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5.1: Vector-Valued Functions and Space Curves


Our study of vector-valued functions combines ideas from our earlier examination of single-variable calculus with our description of vectors in three dimensions from the preceding chapter. These definitions and theorems support the presentation of material in the rest of this chapter and also in the remaining chapters of the text.

Definition of a Vector-Valued Function

Our first step in studying the calculus of vector-valued functions is to define what exactly a vector-valued function is. We can then look at graphs of vector-valued functions and see how they define curves in both two and three dimensions.

Definition: Vector-valued Functions

A vector-valued function is a function of the form

[vecs r(t)=f(t),hat{mathbf{i}}+g(t),hat{mathbf{j}} ; ; ext{or} ; ;vecs r(t)=f(t),hat{mathbf{i}}+g(t),hat{mathbf{j}}+h(t),hat{mathbf{k}},]

where the component functions (f), (g), and (h), are real-valued functions of the parameter (t). Vector-valued functions are also written in the form

[vecs r(t)=⟨f(t),,g(t)⟩ ; ; ext{or} ; ; vecs r(t)=⟨f(t),,g(t),,h(t)⟩.]

In both cases, the first form of the function defines a two-dimensional vector-valued function; the second form describes a three-dimensional vector-valued function.

The parameter (t) can lie between two real numbers: (a≤t≤b). Another possibility is that the value of (t) might take on all real numbers. Last, the component functions themselves may have domain restrictions that enforce restrictions on the value of (t). We often use (t) as a parameter because (t) can represent time.

Example (PageIndex{1}): Evaluating Vector-Valued Functions and Determining Domains

For each of the following vector-valued functions, evaluate (vecs r(0)), (vecs r(frac{pi}{2})), and (vecs r(frac{2pi}{3})). Do any of these functions have domain restrictions?

  1. (vecs r(t)=4cos t,hat{mathbf{i}}+3sin t,hat{mathbf{j}})
  2. (vecs r(t)=3 an t,hat{mathbf{i}}+4 sec t,hat{mathbf{j}}+5t,hat{mathbf{k}})

Solution

  1. To calculate each of the function values, substitute the appropriate value of t into the function:

    egin{align*}vecs r(0) ; & = 4cos(0) hat{mathbf{i}}+3sin(0) hat{mathbf{j}} [5pt] & =4hat{mathbf{i}}+0 hat{mathbf{j}}=4hat{mathbf{i}} [5pt] vecs rleft(frac{pi}{2} ight) ; & = 4cosleft(frac{π}{2} ight)hat{mathbf{i}}+3sinleft(frac{π}{2} ight) hat{mathbf{j}} [5pt] & = 0hat{mathbf{i}}+3 hat{mathbf{j}}=3 hat{mathbf{j}} [5pt] vecs rleft(frac{2pi}{3} ight) ; & =4cosleft(frac{2π}{3} ight)hat{mathbf{i}}+3sinleft(frac{2π}{3} ight) hat{mathbf{j}} [5pt] & =4(−frac{1}{2})hat{mathbf{i}}+3(frac{sqrt{3}}{2}) hat{mathbf{j}}=−2 hat{mathbf{i}}+frac{3 sqrt{3}}{2} hat{mathbf{j}}end{align*}

    To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is (f(t)=4 cos t) and the second component function is (g(t)=3sin t). Neither of these functions has a domain restriction, so the domain of (vecs r(t)=4cos t,hat{mathbf{i}}+3 sin t ,hat{mathbf{j}}) is all real numbers.
  2. To calculate each of the function values, substitute the appropriate value of t into the function:[egin{align*}vecs r(0) ; & = 3 an(0)hat{mathbf{i}}+4sec(0) hat{mathbf{j}}+5(0) hat{mathbf{k}} [5pt] & = 0hat{mathbf{i}}+4j+0 hat{mathbf{k}}=4 hat{mathbf{j}} [5pt] vecs rleft(frac{pi}{2} ight) ; & = 3 anleft(frac{pi}{2} ight)hat{mathbf{i}}+4secleft(frac{pi}{2} ight) hat{mathbf{j}}+5left(frac{pi}{2} ight) hat{mathbf{k}},, ext{which does not exist} [5pt] vecs rleft(frac{2pi}{3} ight) ; & =3 anleft(frac{2 pi}{3} ight)hat{mathbf{i}}+4secleft(frac{2pi}{3} ight) hat{mathbf{j}}+5left(frac{2pi}{ 3} ight) hat{mathbf{k}} [5pt] & =3(−sqrt{3})hat{mathbf{i}}+4(−2)hat{mathbf{j}}+frac{10π}{3} hat{mathbf{k}} [5pt] & =-3sqrt{3})hat{mathbf{i}}−8hat{mathbf{j}}+frac{10π}{3} hat{mathbf{k}}end{align*}]To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is (f(t)=3 an t), the second component function is (g(t)=4sec t), and the third component function is (h(t)=5t). The first two functions are not defined for odd multiples of (frac{pi}{2}), so the function is not defined for odd multiples of (frac{pi}{2}). Therefore, [ ext{D}_{vecs r}={t,|,t≠ frac{(2n+1)pi}{2}}, onumber] where (n) is any integer.

