Unit 3 Derivative Rules Of Compositesap Calculus

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Course Introduction
This course is designed to introduce you to the study of calculus. You will learn concrete applications of how calculus is used and, more importantly, why it works. Calculus is not a new discipline; it has been around since the days of Archimedes. However, Isaac Newton and Gottfried Leibniz, two seventeenth-century European mathematicians concurrently working on the same intellectual discovery hundreds of miles apart, were responsible for developing the field as we know it today. This brings us to our first question, what is today's calculus? In its simplest terms, calculus is the study of functions, rates of change, and continuity. While you may have cultivated a basic understanding of functions in previous math courses, in this course you will come to a more advanced understanding of their complexity, learning to take a closer look at their behaviors and nuances. In this course, we will address three major topics: limits, derivatives, and integrals, as well as study their respective foundations and applications. By the end of this course, you will have a solid understanding of the behavior of functions and graphs. Whether you are entirely new to calculus or just looking for a refresher on a particular topic, this course has something to offer, balancing computational proficiency with conceptual depth.
Unit 1: Analytic Geometry
Most of the material in this unit will be review. However, the notions of points, lines, circles, distance, and functions will be central in everything that follows. Lines are basic geometric objects which will be of great importance in the study of differential calculus, particularly in the study of tangent lines and linear approximations.
We will also take a look at the practical uses of mathematical functions. This course will use mathematical models, or structures, that predict practical situations in order to describe and study a number of real-life problems and situations. They are essential to the development of every major business and every scientific field in the modern world.
Completing this unit should take you approximately 9 hours.
Unit 2: Instantaneous Rate of Change: The Derivative
In this unit, you will study the instantaneous rate of change of a function. Motivated by this concept, you will develop the notion of limits, continuity, and the derivative. The limit asks the question, 'What does the function do as the independent variable becomes closer and closer to a certain value?' In simpler terms, the limit is the natural tendency of a function. The limit is incredibly important due to its relationship to the derivative, the integral, and countless other key mathematical concepts. A strong understanding of the limit is essential to success in the field of mathematics.
A derivative is a description of how a function changes as its input varies. In the case of a straight line, this derivative, or slope, is the same at every point, which is why we can describe the slope of an entire function with one number when it is linear. You will learn that we can do the same for nonlinear functions. The slope, however, will not be constant; it will change as the independent variable changes.
Completing this unit should take you approximately 16 hours.
Unit 3: Rules for Finding Derivatives
Computing a derivative requires the computation of a limit. Because limit computations can be rather involved, we like to minimize the amount of work we have to do in practice. In this unit, we build up some rules for differentiation which will speed up our calculations of derivatives. In particular, you will see how to differentiate the sum, difference, product, quotient, and composition of two (or more) functions. You will also learn rules for differentiating power functions (including polynomial and root functions).
Completing this unit should take you approximately 12 hours.
Unit 4: Transcendental Functions
In this unit, you will investigate the derivatives of trigonometric, inverse trigonometric, exponential, and logarithmic functions. Along the way, you will develop a technique of differentiation called implicit differentiation. Aside from allowing you to compute derivatives of inverse function, implicit differentiation will also be important in studying related rates problems later on.
Completing this unit should take you approximately 17 hours.
Unit 5: Curve Sketching
This unit will ask you to apply a little critical thinking to the topics this course has covered thus far. To properly sketch a curve, you must analyze the function and its first and second derivatives in order to obtain information about how the function behaves, taking into account its intercepts, asymptotes (vertical and horizontal), maximum values, minimum values, points of inflection, and the respective intervals between each of the above. After collecting this information, you will need to piece it all together in order to sketch an approximation to the graph of the original function.
Completing this unit should take you approximately 10 hours.
Unit 6: Applications of the Derivative
With a sufficient amount of sophisticated machinery under your belt, you will now start to look at how differentiation can be used to solve problems in various applied settings. Optimization is an important notion in fields like biology, economics, and physics when we want to know when growth is maximized, for example.
In addition to providing methods to solve problems directly, the derivative can also be used to find approximate solutions to problems. You will explore two such methods in this section: Newton's method and the method of differentials.
Completing this unit should take you approximately 13 hours.
Unit 7: Integration
In this unit of the course, you will learn about integral calculus, a subfield of calculus that studies the area formed under the curve of a function. Though not necessarily intuitive, this concept is closely related to the derivative, which you will revisit in this unit.
Completing this unit should take you approximately 21 hours.
Unit 8: Applications of Integration
In this unit, we will take a first look at how integration can and has been used to solve various types of problems. Now that you have conceptualized the relationship between integration and areas and distances, you are ready to take a closer look at various applications; these range from basic geometric identities to more advanced situations in physics and engineering.
Completing this unit should take you approximately 19 hours.
Final Exam

Learning Objectives

Write an expression for the derivative of a vector-valued function.
Find the tangent vector at a point for a given position vector.
Find the unit tangent vector at a point for a given position vector and explain its significance.
Calculate the definite integral of a vector-valued function.

