In the next few sections, we’ll get the derivative rules that will let us find formulas for derivatives when our function comes to us as a formula. This is a very algebraic section, and you should get lots of practice. When you tell someone you have studied calculus, this is the one skill they will expect you to have.

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This AP Calculus AB class covers a review of Unit 3 on derivatives of composites and the Chain Rule.Skill:1.E Apply appropriate mathematical rules or procedu. Unit 3 - Derivatives Student Learning Objectives for Unit 3: Upon completion of Unit 3, students will be able to Conceptualize derivative presented graphically, numerically, and analytically. Interpret derivative as an instantaneous rate of change Calculate slopes and derivatives using the definition of a derivative.

Building Blocks

These are the simplest rules – rules for the basic functions. We won't prove these rules; we'll just use them. But first, let's look at a few so that we can see they make sense.

Example 1

Find the derivative of ( y=f(x)=mx+b )

This is a linear function, so its graph is its own tangent line! The slope of the tangent line, the derivative, is the slope of the line: [f'(x)=m]

Rule:

The derivative of a linear function is its slope.

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Example 2

Find the derivative of ( f(x)=135 ).

Think about this one graphically, too. The graph of f(x) is a horizontal line. So its slope is zero: [f'(x)=0]

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Example 3

Find the derivative of ( f(x)=x^2 ).

Recall the formal definition of the derivative: [f'(x)=limlimits_{hto 0} frac{f(x+h)-f(x)}{h}.]

Using our function ( f(x)=x^2 ), ( f(x+h)=(x+h)^2=x^2+2xh+h^2 ).

Then [ begin{align*} f'(x)=& limlimits_{hto 0} frac{f(x+h)-f(x)}{h} =& limlimits_{hto 0} frac{x^2+2xh+h^2-x^2}{h} =& limlimits_{hto 0} frac{2xh+h^2}{h} =& limlimits_{hto 0} frac{h(2x+h)}{h} =& limlimits_{hto 0} (2x+h) =& 2x end{align*} ]

From all that, we find that ( f'(x)=2x ).

Luckily, there is a handy rule we use to skip using the limit:

Power Rule

The derivative of ( f(x)=x^n ) is [f'(x)=nx^{n-1}.]

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Example 4

Find the derivative of ( g(x)=4x^3 )

Using the power rule, we know that if ( f(x)=x^3 ), then ( f'(x)=3x^2 ). Notice that (g) is 4 times the function (f). Think about what this change means to the graph of (g) – it’s now 4 times as tall as the graph of (f). If we find the slope of a secant line, it will be ( frac{Delta g}{Delta x}= frac{4Delta f}{Delta x} =4frac{Delta f}{Delta x} ); each slope will be 4 times the slope of the secant line on the (f) graph. This property will hold for the slopes of tangent lines, too: [frac{d}{dx}left(4x^3right)=4frac{d}{dx}left(x^3right)=4cdot 3x^2=12x^2.]

Rule:

Constants come along for the ride, i.e., ( frac{d}{dx}left( kfright)=kf'.)

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Here are all the basic rules in one place.

Derivative Rules: Building Blocks

In what follows, (f) and (g) are differentiable functions of (x).

Constant Multiple Rule

[ frac{d}{dx}left( kfright)=kf']

Sum and Difference Rule

[frac{d}{dx}left(fpm gright)=f' pm g']

Power Rule

[frac{d}{dx}left(x^nright)=nx^{n-1}]

Special cases: [frac{d}{dx}left(kright)=0 quad text{(Because ( k=kx^0 ).)}] [frac{d}{dx}left(xright)=1 quad text{(Because ( x=x^1 ).)}]

Exponential Functions

[frac{d}{dx}left(e^xright)=e^x] [frac{d}{dx}left(a^xright)=ln(a),a^x]

Natural Logarithm

[frac{d}{dx}left(ln(x)right)=frac{1}{x}]

The sum, difference, and constant multiple rule combined with the power rule allow us to easily find the derivative of any polynomial.

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Example 5

Find the derivative of ( p(x)=17x^{10}+13x^8-1.8x+1003 ).

[ begin{align*} frac{d}{dx}left( 17x^{10}+13x^8-1.8x+1003 right)=& frac{d}{dx}left( 17x^{10} right)+frac{d}{dx}left( 13x^8 right)-frac{d}{dx}left( 1.8x right)+frac{d}{dx}left( 1003 right) =& 17frac{d}{dx}left( x^{10} right)+13frac{d}{dx}left( x^8 right)-1.8frac{d}{dx}left( x right)+frac{d}{dx}left( 1003 right) =& 17left(10x^9right)+13left(8x^7right)-1.8left(1right)+0 =& 170x^9+104x^7-1.8 end{align*} ]

You don't have to show every single step. Do be careful when you're first working with the rules, but pretty soon you’ll be able to just write down the derivative directly:

Example 6

Find (frac{d}{dx}left( 17x^2-33x+12 right)).

