Defining and Differentiating Vector-Valued Functions - AP Calculus BC Study Guide

The Core Idea: Defining and Differentiating Vector-Valued Functions

Vector-valued functions provide a powerful way to describe the motion of a particle in a two-dimensional plane. Instead of defining position with a single function $y=f(x)$, we define the particle's $x$ and $y$ coordinates independently as functions of a third parameter, typically time $t$. This results in a function of the form $r(t)=<x(t),y(t)>$, which gives the position vector of the particle at any given time.

The core idea of this topic is to extend the principles of differential calculus to these vector-valued functions. By differentiating the function with respect to the parameter $t$, we can determine key characteristics of the particle's motion. The first derivative, $r^{\prime }(t)$, yields the velocity vector, which tells us both the direction and rate of change of the particle's position. The second derivative, $r^{\prime \prime }(t)$, gives the acceleration vector. This framework allows us to analyze the instantaneous velocity, acceleration, and speed of an object moving along a curve in the plane.

Key Definitions and Formulas

The analysis of motion for vector-valued functions relies on a few fundamental definitions and formulas derived directly from the principles of differentiation.

1. Vector-Valued Function

A function that describes the position of a particle in the plane at time $t$ is given in one of two equivalent forms:

• Component form: $r(t)=<x(t),y(t)>$

• Standard unit vector form: $r(t)=x(t)i+y(t)j$

2. The Derivative (Velocity Vector)

The derivative of a vector-valued function is found by differentiating each of its component functions. The result is the velocity vector, $v(t)$.

• Component form:

  $$v(t)=r^{\prime }(t)=⟨\frac{d}{dt}[x(t)],\frac{d}{dt}[y(t)]⟩=<x^{\prime }(t),y^{\prime }(t)>$$

• Standard unit vector form:

  $$v(t)=r^{\prime }(t)=x^{\prime }(t)i+y^{\prime }(t)j$$

3. The Second Derivative (Acceleration Vector)

The second derivative of the position function $r(t)$ (or the first derivative of the velocity function $v(t)$) is the acceleration vector, $a(t)$.

• Component form:

  $$a(t)=v^{\prime }(t)=r^{\prime \prime }(t)=<x^{\prime \prime }(t),y^{\prime \prime }(t)>$$

• Standard unit vector form:

  $$a(t)=v^{\prime }(t)=r^{\prime \prime }(t)=x^{\prime \prime }(t)i+y^{\prime \prime }(t)j$$

4. Speed

Speed is the magnitude (or norm) of the velocity vector. It is a scalar quantity, not a vector.

• Formula:

  $$\text{speed}=||v(t)||=||r^{\prime }(t)||=\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}$$

Understanding Vector vs. Scalar Quantities

A critical nuance in this topic is the distinction between vector and scalar quantities. A vector, such as velocity or acceleration, has both magnitude and direction. A scalar, such as speed, has only magnitude.

The velocity vector $v(t)=<x^{\prime }(t),y^{\prime }(t)>$ tells you two things at once: how fast the particle is moving in the horizontal direction ($x^{\prime }(t)$) and how fast it is moving in the vertical direction ($y^{\prime }(t)$). The direction of this vector is tangent to the particle's path at that instant.

Speed, on the other hand, is the overall rate of motion, irrespective of direction. It is calculated by applying the Pythagorean theorem to the components of the velocity vector. Imagine a right triangle where the legs are the horizontal velocity ($x^{\prime }(t)$) and the vertical velocity ($y^{\prime }(t)$). The hypotenuse of this triangle represents the overall speed of the particle. This is why the formula $speed=\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}$ is identical to the formula for the magnitude of a vector. A particle can have a constant speed while its velocity vector is constantly changing (as in uniform circular motion).

Core Concepts & Rules

• Position: A particle's position in the $xy$-plane is defined by a vector-valued function $r(t)=<x(t),y(t)>$.

• Component-wise Differentiation: To differentiate a vector-valued function, you differentiate each of its component functions independently with respect to the parameter $t$.

• Velocity Vector: The first derivative of the position vector, $r^{\prime }(t)$, is the velocity vector $v(t)$. Its components, $x^{\prime }(t)$ and $y^{\prime }(t)$, represent the instantaneous rates of change of the horizontal and vertical positions, respectively.

