How do you find a higher-order derivative of a vector-valued function?

Keep differentiating each component. Differentiating twice gives the second derivative, and the process can be repeated as often as needed.

True or False? The derivative of a vector-valued function is a scalar.

False. Differentiating each component gives two results, which are grouped back into a vector. The derivative of a vector-valued function is itself a vector-valued function.

Which of the three motion vectors points along the direction of motion?

The velocity vector. It is tangent to the path, so it always points the way the particle is heading. The displacement and acceleration vectors generally do not.

How do you get from a particle's acceleration vector to its displacement vector?

Integrate twice. The first integration gives the velocity vector and needs two constants, found from a known velocity. The second gives the displacement vector and needs two more, found from a known position.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 9: Parametric Equations, Vector-Valued Functions & Polar CoordinatesVector-Valued Functions

Vector-Valued Functions (College Board AP® Calculus BC): Flashcards

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FrontDefining Vector-Valued Functions
Define a vector-valued function.

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Cards in this collection (22)

Define a vector-valued function.
A pair of functions $x (t)$ and $y (t)$ grouped together in vector notation as $r (t) = ⟨ x (t), y (t) ⟩$ .
One input gives two outputs.
A vector-valued function takes one input $t$ and returns two outputs grouped in a vector. It is written
$r (t) = ⟨ x (t),______⟩$
The completed function is $r (t) = ⟨ x (t), y (t) ⟩$ .
The first component is $x (t)$ and the second component is $y (t)$ .
What are the two outputs of a vector-valued function called?
The first component and the second component.
For $r (t_{0})$ they are $x (t_{0})$ and $y (t_{0})$ .
True or False?
A vector-valued function describes something different from a pair of parametric equations.
False.
The components $x (t)$ and $y (t)$ are parametric equations, so the two describe the same curve.
The vector notation simply groups them together.
Evaluate $r (t) = ⟨ t^{2} + 2, \frac{20}{\sqrt{t}} ⟩$ at $t = 4$ .
Substitute into both components.
$r (4) = ⟨ 16 + 2, \frac{20}{2} ⟩ = ⟨ 18, 10 ⟩$
The derivative of a vector-valued function is found by differentiating each component separately. For $r (t) = ⟨ x (t), y (t) ⟩$ it is
$\frac{d r}{d t} = ⟨ x' (t),______⟩$
The completed derivative is $\frac{d r}{d t} = ⟨ x' (t), y' (t) ⟩$ .
Each component is an ordinary function of $t$ , so each is differentiated on its own and the results are regrouped as a vector.
Why can a vector-valued function be differentiated component by component?
Each component is an ordinary real-valued function of $t$ , so every differentiation rule already met applies to it unchanged.
The vector notation only groups the results.
How do you find a higher-order derivative of a vector-valued function?
Keep differentiating each component.
Differentiating twice gives the second derivative, and the process can be repeated as often as needed.
True or False?
The derivative of a vector-valued function is a scalar.
False.
Differentiating each component gives two results, which are grouped back into a vector.
The derivative of a vector-valued function is itself a vector-valued function.
Find the second derivative of $⟨ 3 t^{2}, 5 e^{t} + \sin t ⟩$ at $t = 0$ .
Differentiating once gives $⟨ 6 t, 5 e^{t} + \cos t ⟩$ and again gives $⟨ 6, 5 e^{t} - \sin t ⟩$ .
At $t = 0$ this is $⟨ 6, 5 ⟩$ .
A vector-valued function is integrated by integrating each component separately. The rule is
$\int ⟨ x (t), y (t) ⟩ d t = ⟨ \int x (t) d t,______⟩$
The completed rule is $\int ⟨ x (t), y (t) ⟩ d t = ⟨ \int x (t) d t, \int y (t) d t ⟩$ .
Each component is integrated on its own, and the two results are regrouped as a vector.
True or False?
One constant of integration is enough when integrating a vector-valued function.
False.
Each component is integrated separately, so each needs its own constant.
Give them different letters, such as $C$ and $D$ , to keep them apart.
How do you find the two constants of integration?
The question gives a known vector value at a particular $t$ .
Substitute that $t$ , set the result equal to the known vector, and equate the components to get two equations.
A vector-valued function has derivative $⟨ 12 t^{3} - 2 t, 1 + 3 \sqrt{t} ⟩$ and equals $⟨ - 3, 7 ⟩$ at $t = 1$ . Find it.
Integrating gives $⟨ 3 t^{4} - t^{2} + C, t + 2 t^{\frac{3}{2}} + D ⟩$ .
At $t = 1$ this is $⟨ 2 + C, 3 + D ⟩$ , so $C = - 5$ and $D = 4$ .
Which integration techniques work on a vector-valued function?
All of them.
Each component is an ordinary function of $t$ , so substitution, parts and every other method apply to it exactly as before.
A particle's velocity vector is $⟨ x' (t), y' (t) ⟩$ . Differentiating it again gives the acceleration vector
$⟨ x'' (t),______⟩$
The completed acceleration vector is $⟨ x'' (t), y'' (t) ⟩$ .
The acceleration vector is the second derivative of the displacement vector, reached by differentiating each component twice.
Which of the three motion vectors points along the direction of motion?
The velocity vector.
It is tangent to the path, so it always points the way the particle is heading.
The displacement and acceleration vectors generally do not.
True or False?
The displacement vector of a particle is the same thing as its position vector.
True.
Both names describe the vector from the origin to the particle at time $t$ .
Its magnitude, $\sqrt{{(x (t))}^{2} + {(y (t))}^{2}}$ , is the particle's distance from the origin.
What is the magnitude of the acceleration vector?
It is $\sqrt{{(x'' (t))}^{2} + {(y'' (t))}^{2}}$ .
The magnitude of the velocity vector is found the same way and is the particle's speed.
A particle has velocity vector $⟨ 3 t^{2} - 12, 10 t^{4} ⟩$ . When does it travel parallel to the $y$ -axis?
When the $x$ -component is zero, so $3 t^{2} - 12 = 0$ and $t = \pm 2$ .
Time cannot be negative, so $t = 2$ .
How do you get from a particle's acceleration vector to its displacement vector?
Integrate twice.
The first integration gives the velocity vector and needs two constants, found from a known velocity.
The second gives the displacement vector and needs two more, found from a known position.
A particle has acceleration $⟨ 12 t + 2, - 4 \sin t ⟩$ , is at rest at $t = 0$ , and starts at $⟨ - 2, 1 ⟩$ . Find its displacement vector.
Integrating gives velocity $⟨ 6 t^{2} + 2 t, 4 \cos t - 4 ⟩$ , since being at rest at $t = 0$ forces $C = 0$ and $D = - 4$ .
Integrating again and using the start point gives $⟨ 2 t^{3} + t^{2} - 2, 4 \sin t - 4 t + 1 ⟩$ .

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