Exercise (PageIndex{1})

For the vector-valued function (vecs r(t)=(t^2−3t) ,hat{mathbf{i}}+(4t+1) ,hat{mathbf{j}}), evaluate (vecs r(0),, vecs r(1)), and (vecs r(−4)). Does this function have any domain restrictions?

Hint

Substitute the appropriate values of (t) into the function.

Answer

(vecs r(0) = hat{mathbf{j}},, vecs r(1)=−2 hat{mathbf{i}}+5 hat{mathbf{j}},, vecs r(−4)=28 hat{mathbf{i}}−15 hat{mathbf{j}})

The domain of (vecs r(t)=(t^2−3t)hat{mathbf{i}}+(4t+1)hat{mathbf{j}}) is all real numbers.

Example (PageIndex{1}) illustrates an important concept. The domain of a vector-valued function consists of real numbers. The domain can be all real numbers or a subset of the real numbers. The range of a vector-valued function consists of vectors. Each real number in the domain of a vector-valued function is mapped to either a two- or a three-dimensional vector.

Graphing Vector-Valued Functions

Recall that a plane vector consists of two quantities: direction and magnitude. Given any point in the plane (the initial point), if we move in a specific direction for a specific distance, we arrive at a second point. This represents the terminal point of the vector. We calculate the components of the vector by subtracting the coordinates of the initial point from the coordinates of the terminal point.

A vector is considered to be in standard position if the initial point is located at the origin. When graphing a vector-valued function, we typically graph the vectors in the domain of the function in standard position, because doing so guarantees the uniqueness of the graph. This convention applies to the graphs of three-dimensional vector-valued functions as well. The graph of a vector-valued function of the form

[vecs r(t)=f(t), hat{mathbf{i}}+g(t),hat{mathbf{j}} onumber ]

consists of the set of all points ((f(t),,g(t))), and the path it traces is called a plane curve. The graph of a vector-valued function of the form

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

consists of the set of all points ((f(t),,g(t),,h(t))), and the path it traces is called a space curve. Any representation of a plane curve or space curve using a vector-valued function is called a vector parameterization of the curve.

Each plane curve and space curve has an orientation, indicated by arrows drawn in on the curve, that shows the direction of motion along the curve as the value of the parameter (t) increases.

Example (PageIndex{2}) : Graphing a Vector-Valued Function

Create a graph of each of the following vector-valued functions:

  1. The plane curve represented by (vecs r(t)=4 cos t ,hat{mathbf{i}}+3 sin t ,hat{mathbf{j}}), (0≤t≤2pi)
  2. The plane curve represented by (vecs r(t)=4 cos(t^3) ,hat{mathbf{i}}+3 sin(t^3) ,hat{mathbf{j}}), (0≤t≤sqrt[3]{2pi})
  3. The space curve represented by (vecs r(t)=4 cos t ,hat{mathbf{i}}+4 sin t ,hat{mathbf{j}}+t ,hat{mathbf{k}}), (0≤t≤4pi)

Solution

1. As with any graph, we start with a table of values. We then graph each of the vectors in the second column of the table in standard position and connect the terminal points of each vector to form a curve (Figure (PageIndex{1})). This curve turns out to be an ellipse centered at the origin.

Table (PageIndex{1}): Table of Values for (vecs r(t)=4 cos t ,hat{mathbf{i}}+3 sin t ,hat{mathbf{j}}), (0≤t≤2pi)
(t)(vecs r(t))(t)(vecs r(t))
(0)(4hat{mathbf{i}})(pi)(-4hat{mathbf{i}})
(dfrac{pi}{4})(2 sqrt{2} hat{mathbf{i}} + frac{3 sqrt{2}}{2}hat{mathbf{j}})(dfrac{5pi}{4})(-2 sqrt{2} hat{mathbf{i}} - frac{3 sqrt{2}}{2}hat{mathbf{j}})
(dfrac{pi}{2})(mathrm{3hat{mathbf{j}}})(dfrac{3pi}{2})(mathrm{-3hat{mathbf{j}}})
(dfrac{3pi}{4})( -2 sqrt{2} hat{mathbf{i}} + frac{3 sqrt{2}}{2}hat{mathbf{j}})(dfrac{7pi}{4})( 2 sqrt{2} hat{mathbf{i}} - frac{3 sqrt{2}}{2}hat{mathbf{j}})
(2pi)(4hat{mathbf{i}})

