For the parametric equations x=t^2 and y=t+1, which rectangular equation represents the curve?
Parametric Equations and Motion Online Quiz Questions
Use this free practice quiz with 20 questions to review Parametric Equations and Motion, test your knowledge, and prepare for your next test or exam.
For x=t^2+1 and y=t^3-2t, what is the slope of the tangent line at t=1?
- A
-2
- B
-1/2
- C
1/2
- D
2
A projectile has velocity v(t)=⟨3,10-9.8t⟩ in metres per second. What is its speed at t=1 second?
- A
Approximately 3.01 m/s
- B
Approximately 3.20 m/s
- C
0.20 m/s
- D
3.00 m/s exactly
Select all conditions that correctly describe regular horizontal or vertical tangents of a parametric curve.
- A
A horizontal tangent can occur when y'(t)=0 and x'(t)≠0.
- B
A horizontal tangent requires both x'(t) and y'(t) to be zero.
- C
A vertical tangent can occur when x'(t)=0 and y'(t)≠0.
- D
Whenever x'(t)=0, the curve necessarily has a vertical tangent.
Select all statements that correctly describe the arc-length or distance-traveled integral for a parametrized motion.
- A
It integrates the magnitude of velocity over the interval.
- B
It can count repeated traversals of a curve.
- C
It is always equal to the magnitude of the particle's displacement.
- D
It becomes negative whenever the particle reverses direction.
True or false: If r(t)=⟨x(t),y(t)⟩ is a particle's position, then its velocity is v(t)=⟨x'(t),y'(t)⟩.
- A
True
- B
False
True or false: The signed area under a parametric curve from t=a to t=b is always calculated by integrating y(t)y'(t) with respect to t.
- A
True
- B
False
What is the standard term for the derivative of a particle's position with respect to time?
For the projectile x(t)=3t and y(t)=10t-4.9t^2, at what time does the projectile reach its highest point? Enter the time in seconds; an absolute tolerance of 0.01 s is accepted.
At a regular point of a parametric curve, a horizontal tangent requires and .
The magnitude of a particle's velocity vector is its .
Show how to find the tangent line at t=1 for the parametric curve x=t^2+1, y=t^3-2t. Include the point on the curve, the derivative calculations, the slope, and the final tangent-line equation.
For x=t^2+1 and y=t^3-2t, what conclusion follows at t=1 from the value of the second derivative?
- A
The curve is concave down because the slope is positive.
- B
The curve is concave up because d²y/dx²=5/4>0.
- C
The curve has no concavity because x'(1) is nonzero.
- D
The curve is vertical because d²y/dx² is positive.
True or false: A particle can have a nonzero velocity at a point where its path has a horizontal tangent.
- A
True
- B
False
Eliminate the parameter from x=t² and y=t+1. Which rectangular equation describes the curve?
- A
x=(y+1)²
- B
x=(y−1)²
- C
y=(x−1)²
- D
x=y²−1
A particle has velocity v(t)=⟨3,4⟩ metres per second at a particular instant. What is its speed at that instant?
- A
1
- B
4
- C
5
- D
7
What is the name of the quantity defined as the magnitude of a particle’s velocity?
For x=t³+t and y=t², what type of tangent does the curve have at t=0?
- A
A vertical tangent, because x'(0)=0
- B
A horizontal tangent, because y'(0)=0 and x'(0)≠0
- C
No tangent, because both derivatives are zero
- D
A horizontal tangent, because x'(0)=0 and y'(0)≠0
For the projectile x(t)=3t and y(t)=10t−4.9t², at what exact time does the projectile reach its highest point? Enter the time as a fraction in seconds; do not round.
A projectile has position x(t)=3t and y(t)=10t−4.9t², with distances in metres and time in seconds. Which vector gives its acceleration?
- A
⟨3,10−9.8t⟩ m/s²
- B
⟨0,−9.8⟩ m/s²
- C
⟨0,−4.9⟩ m/s²
- D
⟨3,−9.8t⟩ m/s²