To study the calculus of vector-valued functions, we follow a similar path to the one we took in studying real-valued functions. First, we define the derivative, then we examine applications of the derivative, then we move on to defining integrals. However, we will find some interesting new ideas along the way as a result of the vector nature of these functions and the properties of space curves.

Unit 3: Derivatives Functions can be analyzed graphically by their limiting behavior and rates of change. At the conclusion of this unit, students will be able to: 1. Evaluate Derivatives 2. Find the equation of a tangent line. 4 weeks Unit 4: Applications of Derivatives.
Many of the rules for calculating derivatives of real-valued functions can be applied to calculating the derivatives of vector-valued functions as well. Recall that the derivative of a real-valued function can be interpreted as the slope of a tangent line or the instantaneous rate of change of the function.

Derivatives of Vector-Valued Functions

The limit as x goes to negative four of x cubed plus 1. Convert “The limit of the square root of xplus 1, plus x, minus 3, as xgoes to 17” into the mathematical notation for limits. Evaluate lim𝑥→−4(𝑥2+ 8𝑥−1) 4. Evaluate lim𝑥→1(−5𝑥3+ 𝑥2+ 2) In problems 5 – 8, state the problem in limit notation and what it seems to be approaching.

Now that we have seen what a vector-valued function is and how to take its limit, the next step is to learn how to differentiate a vector-valued function. The definition of the derivative of a vector-valued function is nearly identical to the definition of a real-valued function of one variable. However, because the range of a vector-valued function consists of vectors, the same is true for the range of the derivative of a vector-valued function.

Definition: Derivative of Vector-Valued Functions

The derivative of a vector-valued function (vecs{r}(t)) is

[vecs{r}′(t) = lim limits_{Delta t to 0} dfrac{vecs{r}(t+Delta t)−vecs{r}(t)}{ Delta t} label{eq1} ]

provided the limit exists. If (vecs{r}'(t)) exists, then (vecs{r}(t)) is differentiable at (t). If (vecs{r}′(t)) exists for all (t) in an open interval ((a,b)) then (vecs{r}(t)) is differentiable over the interval ((a,b)). For the function to be differentiable over the closed interval ([a,b]), the following two limits must exist as well:

[vecs{r}′(a) = lim limits_{Delta t to 0^+} dfrac{vecs{r}(a+Delta t)−vecs{r}(a)}{ Delta t} ]

and

[vecs{r}′(b) = lim limits_{Delta t to 0^-} dfrac{vecs{r}(b+Delta t)−vecs{r}(b)}{ Delta t}]

Many of the rules for calculating derivatives of real-valued functions can be applied to calculating the derivatives of vector-valued functions as well. Recall that the derivative of a real-valued function can be interpreted as the slope of a tangent line or the instantaneous rate of change of the function. The derivative of a vector-valued function can be understood to be an instantaneous rate of change as well; for example, when the function represents the position of an object at a given point in time, the derivative represents its velocity at that same point in time.

We now demonstrate taking the derivative of a vector-valued function.

Example (PageIndex{1}): Finding the Derivative of a Vector-Valued Function

Use the definition to calculate the derivative of the function

[vecs{r}(t)=(3t+4) ,mathbf{hat{i}}+(t^2−4t+3) ,mathbf{hat{j}} .nonumber]

Solution

Let’s use Equation ref{eq1}:

[begin{align*} vecs{r}′(t) &= lim limits_{Delta t to 0} dfrac{vecs{r}(t+Δt)− vecs{r} (t)}{Δt} [4pt]
&= lim limits_{Delta t to 0} dfrac{[(3(t+Δt)+4),hat{mathbf{i}}+((t+Δt)^2−4(t+Δt)+3),hat{mathbf{j}}]−[(3t+4) ,hat{mathbf{i}}+(t^2−4t+3) ,hat{mathbf{j}}]}{Δt} [4pt]
&= lim limits_{Delta t to 0} dfrac{(3t+3Δt+4),hat{mathbf{i}}−(3t+4) ,hat{mathbf{i}}+(t^2+2tΔt+(Δt)^2−4t−4Δt+3) ,hat{mathbf{j}}−(t^2−4t+3),hat{mathbf{j}}}{Δt} [4pt]
&= lim limits_{Delta t to 0} dfrac{(3Δt),hat{mathbf{i}}+(2tΔt+(Δt)^2−4Δt),hat{mathbf{j}}}{Δt} [4pt]
&= lim limits_{Delta t to 0} (3 ,hat{mathbf{i}}+(2t+Δt−4),hat{mathbf{j}}) [4pt]
&=3 ,hat{mathbf{i}}+(2t−4) ,hat{mathbf{j}} end{align*} ]

Exercise (PageIndex{1})

Use the definition to calculate the derivative of the function (vecs{r}(t)=(2t^2+3) ,mathbf{hat{i}}+(5t−6) ,mathbf{hat{j}}).