2. The table of values for (vecs r(t)=4 cos(t^3) ,hat{mathbf{i}}+3 sin(t^3) ,hat{mathbf{j}}), (0≤t≤sqrt[3]{2pi}) is as follows:

Table of Values for (vecs r(t)=4 cos(t^3) ,hat{mathbf{i}}+3 sin(t^3) ,hat{mathbf{j}}), (0≤t≤sqrt[3]{2pi})
(t)(vecs r(t))(t)(vecs r(t))
(0)(mathrm{4hat{mathbf{i}}})(displaystylesqrt[3]{pi})(mathrm{-4hat{mathbf{i}}})
(displaystyle sqrt[3]{dfrac{pi}{4}})(mathrm{ 2 sqrt{2} hat{mathbf{i}} + frac{3 sqrt{2}}{2}hat{mathbf{j}}})(displaystyle sqrt[3]{dfrac{5pi}{4}})(mathrm{ -2 sqrt{2} hat{mathbf{i}} - frac{3 sqrt{2}}{2}hat{mathbf{j}}})
(displaystyle sqrt[3]{dfrac{pi}{2}})(mathrm{3hat{mathbf{j}}})(displaystyle sqrt[3]{dfrac{3pi}{2}})(mathrm{-3hat{mathbf{j}}})
(displaystyle sqrt[3]{dfrac{3pi}{4}})(mathrm{ -2 sqrt{2} hat{mathbf{i}} + frac{3 sqrt{2}}{2}hat{mathbf{j}}})(displaystyle sqrt[3]{dfrac{7pi}{4}})(mathrm{ 2 sqrt{2} hat{mathbf{i}} - frac{3 sqrt{2}}{2}hat{mathbf{j}}})
( displaystylesqrt[3]{2pi})(mathrm{4hat{mathbf{i}}})

The graph of this curve is also an ellipse centered at the origin.

3. We go through the same procedure for a three-dimensional vector function.

Table of Values for (mathrm{r(t)= 4 cos t hat{mathbf{i}}+4 sin t hat{mathbf{j}}+t hat{mathbf{k}}}), (mathrm{0≤t≤4pi})
(t)(vecs r(t))(t)(vecs r(t))
(mathrm{0})(mathrm{4hat{mathbf{i}}})(mathrm{pi})(mathrm{-4hat{mathbf{i}}}+ pi hat{mathbf{k}})
(dfrac{pi}{4})(mathrm{2 sqrt{2} hat{mathbf{i}} + 2sqrt{2} hat{mathbf{j}} + frac{pi}{4} hat{mathbf{k}}})(dfrac{5pi}{4})(mathrm{ -2 sqrt{2} hat{mathbf{i}} - 2sqrt{2} hat{mathbf{j}} + frac{5pi}{4} hat{mathbf{k}}})
(dfrac{pi}{2})(mathrm{4hat{mathbf{j}} +frac{pi}{2} hat{mathbf{k}}})(dfrac{3pi}{2})(mathrm{-4hat{mathbf{j}} +frac{3pi}{2} hat{mathbf{k}}})
(dfrac{3pi}{4})(mathrm{ -2 sqrt{2} hat{mathbf{i}} + 2sqrt{2} hat{mathbf{j}} + frac{3pi}{4} hat{mathbf{k}}})(dfrac{7pi}{4})(mathrm{ 2 sqrt{2} hat{mathbf{i}} - 2sqrt{2} hat{mathbf{j}} + frac{7pi}{4} hat{mathbf{k}}})
(mathrm{2pi})(mathrm{4hat{mathbf{j}} + 2pi hat{mathbf{k}}})

The values then repeat themselves, except for the fact that the coefficient of (hat{mathbf{k}}) is always increasing ( (PageIndex{3})). This curve is called a helix. Notice that if the (hat{mathbf{k}}) component is eliminated, then the function becomes (vecs r(t)=4cos t hat{mathbf{i}}+ 4sin t hat{mathbf{j}}), which is a circle of radius 4 centered at the origin.