• Acceleration Vector: The second derivative of the position vector, $r^{\prime \prime }(t)$, is the acceleration vector $a(t)$. It describes how the velocity vector is changing.

• Speed as Magnitude: Speed is the magnitude of the velocity vector, $||v(t)||$. It is a non-negative scalar value that measures the total rate of movement along the path.

Step-by-Step Example 1: Basic Application

Problem: A particle moves in the $xy$-plane so that its position at any time $t$ is given by the vector-valued function $r(t)=<t^3-6t^2,4\ln (t)>$ for $t>0$. Find the velocity vector, acceleration vector, and the speed of the particle at time $t=2$.

Step 1: Find the velocity vector $v(t)$

Differentiate each component of the position vector $r(t)$ with respect to $t$.

• $x(t)=t^3-6t^2⟹x^{\prime }(t)=3t^2-12t$

• $y(t)=4\ln (t)⟹y^{\prime }(t)=4\cdot \frac{1}{t}=\frac{4}{t}$

The velocity vector is $v(t)=r^{\prime }(t)=<3t^2-12t,\frac{4}{t}>$.

Step 2: Find the acceleration vector $a(t)$

Differentiate each component of the velocity vector $v(t)$ with respect to $t$.

• $x^{\prime }(t)=3t^2-12t⟹x^{\prime \prime }(t)=6t-12$

• $y^{\prime }(t)=4t^{-1}⟹y^{\prime \prime }(t)=-4t^{-2}=-\frac{4}{t^2}$

The acceleration vector is $a(t)=r^{\prime \prime }(t)=<6t-12,-\frac{4}{t^2}>$.

Step 3: Evaluate the velocity vector at $t=2$

Substitute $t=2$ into the expression for $v(t)$.

• $x^{\prime }(2)=3(2{)}^2-12(2)=3(4)-24=12-24=-12$

• $y^{\prime }(2)=\frac{4}{2}=2$

So, $v(2)=<-12,2>$.

Step 4: Evaluate the acceleration vector at $t=2$

Substitute $t=2$ into the expression for $a(t)$.

• $x^{\prime \prime }(2)=6(2)-12=12-12=0$

• $y^{\prime \prime }(2)=-\frac{4}{(2{)}^2}=-\frac{4}{4}=-1$

So, $a(2)=<0,-1>$.

Step 5: Calculate the speed at $t=2$

Use the speed formula with the components of the velocity vector at $t=2$, which are $x^{\prime }(2)=-12$ and $y^{\prime }(2)=2$.

• $\text{speed}=||v(2)||=\sqrt{(x^{\prime }(2){)}^2+(y^{\prime }(2){)}^2}$

• $\text{speed}=\sqrt{(-12{)}^2+(2{)}^2}$

• $\text{speed}=\sqrt{144+4}$

• $\text{speed}=\sqrt{148}$

The speed at $t=2$ is $\sqrt{148}$.

Step-by-Step Example 2: Exam-Style Application

Problem: The velocity of a particle moving in the $xy$-plane is given by the vector $v(t)=<\cos (t^2),e^{2t}-5t>$. At time $t=0$, the particle is at the point $(3,-1)$.

(a) Find the acceleration vector of the particle at time $t=1$.

(b) Find the speed of the particle at time $t=1$.

Part (a): Find the acceleration vector at $t=1$

Step 1: Find the general acceleration vector $a(t)$

The acceleration vector $a(t)$ is the derivative of the velocity vector $v(t)$. We must differentiate each component of $v(t)$.

• Let $v_x(t)=\cos (t^2)$. Using the chain rule, $a_x(t)=v_x^{\prime }(t)=-\sin (t^2)\cdot (2t)=-2t\sin (t^2)$.

• Let $v_y(t)=e^{2t}-5t$. The derivative is $a_y(t)=v_y^{\prime }(t)=e^{2t}\cdot 2-5=2e^{2t}-5$.

The acceleration vector is $a(t)=<-2t\sin (t^2),2e^{2t}-5>$.

Step 2: Evaluate $a(t)$ at $t=1$

Substitute $t=1$ into the expression for $a(t)$.

• $a_x(1)=-2(1)\sin (1^2)=-2\sin (1)$

• $a_y(1)=2e^{2(1)}-5=2e^2-5$

The acceleration vector at $t=1$ is $a(1)=<-2\sin (1),2e^2-5>$. (Note: The initial position was extra information not needed for this part of the problem).