Hint

Use Equation ref{eq1}.

Answer

[vecs{r}′(t)=4t ,mathbf{hat{i}}+5 ,mathbf{hat{j}} nonumber]

Notice that in the calculations in Example (PageIndex{1}), we could also obtain the answer by first calculating the derivative of each component function, then putting these derivatives back into the vector-valued function. This is always true for calculating the derivative of a vector-valued function, whether it is in two or three dimensions. We state this in the following theorem. The proof of this theorem follows directly from the definitions of the limit of a vector-valued function and the derivative of a vector-valued function.

Theorem (PageIndex{1}): Differentiation of Vector-Valued Functions

Let (f), (g), and (h) be differentiable functions of (t).

If (vecs{r}(t)=f(t) ,mathbf{hat{i}}+g(t) ,mathbf{hat{j}}) then [vecs{r}′(t)=f′(t) ,mathbf{hat{i}}+g′(t) ,mathbf{hat{j}}.]
If (vecs{r}(t)=f(t) ,mathbf{hat{i}}+g(t) ,mathbf{hat{j}} + h(t) ,mathbf{hat{k}}) then [vecs{r}′(t)=f′(t) ,mathbf{hat{i}}+g′(t) ,mathbf{hat{j}} + h′(t) ,mathbf{hat{k}}.]

Example (PageIndex{2}): Calculating the Derivative of Vector-Valued Functions

Use Theorem (PageIndex{1}) to calculate the derivative of each of the following functions.

(vecs{r}(t)=(6t+8) ,mathbf{hat{i}}+(4t^2+2t−3) ,mathbf{hat{j}})
(vecs{r}(t)=3 cos t ,mathbf{hat{i}}+4 sin t ,mathbf{hat{j}})
(vecs{r}(t)=e^t sin t ,mathbf{hat{i}}+e^t cos t ,mathbf{hat{j}}−e^{2t} ,mathbf{hat{k}})

Solution

We use Theorem (PageIndex{1}) and what we know about differentiating functions of one variable.

The first component of [vecs r(t)=(6t+8) ,mathbf{hat{i}}+(4t^2+2t−3) ,mathbf{hat{j}} nonumber] is (f(t)=6t+8). The second component is (g(t)=4t^2+2t−3). We have (f′(t)=6) and (g′(t)=8t+2), so the Theorem (PageIndex{1}) gives (vecs r′(t)=6 ,mathbf{hat{i}}+(8t+2),mathbf{hat{j}}).
The first component is (f(t)=3 cos t) and the second component is (g(t)=4 sin t). We have (f′(t)=−3 sin t) and (g′(t)=4 cos t), so we obtain (vecs r′(t)=−3 sin t ,mathbf{hat{i}}+4 cos t ,mathbf{hat{j}}).
The first component of (vecs r(t)=e^t sin t ,mathbf{hat{i}}+e^t cos t ,mathbf{hat{j}}−e^{2t} ,mathbf{hat{k}}) is (f(t)=e^t sin t), the second component is (g(t)=e^t cos t), and the third component is (h(t)=−e^{2t}). We have (f′(t)=e^t(sin t+cos t)), (g′(t)=e^t (cos t−sin t)), and (h′(t)=−2e^{2t}), so the theorem gives (vecs r′(t)=e^t(sin t+cos t),mathbf{hat{i}}+e^t(cos t−sin t),mathbf{hat{j}}−2e^{2t} ,mathbf{hat{k}}).

Exercise (PageIndex{2})

Calculate the derivative of the function

[vecs{r}(t)=(t ln t),mathbf{hat{i}}+(5e^t) ,mathbf{hat{j}}+(cos t−sin t) ,mathbf{hat{k}}. nonumber]

Hint

Identify the component functions and use Theorem (PageIndex{1}).

Answer

[vecs{r}′(t)=(1+ ln t) ,mathbf{hat{i}}+5e^t ,mathbf{hat{j}}−(sin t+cos t),mathbf{hat{k}} nonumber]

We can extend to vector-valued functions the properties of the derivative that we presented previously. In particular, the constant multiple rule, the sum and difference rules, the product rule, and the chain rule all extend to vector-valued functions. However, in the case of the product rule, there are actually three extensions:

for a real-valued function multiplied by a vector-valued function,
for the dot product of two vector-valued functions, and
for the cross product of two vector-valued functions.

Theorem: Properties of the Derivative of Vector-Valued Functions

Let (vecs{r}) and (vecs{u}) be differentiable vector-valued functions of (t), let (f) be a differentiable real-valued function of (t), and let (c) be a scalar.

