WorksheetsUnit 9A Review
Total questions: 59
Worksheet time: 2hrs 58mins
Find dx2d2y for the parametric equations given by x(t)=t2+1 and y(t)=t3
4t3
2t3
3t
6t
23
Consider the curve in the xy plane represented by x=et and y=te−t for t≥0 . The slope of the tangent to the curve at the point when x=3 (not t=3) is...
20.086
0.342
-0.005
-0.011
-0.03
The flight of a paper airplane is modeled by the curve given by x=t−3sint and y=4−3cost for time t in seconds and 0≤t≤10 (x and y are in feet). What is the distance traveled by the paper airplane?
31.343 feet
20.076 feet
49.229 feet
68.930 feet
The flight of a paper airplane is modeled by the curve given by x=t−3sint and y=4−3cost for time t in seconds and 0≤t≤10 (x and y are in feet). What describes the motion of the paper airplane at t=7 seconds?
Up and to the right
Down and to the right
Up and to the left
Down and to the left
The flight of a paper airplane is modeled by the curve given by x=t−3sint and y=4−3cost for time t in seconds and 0≤t≤10 (x and y are in feet). What is the slope of the tangent line to the path of the paper airplane at t=7 seconds?
-0.185
-0.640
-1.562
-5.409
Consider the curve given by x=t3+2t−1 and y=t2−t+5 . What is the arc length of the curve for 0≤t≤2 ?
12.278
4.960
13.230
5.619
A curve in the plane is defined by x=t3+t and y=t4+2t2 . An equation of the tangent line to the curve at t=1 is...
y=2x−1
y=2x
y=4x−5
y=8x
y=8x+13
Find dxdy for the curve given by x=(t−1)3 and y=t .
3(t−1)2
6t(t−1)21
t6(t−1)2
6t(t−1)2
The length of the curve determined by x=t2 and y=t from t=0 to t=4 is
∫044t+1dt
2∫04t2+1dt
∫042t2+1dt
∫044t2+1dt
2π∫044t2+1dt
Find the equation of the tangent line for the curve defined by x=2t and y=t2+5 at the point where t=2.
y=2x+1
y=2x+5
y=x+4
y=x−4
Consider the curve given by x=2sint and y=4cos2t for 0≤t≤π . Find dx2d2y at t=4π
−22
−2
22
2
Find the slope at the value t=3 for the parametric equations x=t+10 and y=t2+2t
8
-4
27/2
4
-27/2
Find the concavity at the value t=3 for the parametric equations x=t+10 and y=t2+2t
concave up
concave down
inflection point
Find all points of horizontal tangency to the curve x=7+3cost and y=−5+sint
(7,-5)
(4,-5)
(7,-4)
(7,-6)
If r(t)=<5t2−2t,3−lnt> , then −31r′(2)=
<18,21>
<18,−21>
<−6,−61>
<−6,61>
Given r′(t)=<te−t2,−e−t> and r(0)=<21,−1> find r(t) that satisfies the initial condition.
<2−e−t2,e−t−2>
<2−e−t2,e−t>
<22−e−t2,e−t>
<22−e−t2,e−t−2>
If r(t)=<e−4t+9,7−4t3> , then r"(t)=
<e−4t,−12t2>
<−4e−4t,−12t2>
<16e−4t,−24t>
<e−4t,−24t>
If dtdR=<23t+1,e−t> where R(0)=<0,0> , find R(t) .
<(t+1)23,−e−t>
<(t+1)23−1,−[e−t−1]>
<(t+1)21−1,−[e−t−1]>
<(t+1)23+1,−[e−t+1]>
∫02<et,−tet>dt=
<e2−1,−e2−1>
<e2−1,−e2>
<e2,−e2−1>
<e2−1,e2+1>
For t≥0 , a particle moves along a curve so that its position at time t is given by (x(t),y(t)). At t=3, the particle is at position (6,2). Given dtdx=1−cos2t and dtdy=1.2tt1.5 , is the vertical movement of the particle up or down at t=3?
Up
Down
For t≥0 , a particle moves along a curve so that its position at time t is given by (x(t),y(t)). At t=3, the particle is at position (6,2). Given dtdx=1−cos2t and dtdy=1.2tt1.5 , find the y-coordinate of the particle's position at t=5
9.644
8.352
7.123
8.857
For t≥0 , a particle moves along a curve so that its position at time t is given by (x(t),y(t)). At t=3, the particle is at position (6,2). Given dtdx=1−cos2t and dtdy=1.2tt1.5 , find the speed of the particle at t=5.