Part (b): Find the speed of the particle at $t=1$

Step 1: Identify the velocity components at $t=1$

We are given the velocity vector $v(t)=<\cos (t^2),e^{2t}-5t>$. Substitute $t=1$.

• $x^{\prime }(1)=\cos (1^2)=\cos (1)$

• $y^{\prime }(1)=e^{2(1)}-5(1)=e^2-5$

So, $v(1)=<\cos (1),e^2-5>$.

Step 2: Apply the speed formula

The speed is the magnitude of the velocity vector $v(1)$.

• $\text{speed}=||v(1)||=\sqrt{(x^{\prime }(1){)}^2+(y^{\prime }(1){)}^2}$

• $\text{speed}=\sqrt{(\cos (1){)}^2+(e^2-5{)}^2}$

This is the exact expression for the speed at $t=1$.

Using Your Calculator

For vector-valued functions with components that are difficult or impossible to differentiate by hand, a graphing calculator is essential. The primary use is to find numerical derivatives of the components at a specific point, which is often required for calculating speed or a specific acceleration vector.

Problem: A particle's position is given by $r(t)=<\arctan (t),\sqrt{t^2+3t}>$. Find the speed of the particle at $t=4$.

Calculator Steps (TI-84 Style):

1. Find $x^{\prime }(4)$:

   • Press MATH and select 8: nDeriv(.

   • Enter the expression for the derivative of the x-component: $nDeriv(arctan(X),X,4)$.

   • The result is $x^{\prime }(4)\approx 0.0588$. Store this value, for example, in $A$. (STO->` `ALPHA` $A).

2. Find $y^{\prime }(4)$:

   • Press MATH and select 8: nDeriv(.

   • Enter the expression for the derivative of the y-component: $nDeriv(\surd (X^2+3X),X,4)$.

   • The result is $y^{\prime }(4)\approx 1.0444$. Store this value, for example, in $B$. (STO->` `ALPHA` $B).

3. Calculate Speed:

   • Use the stored values in the speed formula: $\surd (A^2+B^2)$.

   • $\surd ((0.0588{)}^2+(1.0444{)}^2)$

   • The result is approximately $1.046$.

Conclusion: The speed of the particle at $t=4$ is approximately $1.046$.

AP Exam Quick Hit

Common Question Types

• Direct Calculation: Given a particle's position $r(t)=<x(t),y(t)>$, you will be asked to find the velocity vector, acceleration vector, or speed at a specific time $t=c$. This is the most straightforward application of the derivative rules.

  • Example: "For a particle with position $r(t)=<5t^2,\sin (\pi t)>$, find the acceleration vector at $t=3$."

• Analysis of Motion: Given the position or velocity vector, determine when the particle is moving horizontally or vertically.

  • Example: "A particle's velocity is $v(t)=<t^2-9,2t-4>$. Find all times $t$ when the particle is moving vertically." (This requires finding when the horizontal component of velocity, $x^{\prime }(t)$, is zero, but the vertical component, $y^{\prime }(t)$, is not).

• Finding Speed from Velocity: Given the velocity vector $v(t)$, find the speed at a specific time. This is a common sub-part of a larger free-response question.

  • Example: "The velocity of a particle is given by $v(t)=<1-2t,t^3>$. What is the speed of the particle at $t=2$?"

Common Mistakes

• Confusing Velocity and Speed: Velocity is a vector $<x^{\prime }(t),y^{\prime }(t)>$, while speed is a scalar $\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}$. Do not provide the vector when asked for the speed.

• Incorrect Speed Calculation: A frequent error is to forget to square the components or to forget the square root in the speed formula. For example, writing $(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2$ or $\sqrt{x^{\prime }(t)+y^{\prime }(t)}$ instead of the correct formula.

• Errors in Differentiation: Basic derivative mistakes (especially with the chain rule, product rule, or derivatives of trig/log/exp functions) in the component functions will lead to incorrect velocity and acceleration vectors.

• Misinterpreting "At Rest": A particle is "at rest" only when its velocity vector is the zero vector, $<0,0>$. This means both$x^{\prime }(t)$ and $y^{\prime }(t)$ must be zero at the exact same time $t$. A common mistake is to find when $x^{\prime }(t)=0$ and when $y^{\prime }(t)=0$ and list both times, even if they are different.