[begin{array}{lrcll} mathrm{i.} & dfrac{d}{,dt}[cvecs r(t)] & = & cvecs r′(t) & text{Scalar multiple} nonumber mathrm{ii.} & dfrac{d}{,dt}[vecs r(t)±vecs u(t)] & = & vecs r′(t)±vecs u′(t) & text{Sum and difference} nonumber mathrm{iii.} & dfrac{d}{,dt}[f(t)vecs u(t)] & = & f′(t)vecs u(t)+f(t)vecs u′(t) & text{Scalar product} nonumber mathrm{iv.} & dfrac{d}{,dt}[vecs r(t)⋅vecs u(t)] & = & vecs r′(t)⋅vecs u(t)+vecs r(t)⋅vecs u′(t) & text{Dot product} nonumber mathrm{v.} & dfrac{d}{,dt}[vecs r(t)×vecs u(t)] & = & vecs r′(t)×vecs u(t)+vecs r(t)×vecs u′(t) & text{Cross product} nonumber mathrm{vi.} & dfrac{d}{,dt}[vecs r(f(t))] & = & vecs r′(f(t))⋅f′(t) & text{Chain rule} nonumber mathrm{vii.} & text{If} ; vecs r(t)·vecs r(t) & = & c, text{then} ; vecs r(t)⋅vecs r′(t) ; =0 ; . & mathrm{} nonumber end{array} nonumber ]

Proof

The proofs of the first two properties follow directly from the definition of the derivative of a vector-valued function. The third property can be derived from the first two properties, along with the product rule. Let (vecs u(t)=g(t),mathbf{hat{i}}+h(t),mathbf{hat{j}}). Then

[begin{align*} dfrac{d}{,dt}[f(t)vecs u(t)] &=dfrac{d}{,dt}[f(t)(g(t) ,mathbf{hat{i}}+h(t) ,mathbf{hat{j}})] [4pt]
&=dfrac{d}{,dt}[f(t)g(t) ,mathbf{hat{i}}+f(t)h(t) ,mathbf{hat{j}}] [4pt]
&=dfrac{d}{,dt}[f(t)g(t)] ,mathbf{hat{i}}+dfrac{d}{,dt}[f(t)h(t)] ,mathbf{hat{j}} [4pt]
&=(f′(t)g(t)+f(t)g′(t)) ,mathbf{hat{i}}+(f′(t)h(t)+f(t)h′(t)) ,mathbf{hat{j}} [4pt]
&=f′(t)vecs u(t)+f(t)vecs u′(t). end{align*} ]

To prove property iv. let (vecs r(t)=f_1(t) ,mathbf{hat{i}}+g_1(t) ,mathbf{hat{j}}) and (vecs u(t)=f_2(t) ,mathbf{hat{i}}+g_2(t) ,mathbf{hat{j}}). Then

[begin{align*} dfrac{d}{,dt}[vecs r(t)⋅vecs u(t)] &=dfrac{d}{,dt}[f_1(t)f_2(t)+g_1(t)g_2(t)] [4pt]
&=f_1′(t)f_2(t)+f_1(t)f_2′(t)+g_1′(t)g_2(t)+g_1(t)g_2′(t)=f_1′(t)f_2(t)+g_1′(t)g_2(t)+f_1(t)f_2′(t)+g_1(t)g_2′(t) [4pt]
&=(f_1′ ,mathbf{hat{i}}+g_1′ ,mathbf{hat{j}})⋅(f_2 ,mathbf{hat{i}}+g_2 ,mathbf{hat{j}})+(f_1 ,mathbf{hat{i}}+g_1 ,mathbf{hat{j}})⋅(f_2′ ,mathbf{hat{i}}+g_2′ ,mathbf{hat{j}}) [4pt]
&=vecs r′(t)⋅vecs u(t)+vecs r(t)⋅vecs u′(t). end{align*} ]

The proof of property v. is similar to that of property iv. Property vi. can be proved using the chain rule. Last, property vii. follows from property iv:

[begin{align*} dfrac{d}{,dt}[vecs r(t)·vecs r(t)] &=dfrac{d}{,dt}[c] [4pt]
vecs r′(t)·vecs r(t)+vecs r(t)·vecs r′(t) &= 0 [4pt]
2vecs r(t)·vecs r′(t) &= 0 [4pt]
vecs r(t)·vecs r′(t) &= 0 end{align*} ]

Now for some examples using these properties.

Example (PageIndex{3}): Using the Properties of Derivatives of Vector-Valued Functions

Given the vector-valued functions

[vecs{r}(t)=(6t+8),mathbf{hat{i}}+(4t^2+2t−3),mathbf{hat{j}}+5t ,mathbf{hat{k}} nonumber]

and

[vecs{u}(t)=(t^2−3),mathbf{hat{i}}+(2t+4),mathbf{hat{j}}+(t^3−3t),mathbf{hat{k}}, nonumber]

calculate each of the following derivatives using the properties of the derivative of vector-valued functions.