You may notice that the graphs in parts a. and b. are identical. This happens because the function describing curve b is a so-called reparameterization of the function describing curve a. In fact, any curve has an infinite number of reparameterizations; for example, we can replace (t) with (2t) in any of the three previous curves without changing the shape of the curve. The interval over which (t) is defined may change, but that is all. We return to this idea later in this chapter when we study arc-length parameterization. As mentioned, the name of the shape of the curve of the graph in (PageIndex{3}) is a helix. The curve resembles a spring, with a circular cross-section looking down along the (z)-axis. It is possible for a helix to be elliptical in cross-section as well. For example, the vector-valued function (vecs r(t)=4 cos t ,hat{mathbf{i}}+3 sin t ,hat{mathbf{j}}+t ,hat{mathbf{k}}) describes an elliptical helix. The projection of this helix into the (xy)-plane is an ellipse. Last, the arrows in the graph of this helix indicate the orientation of the curve as (t) progresses from (0) to (4π).

Exercise (PageIndex{2})

Create a graph of the vector-valued function (vecs r(t)=(t^2−1)hat{mathbf{i}}+(2t−3) hat{mathbf{j}}), (0≤t≤3).

Hint

Start by making a table of values, then graph the vectors for each value of (t).

Answer

At this point, you may notice a similarity between vector-valued functions and parameterized curves. Indeed, given a vector-valued function (vecs r(t)=f(t),hat{mathbf{i}}+g(t),hat{mathbf{j}}) we can define (x=f(t)) and (y=g(t)). If a restriction exists on the values of (t) (for example, (t) is restricted to the interval ([a,b]) for some constants (a

Limits and Continuity of a Vector-Valued Function

We now take a look at the limit of a vector-valued function. This is important to understand to study the calculus of vector-valued functions.

Definition: limit of a vector-valued function

A vector-valued function (vecs r) approaches the limit (vecs L) as (t) approaches (a), written

[lim limits_{t o a} vecs r(t) = vecs L,]

provided

[lim limits_{t o a} ig| vecs r(t) - vecs L ig| = 0.]

This is a rigorous definition of the limit of a vector-valued function. In practice, we use the following theorem:

Theorem: Limit of a vector-valued function

Let (f), (g), and (h) be functions of (t). Then the limit of the vector-valued function (vecs r(t)=f(t) hat{mathbf{i}}+g(t) hat{mathbf{j}}) as t approaches a is given by

[lim limits_{t o a} vecs r(t) = [lim limits_{t o a} f(t)] hat{mathbf{i}} + [lim limits_{t o a} g(t)] hat{mathbf{j}} , label{Th1}]

provided the limits (lim limits_{t o a} f(t)) and (lim limits_{t o a} g(t)) exist.

Similarly, the limit of the vector-valued function (vecs r(t)=f(t) hat{mathbf{i}}+g(t) hat{mathbf{j}}+h(t) hat{mathbf{k}}) as (t) approaches (a) is given by

[lim limits_{t o a} vecs r(t) = [lim limits_{t o a} f(t)] hat{mathbf{i}} + [lim limits_{t o a} g(t)] hat{mathbf{j}} +[lim limits_{t o a} h(t)] hat{mathbf{k}} , label{Th2}]

provided the limits (lim limits_{t o a} f(t)), (lim limits_{t o a} g(t)) and (lim limits_{t o a} h(t)) exist.

In the following example, we show how to calculate the limit of a vector-valued function.

Example (PageIndex{3}): Evaluating the Limit of a Vector-Valued Function

For each of the following vector-valued functions, calculate (lim limits_{t o 3}vecs r(t)) for

  1. (vecs r(t)=(t^2−3t+4) hat{mathbf{i}}+(4t+3)hat{mathbf{j}})
  2. (vecs r(t)=frac{2t−4}{t+1}hat{mathbf{i}}+frac{t}{t^2+1} hat{mathbf{j}}+(4t−3) hat{mathbf{k}})

Solution

  1. Use Equation ef{Th1} and substitute the value (t=3) into the two component expressions:

[ egin{align*} lim limits_{t o 3} vecs r(t) ; & = lim limits_{t o 3} [(t^2−3t+4) hat{mathbf{i}} + (4t+3) hat{mathbf{j}}] [5pt] & = [lim limits_{t o 3} (t^2−3t+4)]hat{mathbf{i}}+[lim limits_{t o 3} (4t+3)] hat{mathbf{j}} [5pt] & = 4 hat{mathbf{i}}+15 hat{mathbf{j}} end{align*}]

  1. Use Equation ef{Th2} and substitute the value (t=3) into the three component expressions:

[ egin{align*} lim limits_{t o 3} vecs r(t) ; & = lim limits_{t o 3}(dfrac{2t−4}{t+1}hat{mathbf{i}}+dfrac{t}{t^2+1}hat{mathbf{j}}+(4t−3) hat{mathbf{k}}) [5pt] & = [lim limits_{t o 3} (dfrac{2t−4}{t+1})]hat{mathbf{i}}+[lim limits_{t o 3} (dfrac{t}{t^2+1})] hat{mathbf{j}} [lim limits_{t o 3} (4t−3)] hat{mathbf{k}} [5pt] & = dfrac{1}{2} hat{mathbf{i}}+dfrac{3}{10}hat{mathbf{j}}+9 hat{mathbf{k}} end{align*}]

Exercise (PageIndex{3})

Calculate (lim limits_{t o 2} vecs r(t)) for the function (vecs r(t) = sqrt{t^2 + 3t - 1},hat{mathbf{i}}−(4t-3)hat{mathbf{j}}− sin frac{(t+1)pi}{2}hat{mathbf{k}})

Hint

Use Equation ef{Th2} from the preceding theorem.

