4.

1. slope = epi/4 + 1

2. slope = epi/4

3. slope =1

epi/4 + 1

4. slope = epi/4

5. slope =1

epi/4 1

6. slope = epi/4 1 correct

Explanation:

The graph of a polar curve r = f() canexpressed by the parametric equations

x = f() cos , y = f() sin .

In this form the slope of the polar curve isgiven by

dy

dx=

y()

x().

Now, when

r = e 2 ,

we see that

y() = e sin + (e 2) cos ,

while

x() = e cos (e 2) sin .

But then

y(pi4

)=

12(2epi/4 2) , x

(pi4

)=

22.

Consequently, at = pi/4,

slope =dy

dx

=pi/4

= epi/4 1 .

019 10.0 points

Which one of the following could be thegraph of the polar function

r = 3 2 sin ?


1. correct

2.

3.

4.

5.

6.

Explanation:

The graph of

r = 3 2 sin in Cartesian coordinates

r

3

6

3

6

pi 2pi

can be used to determine the correspondingpolar graph. Indeed, by tracing in polar co-ordinates the value of r as varies, lookingespecially for values, if any, of where r = 0as well as the values of r at

= 0,pi

2, pi,

3pi

2, 2pi ,

we obtain the polar graph


keywords: polar graph, polar function, car-dioid, limacon,

020 10.0 points

Which one of the following could be thegraph of the polar function

r = cos 3 ?

1.

2.

3.

4. correct

5.

6.

Explanation:

The graphs of

r = sin , r = cos

are circles passing through the origin. Thisquestion shows how the graphs change as the


angle variable changes from to multiples2, 3, etc.To determine the polar graph of

r = cos 3

look first at its Cartesian graph

r

12

1

12

1

pi 2pi

Then, by tracing in polar coordinates thevalue of r as varies, looking especially atthe values of where r = 0 and the values ofr at

= 0 ,pi

2, pi ,

3pi

2, 2pi ,

we obtain the polar graph

keywords: polar graph, six-leaved rose, eight-leaved rose, four-leaved rose

021 10.0 points

Find a polar equation for the curve givenby the Cartesian equation

4y2 = x .

1. r = 4 sec tan

2. r = 4 csc tan

3. 4r = sec cot

4. 4r = csc cot correct

5. 4r = sec tan

6. r = 4 csc cot

Explanation:

We have to substitute for x, y in

4y2 = x

using the relations

x = r cos , y = r sin .

In this case the Cartesian equation becomes

4r2 sin2 = r cos .

Consequently, the polar form of the equationis

4r = csc cot .

m 408m - calculus quest solutions

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