Quiz Review - Parametric/Vector-Valued Functions

The smooth curve C is represented by the parametric equations

 $x\left(t\right)=\tan t$  
 $y\left(t\right)=t^2-3t+8$  
for the interval  $-\frac{\pi}{4}\le t\le\frac{\pi}{2}$  .
The slope of the tangent line to C at the point (0, 8) is

• -3

• -3/2

• 0

• 2

Which of the following is an equation of the tangent line to the plane curve represented by the parametric equations

 $x\left(t\right)=2t-5$  ,  $y\left(t\right)=t^3$  
when t = 2?

• y = -6x + 14

• y = 6x 

• y = 6x + 2

• y = 6x + 14

For the smooth curve, 
 $x\left(t\right)=t^3-12t\ and\ y\left(t\right)=4t^2+t$  

find all the points where the tangent line is either horizontal or vertical.  CHECK ALL THAT APPLY

• horizontal: t = -1/8

• horizontal: t = 1/8

• vertical: t = -2

• vertical: t = 2

• vertical: t = 8

The arc length of the curve represented by the parametric equations

 $x\left(t\right)=2+4t^3$  and  $y\left(t\right)\ =\ 1\ +\ 6t^2$  

from t = 0 to t = 1 is

•  $8\sqrt{2}-4$  

•  $2^{\frac{7}{2}}-1$  

• 4

•  $16\sqrt{2}-8$  

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

 $\frac{dx}{dt}=\frac{\sqrt{t+2}}{e^t}\ and\ \frac{dy}{dt}=\sin^2t$  
What direction is the particle moving at t = 2?

• left and up

• right and up

• right and down

• left and down

Tags

CCSS.8.F.B.4

CCSS.HSF.IF.B.6

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

 $\frac{dx}{dt}=\frac{\sqrt{t+2}}{e^t}\ and\ \frac{dy}{dt}=\sin^2t$  
What is the x-coordinate of the particle at t = 4?

• 0.418

• 1

• 1.253

• 2.144

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

 $\frac{dx}{dt}=\frac{\sqrt{t+2}}{e^t}\ and\ \frac{dy}{dt}=\sin^2t$  
What is the speed of the particle at t = 4?

• -0.413

• 0.575

• 1.253

• 2.841

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

 $\frac{dx}{dt}=\frac{\sqrt{t+2}}{e^t}\ and\ \frac{dy}{dt}=\sin^2t$  
What is the acceleration vector at t = 4?

• -0.413

• 0.583

• (-0.041, 0.989)

• (0.429, 2.194)

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

 $\frac{dx}{dt}=\frac{\sqrt{t+2}}{e^t}\ and\ \frac{dy}{dt}=\sin^2t$  
What is the distance travelled by the particle from t = 2 to t=4?

• 0.253

• 0.583

• 0.651

• 1.432

If  $x=e^{-t}$  and  $y=\cos\left(2t\right)$  then $\frac{dy}{dx}=$  

•  $\frac{e^{-t}}{2\sin\left(2t\right)}$  

•  $\frac{2\sin\left(2t\right)}{e^{-t}}$  

•  $\frac{-e^{-t}}{2\sin\left(2t\right)}$  

•  $\frac{-2\sin\left(2t\right)}{e^{-t}}$