(dfrac{d}{,dt}[vecs{r}(t)⋅ vecs{u}(t)])
(dfrac{d}{,dt}[ vecs{u} (t) times vecs{u}′(t)])

Solution

We have (vecs{r}′(t)=6 ,mathbf{hat{i}}+(8t+2) ,mathbf{hat{j}}+5 ,mathbf{hat{k}}) and (vecs{u}′(t)=2t ,mathbf{hat{i}}+2 ,mathbf{hat{j}}+(3t^2−3) ,mathbf{hat{k}}). Therefore, according to property iv:

[begin{align*} dfrac{d}{,dt}[vecs r(t)⋅vecs u(t)] &= vecs r′(t)⋅vecs u(t)+vecs r(t)⋅vecs u′(t) [4pt]
&= (6 ,mathbf{hat{i}}+(8t+2) ,mathbf{hat{j}}+5 ,mathbf{hat{k}})⋅((t^2−3) ,mathbf{hat{i}}+(2t+4) ,mathbf{hat{j}}+(t^3−3t) ,mathbf{hat{k}}) [4pt]
& ; +((6t+8) ,mathbf{hat{i}}+(4t^2+2t−3) ,mathbf{hat{j}}+5t ,mathbf{hat{k}})⋅(2t ,mathbf{hat{i}}+2 ,mathbf{hat{j}}+(3t^2−3),mathbf{hat{k}}) [4pt]
&= 6(t^2−3)+(8t+2)(2t+4)+5(t^3−3t) [4pt]
& ; +2t(6t+8)+2(4t^2+2t−3)+5t(3t^2−3) [4pt]
&= 20t^3+42t^2+26t−16. end{align*}]
First, we need to adapt property v for this problem:
[dfrac{d}{,dt}[ vecs{u}(t) times vecs{u}′(t)]=vecs{u}′(t)times vecs{u}′(t)+ vecs{u}(t) times vecs{u}′′(t). nonumber]
Recall that the cross product of any vector with itself is zero. Furthermore,(vecs u′′(t)) represents the second derivative of (vecs u(t):)
[vecs u′′(t)=dfrac{d}{,dt}[vecs u′(t)]=dfrac{d}{,dt}[2t ,mathbf{hat{i}}+2 ,mathbf{hat{j}}+(3t^2−3) ,mathbf{hat{k}}]=2 ,mathbf{hat{i}}+6t ,mathbf{hat{k}}. nonumber]
Therefore,
[begin{align*} dfrac{d}{,dt}[vecs u(t) times vecs u′(t)] &=0+((t^2−3),hat{mathbf{i}}+(2t+4),hat{mathbf{j}}+(t^3−3t),hat{mathbf{k}})times (2 ,hat{mathbf{i}}+6t ,hat{mathbf{k}}) [4pt]
&= begin{vmatrix} ,hat{mathbf{i}} & ,hat{mathbf{j}} & ,hat{mathbf{k}} t^2-3 & 2t+4 & t^3 -3t 2 & 0 & 6t end{vmatrix} [4pt]
& =6t(2t+4) ,hat{mathbf{i}}−(6t(t^2−3)−2(t^3−3t)) ,hat{mathbf{j}}−2(2t+4) ,hat{mathbf{k}} [4pt]
& =(12t^2+24t) ,hat{mathbf{i}}+(12t−4t^3) ,hat{mathbf{j}}−(4t+8),hat{mathbf{k}}. end{align*}]

Exercise (PageIndex{3})

Calculate (dfrac{d}{,dt}[vecs{r}(t)⋅ vecs{r}′(t)]) and ( dfrac{d}{,dt}[vecs{u}(t) times vecs{r}(t)]) for the vector-valued functions:

(vecs{r}(t)=cos t ,mathbf{hat{i}}+ sin t ,mathbf{hat{j}}−e^{2t} ,mathbf{hat{k}})
(vecs{u}(t)=t ,mathbf{hat{i}}+ sin t ,mathbf{hat{j}}+ cos t ,mathbf{hat{k}}),

Hint

Follow the same steps as in Example (PageIndex{3}).