Answer

[lim limits_{t o 2} vecs r(t) = 3hat{mathbf{i}}−5hat{mathbf{j}}+hat{mathbf{k}}]

Now that we know how to calculate the limit of a vector-valued function, we can define continuity at a point for such a function.

Definitions

Let (f), (g), and (h) be functions of (t). Then, the vector-valued function (vecs r(t)=f(t) hat{mathbf{i}}+g(t)hat{mathbf{j}}) is continuous at point (t=a) if the following three conditions hold:

  1. (vecs r(a)) exists
  2. (lim limits_{t o a} vecs r(t)) exists
  3. (lim limits_{t o a} vecs r(t) = vecs r(a))

Similarly, the vector-valued function (vecs r(t)=f(t) hat{mathbf{i}}+g(t)hat{mathbf{j}}+h(t)hat{mathbf{k}}) is continuous at point (t=a) if the following three conditions hold:

  1. (vecs r(a)) exists
  2. (lim limits_{t o a} vecs r(t)) exists
  3. (lim limits_{t o a} vecs r(t) = vecs r(a))

Summary

  • A vector-valued function is a function of the form (vecs r(t)=f(t) hat{mathbf{i}}+ g(t) hat{mathbf{j}}) or (vecs r(t)=f(t) hat{mathbf{i}}+g(t) hat{mathbf{j}}+h(t) hat{mathbf{k}}), where the component functions (f), (g), and (h) are real-valued functions of the parameter (t).
  • The graph of a vector-valued function of the form (vecs r(t)=f(t) hat{mathbf{i}}+g(t) hat{mathbf{j}}) is called a plane curve. The graph of a vector-valued function of the form (vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}}+h(t) hat{mathbf{k}}) is called a space curve.
  • It is possible to represent an arbitrary plane curve by a vector-valued function.
  • To calculate the limit of a vector-valued function, calculate the limits of the component functions separately.

Key Equations

  • Vector-valued function
    (vecs r(t)=f(t) hat{mathbf{i}}+g(t) hat{mathbf{j}}) or (vecs r(t)=f(t) hat{mathbf{i}}+g(t) hat{mathbf{j}}+h(t) hat{mathbf{k}}),or (vecs r(t)=⟨f(t),g(t)⟩) or (vecs r(t)=⟨f(t),g(t),h(t)⟩)
  • Limit of a vector-valued function
    (lim limits_{t o a} vecs r(t) = [lim limits_{t o a} f(t)] hat{mathbf{i}} + [lim limits_{t o a} g(t)] hat{mathbf{j}}) or (lim limits_{t o a} vecs r(t) = [lim limits_{t o a} f(t)] hat{mathbf{i}} + [lim limits_{t o a} g(t)] hat{mathbf{j}} + [lim limits_{t o a} h(t)] hat{mathbf{k}})

Glossary

component functions
the component functions of the vector-valued function (vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}}) are (f(t)) and (g(t)), and the component functions of the vector-valued function (vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}}+h(t)hat{mathbf{k}}) are (f(t)), (g(t)) and (h(t))
helix
a three-dimensional curve in the shape of a spiral
limit of a vector-valued function
a vector-valued function (vecs r(t)) has a limit (vecs L) as (t) approaches (a) if (lim limits{t o a} left| vecs r(t) - vecs L ight| = 0)
plane curve
the set of ordered pairs ((f(t),g(t))) together with their defining parametric equations (x=f(t)) and (y=g(t))
reparameterization
an alternative parameterization of a given vector-valued function
space curve
the set of ordered triples ((f(t),g(t),h(t))) together with their defining parametric equations (x=f(t)), (y=g(t)) and (z=h(t))
vector parameterization
any representation of a plane or space curve using a vector-valued function
vector-valued function
a function of the form (vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}}) or (vecs r(t)=f(t)hat{mathbf{i}}+g(t)hat{mathbf{j}}+h(t)hat{mathbf{k}}),where the component functions (f), (g), and (h) are real-valued functions of the parameter (t).