4.586
3.186
3.975
5.184
For t≥0 , a particle moves along a curve so that its position at time t is given by (x(t),y(t)). At t=3, the particle is at position (6,2). Given dtdx=1−cos2t and dtdy=1.2tt1.5 , find the distance traveled by the particle from time t=3 to t=5.
7.746
8.512
9.149
9.867
The position of a particle moving in the xy-plane is given by (x(t),y(t)) for which x′(t)=tsint and y′(t)=5e−3t+2 . What is the slope of the line tangent to the path of the particle at the point at which t=2?
0.904
1.107
1.819
2.012
3.660
The position of a particle moving in the xy-plane is given by x(t)=t3−3t2 and y(t)=12t−3t2 . At which point is the particle at rest?
(-4,12)
(-3,6)
(-2,9)
(0,0)
(3,4)
The position of a particle moving along a curve at time
t≥0 is given by the parametric equations (x(t),y(t)) where x(t)=3t−5 and y(t)=t2+3t−28 . What is the slope of the tangent line to the curve at t=6?2
3
5
15
For t≥0 , a particle is moving along a curve so that its position at time t is given by the parametric equations x(t)=sint and y(t)=t2−t+3 . Which of the following expressions gives the speed of the particle at time t?
sint+(t2−t+3)
cost+(2t−1)
sin2t+(t2−t+3)2
cos2t+(2t−1)2
At time t, a particle moving along a curve in the xy-plane has position (x(t), y(t)) where x(t)=sin(3πt) and y(t)=t2−5 . Which of the following gives the direction of motion for the particle at t=2?
Up and to the right
Up and to the left
Down and to the right
Down and to the left
For t≥0 , a particle is moving along a curve so that its position at time t is given by the parametric relations x(t) (the graph) and y(t)=t2−3t+1 . Which of the following expressions gives the total distance the particle travels on the interval [0,4]
∫04(−1)2+(2t−3)2dt
∫04(−1)2+(t2−3t+1)2dt
∫02(−2)2+(2t−3)2dt + ∫24(0)2+(2t−3)2dt
∫02(−2)2+(t2−3t+1)2dt + ∫24(0)2+(t2−3t+1)2dt
A curve is defined parametrically by the equations x(t)=3t and y(t)=41t2 . Which of the following represents the equation of the line tangent to the curve at t=4?
y=38x
y=4+38(x−6)
y=4+61(x−6)
y=4+34(x−6)
For what value(s) of t does the curve defined by the parametric equations x=t3−2t2−44 and y=t2+81t2 have a horizontal tangent?
0
2
0 and 4/3
4/3 and 2
The position of a particle moving in the xy-plane is given by the parametric equations x(t)=2cost and y(t)=2sint for time t≥0 . What is the speed of the particle when t=1.2?
0.919
0.959
2.301
5.293
At t≥0 , a particle moving in the xy-plane has a position vector given by r(t)=<t2−t+4,e2t−10> . What is the speed of the particle at time t=5?
25
82
85
577
An object moves in the xy-plane so that its position at any time t is given by the parametric equations x(t)=e2t and y(t)=2t3−t2+4 . What is the slope of the tangent line at t=2?
e410
10e4
e420
e416
Find the arc length of the curve defined by x=arcsint, y=ln1−t2 on the interval 0≤t≤21
0.549
0.836
1.247
0.333
Find dy/dx for the graph of the parametric equations
x=t2,y=t2+6t+5t+6t
1+t3
1+3t
tt+6
Find dx2d2y for the graph of the parametric equations
x=t2,y=t2+6t+5
2t3−3
−t6
−t23
−2t3
Find dxdy for the graph of the parametric equations
x=t,y=3t2+2t3t+t1
12t3+4t
t+1t
6t3+2t
Find dx2d2y for the graph of the parametric equations
x=t,y=3t2+2t36t+4
9+t1
18t+t2
9t+1
Find dxdy for the graph of the parametric equations
x=lnt ,y=t2+t2t2+t
2+t1
2t2+t1
2t+1
Find dx2d2y for the graph of the parametric equations
x=lnt ,y=t2+t4t+1
4+t1
4t2+t
lnt4t+1
For the curve defined by the parametric equations x=2cost, y=3sint , find the tangent line where t=4π
(y−232)=−23(x−2)
(y−2)=−23(x−232)
(y−232)=23(x−2)
(y−2)=23(x−232)
Given the parametric equations x=t2−t+1, y=t3−3t , find the point(s) where the curve has a horizontal tangent.