Answer

(dfrac{d}{,dt}[vecs{r}(t)⋅ vecs{r}′(t)]=8e^{4t})

( dfrac{d}{,dt}[ vecs{u}(t) times vecs{r}(t)] =−(e^{2t}(cos t+2 sin t)+ cos 2t) ,mathbf{hat{i}}+(e^{2t}(2t+1)− sin 2t) ,mathbf{hat{j}}+(t cos t+ sin t− cos 2t) ,mathbf{hat{k}})

Tangent Vectors and Unit Tangent Vectors

Recall that the derivative at a point can be interpreted as the slope of the tangent line to the graph at that point. In the case of a vector-valued function, the derivative provides a tangent vector to the curve represented by the function. Consider the vector-valued function

[vecs{r}(t)=cos t ,mathbf{hat{i}} + sin t ,mathbf{hat{j}} label{eq10}]

The derivative of this function is

[vecs{r}′(t)=−sin t ,mathbf{hat{i}} + cos t ,mathbf{hat{j}} nonumber ]

If we substitute the value (t=π/6) into both functions we get

[vecs{r} left(dfrac{π}{6}right)=dfrac{sqrt{3}}{2} ,mathbf{hat{i}}+dfrac{1}{2},mathbf{hat{j}} nonumber]

and

[ vecs{r}′ left(dfrac{π}{6} right)=−dfrac{1}{2},mathbf{hat{i}}+dfrac{sqrt{3}}{2},mathbf{hat{j}}. nonumber]

The graph of this function appears in Figure (PageIndex{1}), along with the vectors (vecs{r}left(dfrac{π}{6}right)) and (vecs{r}' left(dfrac{π}{6}right)).

Notice that the vector (vecs{r}′left(dfrac{π}{6}right)) is tangent to the circle at the point corresponding to (t=dfrac{π}{6}). This is an example of a tangent vector to the plane curve defined by Equation ref{eq10}.

Definition: principal unit tangent vector

Let (C) be a curve defined by a vector-valued function (vecs{r}), and assume that (vecs{r}′(t)) exists when (mathrm{t=t_0}) A tangent vector (vecs{r}) at (t=t_0) is any vector such that, when the tail of the vector is placed at point (vecs r(t_0)) on the graph, vector (vecs{r}) is tangent to curve (C). Vector (vecs{r}′(t_0)) is an example of a tangent vector at point (t=t_0). Furthermore, assume that (vecs{r}′(t)≠0). The principal unit tangent vector at (t) is defined to be

[vecs{T}(t)=dfrac{ vecs{r}′(t)}{‖vecs{r}′(t)‖},]

provided (‖vecs{r}′(t)‖≠0).

The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. To calculate a unit tangent vector, first find the derivative (vecs{r}′(t)). Second, calculate the magnitude of the derivative. The third step is to divide the derivative by its magnitude.

Example (PageIndex{4}): Finding a Unit Tangent Vector

Find the unit tangent vector for each of the following vector-valued functions:

(vecs{r}(t)=cos t ,mathbf{hat{i}}+sin t ,mathbf{hat{j}})
(vecs{u}(t)=(3t^2+2t) ,mathbf{hat{i}}+(2−4t^3),mathbf{hat{j}}+(6t+5),mathbf{hat{k}})

Solution

(begin{array}{lrcl} text{First step:} & vecs r′(t) & = & − sin t ,hat{mathbf{i}}+ cos t ,hat{mathbf{j}} text{Second step:} & ‖vecs r′(t)‖ & = & sqrt{(− sin t)^2+( cos t)^2} = 1 text{Third step:} & vecs T(t) & = & dfrac{vecs r′(t)}{‖vecs r′(t)‖}=dfrac{− sin t ,hat{mathbf{i}}+ cos t ,hat{mathbf{j}}}{1}=− sin t ,hat{mathbf{i}}+ cos t ,hat{mathbf{j}} end{array})
(begin{array}{lrcl} text{First step:} & vecs r′(t) & = & (6t+2) ,hat{mathbf{i}}−12t^2 ,hat{mathbf{j}}+6 ,hat{mathbf{k}} text{Second step:} & ‖vecs r′(t)‖ & = & sqrt{(6t+2)^2+(−12t^2)^2+6^2} text{} & text{} & = & sqrt{144t^4+36t^2+24t+40} text{} & text{} & = & 2 sqrt{36t^4+9t^2+6t+10} text{Third step:} & vecs T(t) & = & dfrac{vecs r′(t)}{‖vecs r′(t)‖}=dfrac{(6t+2) ,hat{mathbf{i}}−12t^2 ,hat{mathbf{j}}+6 ,hat{mathbf{k}}}{2 sqrt{36t^4+9t^2+6t+10}} text{} & text{} & = & dfrac{3t+1}{sqrt{36t^4+9t^2+6t+10}} ,hat{mathbf{i}} - dfrac{6t^2}{sqrt{36t^4+9t^2+6t+10}} ,hat{mathbf{j}} + dfrac{3}{sqrt{36t^4+9t^2+6t+10}} ,hat{mathbf{k}} end{array})

Exercise (PageIndex{4})

Find the unit tangent vector for the vector-valued function

[vecs r(t)=(t^2−3),mathbf{hat{i}}+(2t+1) ,mathbf{hat{j}}+(t−2) ,mathbf{hat{k}}. nonumber]

Hint

Follow the same steps as in Example (PageIndex{4}).