APEX Calculus

The previous section introduced us to a new mathematical object, the vector-valued function. We now apply calculus concepts to these functions. We start with the limit, then work our way through derivatives to integrals.

Subsection 12.2.1 Limits of Vector-Valued Functions

The initial definition of the limit of a vector-valued function is a bit intimidating, as was the definition of the limit in Definition 1.2.2. The theorem following the definition shows that in practice, taking limits of vector-valued functions is no more difficult than taking limits of real-valued functions.

We can define one-sided limits in a manner very similar to Definition 12.2.1.

Definition 12.2.1 . Limits of Vector-Valued Functions.

Let (I) be an open interval containing (c ext<,>) and let (vec r(t)) be a vector-valued function defined on (I ext<,>) except possibly at (c ext<.>) The , expressed as

means that given any (varepsilon gt 0 ext<,>) there exists a (delta gt 0) such that for all (t eq c ext<,>) if (abs lt delta ext<,>) we have ( orm lt varepsilon ext<.>)

Note how the measurement of distance between real numbers is the absolute value of their difference the measure of distance between vectors is the vector norm, or magnitude, of their difference.

Theorem 12.2.2 states that we can compute limits of vector-valued functions component-wise.

Theorem 12.2.2 . Limits of Vector-Valued Functions.

Let (vec r(t) = la ,f(t),g(t), a) be a vector-valued function in (mathbb^2) defined on an open interval (I) containing (c ext<,>) except possibly at (c ext<.>) Then

Let (vec r(t) = la ,f(t),g(t),h(t), a) be a vector-valued function in (mathbb^3) defined on an open interval (I) containing (c ext<,>) except possibly at (c ext<.>) Then

Example 12.2.3 . Finding limits of vector-valued functions.

Let (dsvec r(t) = la frac,, t^2-3t+3,,cos(t) a ext<.>) Find (limlimits_vec r(t) ext<.>)

We apply the theorem and compute limits component-wise.

Subsection 12.2.2 Continuity

Definition 12.2.4 . Continuity of Vector-Valued Functions.

Let (vec r(t)) be a vector-valued function defined on an open interval (I) containing (c ext<.>)

(vec r(t)) is continuous at (c) if (limlimits_ vec r(t) = r(c) ext<.>)

If (vec r(t)) is continuous at all (c) in (I ext<,>) then (vec r(t)) is continuous on (I ext<.>)

Using one-sided limits, we can also define continuity on closed intervals as done before.

We again have a theorem that lets us evaluate continuity component-wise.

Theorem 12.2.5 . Continuity of Vector-Valued Functions.

Let (vec r(t)) be a vector-valued function defined on an open interval (I) containing (c ext<.>) Then (vec r(t)) is continuous at (c) if, and only if, each of its component functions is continuous at (c ext<.>)

Example 12.2.6 . Evaluating continuity of vector-valued functions.

Let (dsvec r(t) = la frac,, t^2-3t+3,,cos(t) a ext<.>) Determine whether (vec r) is continuous at (t=0) and (t=1 ext<.>)

While the second and third components of (vec r(t)) are defined at (t=0 ext<,>) the first component, ((sin(t) )/t ext<,>) is not. Since the first component is not even defined at (t=0 ext<,>) (vec r(t)) is not defined at (t=0 ext<,>) and hence it is not continuous at (t=0 ext<.>)

At (t=1) each of the component functions is continuous. Therefore (vec r(t)) is continuous at (t=1 ext<.>)

Subsection 12.2.3 Derivatives

Consider a vector-valued function (vec r) defined on an open interval (I) containing (t_0) and (t_1 ext<.>) We can compute the displacement of (vec r) on ([t_0,t_1] ext<,>) as shown in Figure 12.2.7.(a). Recall that dividing the displacement vector by (t_1-t_0) gives the average rate of change on ([t_0,t_1] ext<,>) as shown in Figure 12.2.7.(b).

The derivative of a vector-valued function is a measure of the instantaneous rate of change, measured by taking the limit as the length of ([t_0,t_1]) goes to 0. Instead of thinking of an interval as ([t_0,t_1] ext<,>) we think of it as ([c,c+h]) for some value of (h) (hence the interval has length (h)). The average rate of change is

for any value of (h eq0 ext<.>) We take the limit as (h o0) to measure the instantaneous rate of change this is the derivative of (vec r ext<.>)

Definition 12.2.8 . Derivative of a Vector-Valued Function.

Let (vec r(t)) be continuous on an open interval (I) containing (c ext<.>)

Alternate notations for the derivative of (vec r) include:

If a vector-valued function has a derivative for all (c) in an open interval (I ext<,>) we say that (vec r(t)) is on (I ext<.>)

Once again we might view this definition as intimidating, but recall that we can evaluate limits component-wise. The following theorem verifies that this means we can compute derivatives component-wise as well, making the task not too difficult.