(1,-2)
(3,2)
(-3,2)
(1,2)
Given the parametric equations x=t2−t+1, y=t3−3t , find the point(s) where the curve has a vertical tangent.
(43,−811)
(43,813)
(47,−811)
(47,813)
Given the parametric equations x=3+2cost, y=−1+4sint , find the point(s) where the curve has a vertical tangent.
(5,-1)
(1,-1)
(3,3)
(3,-5)
What is the arc length of the curve created by the parametric equations x=t2, y=t3, 0≤t≤2
9.073
12
5.506
15.027
What is the arc length of the curve created by the parametric equations x=e2t+1, y=3t−1, −2≤t≤2
∫−224e4t+9dt
∫−22e4t+9dt
∫−222e4t+9dt
∫−224e2t+9dt
Identify the curve of x=2t+1, y=t2+2 .
A
B
C
D
At time t on [0,2], the velocity of a particle moving along the x-axis is given by v(t)=et2−2 . What is the total distance traveled by the particle during the time interval [0,2]?
12.453
13.638
51.598
53.598
The velocity of a particle moving in a straight line for t≥0 is given by v(t)=ln(t3+1) . What is the acceleration of the particle at t=4?
0.738
3.436
4.174
8.232
At time t≥0 , a particle moving in the xy-plane has velocity vector given by v(t)=<4e−t,sin(1+t)> . What is the total distance the particle travels between t=1 and t=3?
1.861
1.983
2.236
4.851
For time t≥0 seconds, the position of an object traveling along a curve in the xy-plane is given by the parametric equations x(t) and y(t), where dtdx=t2+3 and dtdy=t3+t . At what time t is the speed of the object 10 units per second?
1.675
1.813
4.217
10.191
At time t>0, the position of a particle moving along a curve in the xy-plane is (x(t),y(t)) where dtdx=t−5cost and dtdy=6cos(1+sint) . At time t=3, the particle is at position (-1,2). What is the equation of the tangent line to the path of the particle at t=3?
y=0.314x+2.314
y=0.314x+2.686
y=−3.010x−1.010
y=−3.010x+5.010
At time t>0, the position of a particle moving along a curve in the xy-plane is (x(t),y(t)) where dtdx=t−5cost and dtdy=6cos(1+sint) . At time t=3, the particle is at position (-1,2).
Find the time where the path of the particle is vertical. Is the motion up or down at that moment?
t=1.306, moving up
t=1.306, moving down
t=0.607, moving up
t=0.607, moving down
At time t>0, the position of a particle moving along a curve in the xy-plane is (x(t),y(t)) where dtdx=t−5cost and dtdy=6cos(1+sint) . At time t=3, the particle is at position (-1,2). Find the y-coordinate of the particle's position at t=0.
3.634
-1.794
0.634
-4.794
At time t>0, the position of a particle moving along a curve in the xy-plane is (x(t),y(t)) where dtdx=t−5cost and dtdy=6cos(1+sint) . At time t=3, the particle is at position (-1,2).
Find the total distance traveled by the particle for [0,3].
13.453
16.812
17.995
15.002
For t>0, a particle is moving along a curve so that its position at time t is ((x(t),y(t)). At time t=2, the particle is at position (1,5). It is known that dtdx=ett+2 and dtdy=sin2t . Find the x-coordinate of the particle's position at time t=4.
1.253
1.485
1.111
1.321
For t>0, a particle is moving along a curve so that its position at time t is ((x(t),y(t)). At time t=2, the particle is at position (1,5). It is known that dtdx=ett+2 and dtdy=sin2t . Find the speed of the particle at t=4.
0.575
1.899
0.332
0.136
For t>0, a particle is moving along a curve so that its position at time t is ((x(t),y(t)). At time t=2, the particle is at position (1,5). It is known that dtdx=ett+2 and dtdy=sin2t . Find the y component of the acceleration vector at t=4.
-0.041
0.989
1.234
3.005