Answer

[vecs T(t)=dfrac{2t}{sqrt{4t^2+5}},mathbf{hat{i}}+dfrac{2}{sqrt{4t^2+5}},mathbf{hat{j}}+dfrac{1}{sqrt{4t^2+5}},mathbf{hat{k}} nonumber]

Unit 3 derivative rules of compositesap calculus 2nd edition

Integrals of Vector-Valued Functions

We introduced antiderivatives of real-valued functions in Antiderivatives and definite integrals of real-valued functions in The Definite Integral. Each of these concepts can be extended to vector-valued functions. Also, just as we can calculate the derivative of a vector-valued function by differentiating the component functions separately, we can calculate the antiderivative in the same manner. Furthermore, the Fundamental Theorem of Calculus applies to vector-valued functions as well.

The antiderivative of a vector-valued function appears in applications. For example, if a vector-valued function represents the velocity of an object at time t, then its antiderivative represents position. Or, if the function represents the acceleration of the object at a given time, then the antiderivative represents its velocity.

Definition: Definite and Indefinite Integrals of Vector-Valued Functions

Let (f), (g), and (h) be integrable real-valued functions over the closed interval ([a,b].)

The indefinite integral of a vector-valued function (vecs{r}(t)=f(t) ,hat{mathbf{i}}+g(t) ,hat{mathbf{j}}) is
[int [f(t) ,hat{mathbf{i}}+g(t) ,hat{mathbf{j}}],dt= left[ int f(t),dt right] ,hat{mathbf{i}}+ left[ int g(t),dt right] ,hat{mathbf{j}}.]
The definite integral of a vector-valued function is
[int_a^b [f(t) ,hat{mathbf{i}}+g(t) ,hat{mathbf{j}}],dt = left[ int_a^b f(t),dt right] ,hat{mathbf{i}}+ left[ int_a^b g(t),dt right] ,hat{mathbf{j}}.]
The indefinite integral of a vector-valued function (vecs r(t)=f(t) ,hat{mathbf{i}}+g(t) ,hat{mathbf{j}}+h(t) ,hat{mathbf{k}}) is
[int [f(t) ,hat{mathbf{i}}+g(t),hat{mathbf{j}} + h(t) ,hat{mathbf{k}}],dt= left[ int f(t),dt right] ,hat{mathbf{i}}+ left[ int g(t),dt right] ,hat{mathbf{j}} + left[ int h(t),dt right] ,hat{mathbf{k}}.]
The definite integral of the vector-valued function is
[int_a^b [f(t) ,hat{mathbf{i}}+g(t) ,hat{mathbf{j}} + h(t) ,hat{mathbf{k}}],dt= left[ int_a^b f(t),dt right] ,hat{mathbf{i}}+ left[ int_a^b g(t),dt right] ,hat{mathbf{j}} + left[ int_a^b h(t),dt right] ,hat{mathbf{k}}.]

Since the indefinite integral of a vector-valued function involves indefinite integrals of the component functions, each of these component integrals contains an integration constant. They can all be different. For example, in the two-dimensional case, we can have

[int f(t),dt=F(t)+C_1 ; and ; int g(t),dt=G(t)+C_2, nonumber]

where (F) and (G) are antiderivatives of (f) and (g), respectively. Then

[begin{align*} int [f(t) ,hat{mathbf{i}}+g(t) ,hat{mathbf{j}}],dt &= left[ int f(t),dt right] ,hat{mathbf{i}}+ left[ int g(t),dt right] ,hat{mathbf{j}} [4pt]
&= (F(t)+C_1) ,hat{mathbf{i}}+(G(t)+C_2) ,hat{mathbf{j}} [4pt]
&=F(t) ,hat{mathbf{i}}+G(t) ,hat{mathbf{j}}+C_1 ,hat{mathbf{i}}+C_2 ,hat{mathbf{j}} [4pt]
&= F(t) ,hat{mathbf{i}}+G(t) ,hat{mathbf{j}}+vecs{C} end{align*}]

where (vecs{C}=C_1 ,hat{mathbf{i}}+C_2 ,hat{mathbf{j}}). Therefore, the integration constants becomes a constant vector.