Writing out the rules, we'd write [frac{d}{dx}left( 17x^2-33x+12 right)=17(2x)-33(1)+0=34x-33.]

Once you're familiar with the rules, you can, in your head, multiply the 2 times the 17 and the 33 times 1, and just write [frac{d}{dx}left( 17x^2-33x+12 right)=34x-33.]

The power rule works even if the power is negative or a fraction. In order to apply it, first translate all roots and basic rational expressions into exponents:

Example 7

Find the derivative of ( y=3sqrt{t}-frac{4}{t^4}+5e^t ).

The first step is translate into exponents: [y=3sqrt{t}-frac{4}{t^4}+5e^t=3t^{1/2}-4t^{-4}+5e^t]

Now you can take the derivative: [ begin{align*} frac{d}{dt}left( 3t^{1/2}-4t^{-4}+5e^t right)=& 3left(frac{1}{2}t^{-1/2}right)-4left(-4t^{-5}right)+5left(e^tright) =& frac{3}{2}t^{-1/2}+16t^{-5}+5e^t end{align*} ]

If there is a reason to, you can rewrite the answer with radicals and positive exponents: [y'= frac{3}{2}t^{-1/2}+16t^{-5}+5e^t= frac{3}{2sqrt{t}}+frac{16}{t^5}+5e^t]

Be careful when finding the derivatives with negative exponents.

We can immediately apply these rules to solve the problem we started the chapter with - finding a tangent line.

Example 8

Find the equation of the line tangent to ( g(t)=10-t^2 ) when (t = 2).

The slope of the tangent line is the value of the derivative. We can compute ( g'(t)=-2t ). To find the slope of the tangent line when (t = 2), evaluate the derivative at that point. The slope of the tangent line is -4.

To find the equation of the tangent line, we also need a point on the tangent line. Since the tangent line touches the original function at (t = 2), we can find the point by evaluating the original function: ( g(2)=10-2^2=6 ). The tangent line must pass through the point (2, 6).

Using the point-slope equation of a line, the tangent line will have equation ( y-6=-4(t-2) ). Simplifying to slope-intercept form, the equation is ( y=-4t+14 ).

Graphing, we can verify this line is indeed tangent to the curve:

We can also use these rules to help us find the derivatives we need to interpret the behavior of a function.

Example 9

In a memory experiment, a researcher asks the subject to memorize as many words from a list as possible in 10 seconds. Recall is tested, then the subject is given 10 more seconds to study, and so on. Suppose the number of words remembered after (t) seconds of studying could be modeled by ( W(t)=4t^{2/5} ). Find and interpret ( W'(20) ).

( W'(t)=4cdot frac{2}{5}t^{-3/5}=frac{8}{5}t^{-3/5} ), so ( W'(20)=frac{8}{5}(20)^{-3/5}approx 0.2652 ).

Since (W) is measured in words, and (t) is in seconds, (W') has units words per second. ( W'(20)approx 0.2652 ) means that after 20 seconds of studying, the subject is learning about 0.27 more words for each additional second of studying.

Business and Economics Terms

Next we will delve more deeply into some business applications. To do that, we first need to review some terminology.

Suppose you are producing and selling some item. The profit you make is the amount of money you take in minus what you have to pay to produce the items. Both of these quantities depend on how many you make and sell. (So we have functions here.) Here is a list of definitions for some of the terminology, together with their meaning in algebraic terms and in graphical terms.

Cost

Your cost is the money you have to spend to produce your items.

Fixed Cost

The Fixed Cost (FC) is the amount of money you have to spend regardless of how many items you produce. FC can include things like rent, purchase costs of machinery, and salaries for office staff. You have to pay the fixed costs even if you don’t produce anything.

Total Variable Cost

The Total Variable Cost (TVC) for (q) items is the amount of money you spend to actually produce them. TVC includes things like the materials you use, the electricity to run the machinery, gasoline for your delivery vans, maybe the wages of your production workers. These costs will vary according to how many items you produce.

Total Cost

The Total Cost (TC, or sometimes just C) for (q) items is the total cost of producing them. It’s the sum of the fixed cost and the total variable cost for producing (q) items.

Average Cost

The Average Cost (AC) for (q) items is the total cost divided by (q), or (frac{TC}{q}). You can also talk about the average fixed cost, (frac{FC}{q}), or the average variable cost, (frac{TVC}{q}).

Marginal Cost

The Marginal Cost (MC) at (q) items is the cost of producing the next item. Really, it’s [MC(q) = TC(q + 1) - TC(q).] In many cases, though, it’s easier to approximate this difference using calculus (see Example 11 below). And some sources define the marginal cost directly as the derivative, [MC(q) = TC'(q).] In this course, we will use both of these definitions as if they were interchangeable.