Again, using one-sided limits, we can define differentiability on closed intervals. We'll make use of this a few times in this chapter.

Theorem 12.2.9 . Derivatives of Vector-Valued Functions.

Let (vec r(t) = la , f(t), g(t), a ext<.>) Then

Let (vec r(t) = la , f(t), g(t), h(t), a ext<.>) Then

Example 12.2.10 . Derivatives of vector-valued functions.

Sketch (vec r(t)) and (vrp(t)) on the same axes.

Compute (vrp(1)) and sketch this vector with its initial point at the origin and at (vec r(1) ext<.>)

Theorem 12.2.9 allows us to compute derivatives component-wise, so

(vec r(t)) and (vrp(t)) are graphed together in Figure 12.2.11.(a). Note how plotting the two of these together, in this way, is not very illuminating. When dealing with real-valued functions, plotting (f(x)) with (fp(x)) gave us useful information as we were able to compare (f) and (fp) at the same (x)-values. When dealing with vector-valued functions, it is hard to tell which points on the graph of (vrp) correspond to which points on the graph of (vec r ext<.>)

We easily compute (vrp(1) = la 2,1 a ext<,>) which is drawn in Figure 12.2.11 with its initial point at the origin, as well as at (vec r(1) = la 1,1 a ext<.>) These are sketched in Figure 12.2.11.(b).

Example 12.2.12 . Derivatives of vector-valued functions.

Let (vec r(t) = la cos(t) , sin(t) , t a ext<.>) Compute (vrp(t)) and (vrp(pi/2) ext<.>) Sketch (vrp(pi/2)) with its initial point at the origin and at (vec r(pi/2) ext<.>)

We compute (vrp) as (vrp(t) = la -sin(t) , cos(t) , 1 a ext<.>) At (t= pi/2 ext<,>) we have (vrp(pi/2) = la -1,0,1 a ext<.>) Figure 12.2.13 shows a graph of (vec r(t) ext<,>) with (vrp(pi/2)) plotted with its initial point at the origin and at (vec r(pi/2) ext<.>)

In Examples 12.2.10 and Example 12.2.12, sketching a particular derivative with its initial point at the origin did not seem to reveal anything significant. However, when we sketched the vector with its initial point on the corresponding point on the graph, we did see something significant: the vector appeared to be tangent to the graph. We have not yet defined what “tangent” means in terms of curves in space in fact, we use the derivative to define this term.

Definition 12.2.14 . Tangent Vector, Tangent Line.

Let (vec r(t)) be a differentiable vector-valued function on an open interval (I) containing (c ext<,>) where (vrp(c) eq vec 0 ext<.>)

A vector (vec v) is tangent to the graph of (vec r(t)) at (t=c) if (vec v) is parallel to (vrp(c) ext<.>)

The tangent line to the graph of (vec r(t)) at (t=c) is the line through (vec r(c)) with direction parallel to (vrp(c) ext<.>) An equation of the tangent line is

Example 12.2.15 . Finding tangent lines to curves in space.

Let (vec r(t) = la t,t^2,t^3 a) on ([-1.5,1.5] ext<.>) Find the vector equation of the line tangent to the graph of (vec r) at (t=-1 ext<.>)

To find the equation of a line, we need a point on the line and the line's direction. The point is given by (vec r(-1) = la -1,1,-1 a ext<.>) (To be clear, (la -1,1,-1 a) is a vector, not a point, but we use the point “pointed to” by this vector.)

The direction comes from (vrp(-1) ext<.>) We compute, component-wise, (vrp(t) = la 1,2t, 3t^2 a ext<.>) Thus (vrp(-1) = la 1,-2,3 a ext<.>)

The vector equation of the line is (ell(t) = la -1,1,-1 a + tla 1,-2,3 a ext<.>) This line and (vec r(t)) are sketched in Figure 12.2.16.

Example 12.2.17 . Finding tangent lines to curves.

Find the equations of the lines tangent to (vec r(t) = la t^3,t^2 a) at (t=-1) and (t=0 ext<.>)

We find that (vrp(t) = la 3t^2,2t a ext<.>) At (t=-1 ext<,>) we have

so the equation of the line tangent to the graph of (vec r(t)) at (t=-1) is

This line is graphed with (vec r(t)) in Figure 12.2.18.

At (t=0 ext<,>) we have (vrp(0) = la 0,0 a=vec 0 ext) This implies that the tangent line “has no direction.” We cannot apply Definition 12.2.14, hence cannot find the equation of the tangent line.