Example (PageIndex{5}): Integrating Vector-Valued Functions

Calculate each of the following integrals:

( displaystyle int [(3t^2+2t) ,hat{mathbf{i}}+(3t−6) ,hat{mathbf{j}}+(6t^3+5t^2−4) ,hat{mathbf{k}}],dt)
( displaystyle int [⟨t,t^2,t^3⟩ times ⟨t^3,t^2,t⟩] ,dt)
( displaystyle int_{0}^{frac{pi}{3}} [sin 2t ,hat{mathbf{i}}+ tan t ,hat{mathbf{j}}+e^{−2t} ,hat{mathbf{k}}],dt)

Solution

We use the first part of the definition of the integral of a space curve:
[begin{align*} int[(3t^2+2t),hat{mathbf{i}}+(3t−6) ,hat{mathbf{j}}+(6t^3+5t^2−4),hat{mathbf{k}}],dt &=left[int 3t^2+2t,dt right],hat{mathbf{i}}+ left[int 3t−6,dt right] ,hat{mathbf{j}}+ left[int 6t^3+5t^2−4,dt right] ,hat{mathbf{k}} [4pt]
&=(t^3+t^2) ,hat{mathbf{i}}+left(frac{3}{2}t^2−6tright) ,hat{mathbf{j}}+left(frac{3}{2}t^4+frac{5}{3}t^3−4tright),hat{mathbf{k}}+vecs C. end{align*}]
First calculate (⟨t,t^2,t^3⟩ times ⟨t^3,t^2,t⟩:)
[begin{align*} ⟨t,t^2,t^3⟩ times ⟨t^3,t^2,t⟩ &= begin{vmatrix} hat{mathbf{i}} & ,hat{mathbf{j}} & ,hat{mathbf{k}} t & t^2 & t^3 t^3 & t^2 & t end{vmatrix} [4pt]
&=(t^2(t)−t^3(t^2)) ,hat{mathbf{i}}−(t^2−t^3(t^3)),hat{mathbf{j}}+(t(t^2)−t^2(t^3)),hat{mathbf{k}} [4pt]
&=(t^3−t^5),hat{mathbf{i}}+(t^6−t^2),hat{mathbf{j}}+(t^3−t^5),hat{mathbf{k}}. end{align*} ]
Next, substitute this back into the integral and integrate:
[begin{align*} int [⟨t,t^2,t^3⟩ times ⟨t^3,t^2,t⟩],dt &= int (t^3−t^5) ,hat{mathbf{i}}+(t^6−t^2) ,hat{mathbf{j}}+(t^3−t^5),hat{mathbf{k}},dt [4pt]
&=left(frac{t^4}{4}−frac{t^6}{6}right),hat{mathbf{i}}+left(frac{t^7}{7}−frac{t^3}{3}right),hat{mathbf{j}}+left(frac{t^4}{4}−frac{t^6}{6}right),hat{mathbf{k}}+vecs C. end{align*}]
Use the second part of the definition of the integral of a space curve:
[begin{align*} int_0^{frac{pi}{3}} [sin 2t ,hat{mathbf{i}}+ tan t ,hat{mathbf{j}}+e^{−2t} ,hat{mathbf{k}}],dt &=left[int_0^{frac{π}{3}} sin 2t ,dt right] ,hat{mathbf{i}}+ left[ int_0^{frac{π}{3}} tan t ,dt right] ,hat{mathbf{j}}+left[int_0^{frac{π}{3}}e^{−2t},dt right] ,hat{mathbf{k}} [4pt]
&= (-tfrac{1}{2} cos 2t) Bigvert_{0}^{π/3} ,hat{mathbf{i}}−( ln |cos t|)Bigvert_{0}^{π/3} ,hat{mathbf{j}}−left(tfrac{1}{2}e^{−2t}right)Bigvert_{0}^{π/3} ,hat{mathbf{k}} [4pt]
&=left(−tfrac{1}{2} cos tfrac{2π}{3}+tfrac{1}{2} cos 0right) ,hat{mathbf{i}}−left( ln left( cos tfrac{π}{3}right)− ln( cos 0)right) ,hat{mathbf{j}}−left( tfrac{1}{2}e^{−2π/3}−tfrac{1}{2}e^{−2(0)}right) ,hat{mathbf{k}} [4pt]
& =left(tfrac{1}{4}+tfrac{1}{2}right) ,hat{mathbf{i}}−(−ln 2) ,hat{mathbf{j}}−left(tfrac{1}{2}e^{−2π/3}−tfrac{1}{2}right) ,hat{mathbf{k}} [4pt]
&=tfrac{3}{4},hat{mathbf{i}}+(ln 2) ,hat{mathbf{j}}+left(tfrac{1}{2}−tfrac{1}{2}e^{−2π/3}right),hat{mathbf{k}}. end{align*}]

Exercise (PageIndex{5})

Calculate the following integral:

[int_1^3 [(2t+4) ,mathbf{hat{i}}+(3t^2−4t) ,mathbf{hat{j}}],dt nonumber]

Hint

Use the definition of the definite integral of a plane curve.

Answer

[int_1^3 [(2t+4) ,mathbf{hat{i}}+(3t^2−4t) ,mathbf{hat{j}}],dt = 16 ,mathbf{hat{i}}+10 ,mathbf{hat{j}} nonumber]