The units on marginal cost is cost per item.

Why is it okay that there are two definitions for Marginal Cost (and Marginal Revenue, and Marginal Profit)?

We have been using slopes of secant lines over tiny intervals to approximate derivatives. In this example, we’ll turn that around – we’ll use the derivative to approximate the slope of the secant line.

Notice that the “cost of the next item” definition is actually the slope of a secant line, over an interval of 1 unit: [MC(q) = C(q + 1) - 1 = frac{C(q+1)-1}{1}.]

So this is approximately the same as the derivative of the cost function at q: [MC(q) = C'(q).]

In practice, these two numbers are so close that there’s no practical reason to make a distinction. For our purposes, the marginal cost is the derivative is the cost of the next item.

Example 10

The table shows the total cost (TC) of producing (q) items.

1. What is the fixed cost?
2. When 200 items are made, what is the total variable cost? The average variable cost?
3. When 200 items are made, estimate the marginal cost.

1. The fixed cost is $20,000, the cost even when no items are made.

2. When 200 items are made, the total cost is $45,000. Subtracting the fixed cost, the total variable cost is $45,000 - $20,000 = $25,000.

   The average variable cost is the total variable cost divided by the number of items, so we would divide the $25,000 total variable cost by the 200 items made. $25,000/200 = $125. On average, each item had a variable cost of $125.

3. We need to estimate the value of the derivative, or the slope of the tangent line at (q = 200). Finding the secant line from (q=100) to (q=200) gives a slope of [ frac{45,000-35,000}{200-100}=100.]

   Finding the secant line from (q=200) to (q=300) gives a slope of [frac{53,000-45,000}{300-200}=80.]

   We could estimate the tangent slope by averaging these secant slopes, giving us an estimate of $90/item.

   This tells us that after 200 items have been made, it will cost about $90 to make one more item.

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Example 11

The cost to produce (x) items is (sqrt{x}) hundred dollars.

1. What is the cost for producing 100 items? 101 items? What is cost of the 101st item?
2. For (f(x) = sqrt{x}), calculate (f '(x)) and evaluate (f ') at (x = 100). How does (f '(100)) compare with the last answer in Part a?

1. Put (f(x) = sqrt{x} = x^{1/2})hundred dollars, the cost for (x) items. Then (f(100) =)$1000 and (f(101) =)$1004.99, so it costs $4.99 for that 101st item. Using this definition, the marginal cost is $4.99.
2. ( f'(x)=frac{1}{2}x^{-1/2}), so ( f'(100)=frac{1}{2sqrt{100}}=frac{1}{20} ) hundred dollars = $5.00.

Note how close these answers are! This shows (again) why it’s OK that we use both definitions for marginal cost.

Demand

Demand is the functional relationship between the price (p) and the quantity (q) that can be sold (that is demanded). Depending on your situation, you might think of (p) as a function of (q), or of (q) as a function of (p)

Revenue

Your revenue is the amount of money you actually take in from selling your products.

The Total Revenue (TR, or just R) for (q) items is the total amount of money you take in for selling (q) items. Total Revenue is price multiplied by quantity, [TR = p cdot q.]

Average Revenue

The Average Revenue (AR) for (q) items is the total revenue divided by (q), or [frac{TR}{q}.]

Marginal Revenue

The Marginal Revenue (MR) at (q) items is the revenue from producing the next item, [MR(q) = TR(q + 1) - TR(q).]

Just as with marginal cost, we will use both this definition and the derivative definition: [MR(q) = TR'(q).]

Profit

Your profit is what’s left over from total revenue after costs have been subtracted.

The Profit (P) for (q) items is [TR(q) - TC(q),] the difference between total revenue and total costs.

The average profit for (q) items is [frac{P}{q}.]

The marginal profit at (q) items is [P(q + 1) – P(q),] or [P'(q)]

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Graphical Interpretations of the Basic Business Math Terms

Illustration

Here are the graphs of TR and TC for producing and selling a certain item. The horizontal axis is the number of items, in thousands. The vertical axis is the number of dollars, also in thousands.

First, notice how to find the fixed cost and variable cost from the graph here. FC is the (y)-intercept of the TC graph. ((FC = TC(0)).) The graph of TVC would have the same shape as the graph of TC, shifted down. ((TVC = TC - FC).)

(MC(q) = TC(q + 1) - TC(q)), but that’s impossible to read on this graph. How could you distinguish between TC(4022) and TC(4023)? On this graph, that interval is too small to see, and our best guess at the secant line is actually the tangent line to the TC curve at that point. (This is the reason we want to have the derivative definition handy.)

(MC(q)) is the slope of the tangent line to the TC curve at ( (q, TC(q))).

(MR(q)) is the slope of the tangent line to the TR curve at ((q, TR(q))).

Profit is the distance between the TR and TC curve. If you experiment with a clear ruler, you’ll see that the biggest profit occurs exactly when the tangent lines to the TR and TC curves are parallel. This is the rule profit is maximized when ( MR = MC) which we'll explore later in the chapter.

Example 12

The demand, (D), for a product at a price of (p) dollars is given by ( D(p)=200-0.2p^2 ). Find the marginal revenue when the price is $10.

First we need to form a revenue equation. Since Revenue = Price( times )Quantity, and the demand equation shows the quantity of product that can be sold, we have [R(p)=D(p)cdot p=left(200-0.2p^2right)p=200p-0.2p^3.]

Now we can find marginal revenue by finding the derivative: [R'(p)=200(1)-0.2(3p^2)=200-0.6p^2]

At a price of $10, ( R'(10)=200-0.6(10)^2=140 ).

Notice the units for (R') are (frac{text{dollars of Revenue}}{text{dollar of price}}), so ( R'(10)=140 ) means that when the price is $10, the revenue will increase by $140 for each dollar that the price was increased.

Learning Objectives The sapper for mac.

• 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.

Derivatives of Vector-Valued Functions

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}}).

Use Equation ref{eq1}.

[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).

1. 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}}.]
2. 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.

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

1. 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}}).
2. 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}}).
3. 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]

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

[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:

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

Theorem: Properties of the Derivative of Vector-Valued Functions

Unit 3 Derivative Rules Of Compositesap Calculus Solver

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.

1. (dfrac{d}{,dt}[vecs{r}(t)⋅ vecs{u}(t)])
2. (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:

1. [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*}]
2. 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}}),

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

(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:

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

Solution

1. (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})
2. (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]

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

[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]

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].)

1. 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}}.]

2. 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:

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

Solution

1. We use the first part of the definition of the integral of a space curve:
2. [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*}]
3. 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*}]

4. 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]

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

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

Summary

• To calculate the derivative of a vector-valued function, calculate the derivatives of the component functions, then put them back into a new vector-valued function.
• Many of the properties of differentiation of scalar functions also apply to vector-valued functions.
• The derivative of a vector-valued function (vecs r(t)) is also a tangent vector to the curve. The unit tangent vector (vecs T(t)) is calculated by dividing the derivative of a vector-valued function by its magnitude.
• The antiderivative of a vector-valued function is found by finding the antiderivatives of the component functions, then putting them back together in a vector-valued function.
• The definite integral of a vector-valued function is found by finding the definite integrals of the component functions, then putting them back together in a vector-valued function.

Key Equations

• Derivative of a vector-valued function[vecs r′(t) = lim limits_{Delta t to 0} dfrac{vecs r(t+Delta t)−vecs r(t)}{ Delta t} nonumber]
• Principal unit tangent vector [vecs T(t)=frac{vecs r′(t)}{‖vecs r′(t)‖} nonumber]
• Indefinite integral of a vector-valued function [int [f(t) ,mathbf{hat{i}}+g(t),mathbf{hat{j}} + h(t) ,mathbf{hat{k}}],dt= left[ int f(t),dt right] ,mathbf{hat{i}}+ left[ int g(t),dt right] ,mathbf{hat{j}} + left[ int h(t),dt right] ,mathbf{hat{k}}nonumber]
• Definite integral of a vector-valued function [int_a^b [f(t) ,mathbf{hat{i}}+g(t) ,mathbf{hat{j}} + h(t) ,mathbf{hat{k}}],dt= left[int_a^b f(t),dt right] ,mathbf{hat{i}}+ left[ int _a^b g(t),dt right] ,mathbf{hat{j}} + left[ int _a^b h(t),dt right] ,mathbf{hat{k}}nonumber]

Glossary

definite integral of a vector-valued function

the vector obtained by calculating the definite integral of each of the component functions of a given vector-valued function, then using the results as the components of the resulting function

Unit 3 Derivative Rules Of Compositesap Calculus 2nd Edition

derivative of a vector-valued function

the derivative of a vector-valued function (vecs{r}(t)) is (vecs{r}′(t) = lim limits_{Delta t to 0} frac{vecs r(t+Delta t)−vecs r(t)}{ Delta t}), provided the limit exists

indefinite integral of a vector-valued function

a vector-valued function with a derivative that is equal to a given vector-valued function

Unit 3 Derivative Rules Of Compositesap Calculus Integrals

principal unit tangent vector

a unit vector tangent to a curve C

Unit 3 Derivative Rules Of Compositesap Calculus Calculator

tangent vector

to (vecs{r}(t)) at (t=t_0) any vector (vecs v) such that, when the tail of the vector is placed at point (vecs r(t_0)) on the graph, vector (vecs{v}) is tangent to curve C

Contributors and Attributions

• Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.