We were unable to compute the equation of the tangent line to (vec r(t)= la t^3,t^2 a) at (t=0) because (vrp(0) = vec 0 ext<.>) The graph in Figure 12.2.18 shows that there is a cusp at this point. This leads us to another definition of smooth, previously defined by Definition 10.2.22 in Section 10.2.

Definition 12.2.19 . Smooth Vector-Valued Functions.

Let (vec r(t)) be a differentiable vector-valued function on an open interval (I) where (vrp(t)) is continuous on (I ext<.>) (vec r(t)) is on (I) if (vrp(t) eq vec 0) on (I ext<.>)

Having established derivatives of vector-valued functions, we now explore the relationships between the derivative and other vector operations. The following theorem states how the derivative interacts with vector addition and the various vector products.

Theorem 12.2.20 . Properties of Derivatives of Vector-Valued Functions.

Let (vec r) and (vec s) be differentiable vector-valued functions, let (f) be a differentiable real-valued function, and let (c) be a real number.

(displaystyle ds frac

Big(vec r(t) pm vec s(t)Big) = vrp(t) pm vec s,'(t))

(displaystyle ds frac

Big(cvec r(t)Big) = cvrp(t))

(ds frac

Big(f(t)vec r(t)Big) = fp(t)vec r(t) + f(t)vrp(t)) Product Rule

(ds frac

Big(vec r(t)cdot vec s(t) Big) = vrp(t)cdot vec s(t) + vec r(t)cdot vec s,'(t)) Product Rule

(ds frac

Big(vec r(t) imes vec s(t) Big) = vrp(t) imes vec s(t) + vec r(t) imes vec s,'(t)) Product Rule

(ds frac

Big(vec rig(f(t)ig)Big) = vrpig(f(t)ig)fp(t)) Chain Rule

Example 12.2.21 . Using derivative properties of vector-valued functions.

Let (vec r(t) = la t, t^2-1 a) and let (vec u(t)) be the unit vector that points in the direction of (vec r(t) ext<.>)

Graph (vec r(t)) and (vec u(t)) on the same axes, on ([-2,2] ext<.>)

Find (vec u,'(t)) and sketch (vec u,'(-2) ext<,>) (vec u,'(-1)) and (vec u,'(0) ext<.>) Sketch each with initial point the corresponding point on the graph of (vec u ext<.>)

To form the unit vector that points in the direction of (vec r ext<,>) we need to divide (vec r(t)) by its magnitude.

(vec r(t)) and (vec u(t)) are graphed in Figure 12.2.22. Note how the graph of (vec u(t)) forms part of a circle this must be the case, as the length of (vec u(t)) is 1 for all (t ext<.>)

To compute (vec u,'(t) ext<,>) we use Theorem 12.2.20, writing

and then take the derivative. It is a matter of preference this latter method requires two applications of the Quotient Rule where our method uses the Product and Chain Rules.) We find (fp(t)) using the Chain Rule:

We now find (vec u,'(t)) using part 3 of Theorem 12.2.20:

This is admittedly very “messy” such is usually the case when we deal with unit vectors. We can use this formula to compute (vec u,'(-2) ext<,>) (vec u,'(-1)) and (vec u,'(0) ext<:>)

Each of these is sketched in Figure 12.2.23. Note how the length of the vector gives an indication of how quickly the circle is being traced at that point. When (t=-2 ext<,>) the circle is being drawn relatively slow when (t=-1 ext<,>) the circle is being traced much more quickly.

It is a basic geometric fact that a line tangent to a circle at a point (P) is perpendicular to the line passing through the center of the circle and (P ext<.>) This is illustrated in Figure 12.2.23 each tangent vector is perpendicular to the line that passes through its initial point and the center of the circle. Since the center of the circle is the origin, we can state this another way: (vec u,'(t)) is orthogonal to (vec u(t) ext<.>)

Recall that the dot product serves as a test for orthogonality: if (vec ucdot vec v = 0 ext<,>) then (vec u) is orthogonal to (vec v ext<.>) Thus in the above example, (vec u(t)cdot vec u,'(t)=0 ext<.>)

This is true of any vector-valued function that has a constant length, that is, that traces out part of a circle. It has important implications later on, so we state it as a theorem (and leave its formal proof as an Exercise.)

Theorem 12.2.24 . Vector-Valued Functions of Constant Length.

Let (vec r(t)) be a vector-valued function of constant length that is differentiable on an open interval (I ext<.>) That is, ( orm = c) for all (t) in (I ext<>) equivalently, (vec r(t)cdot vec r(t) = c^2) for all (t) in (I ext<.>) Then (vec r(t)cdotvrp(t) = 0) for all (t) in (I ext<.>)

Subsection 12.2.4 Integration

Before formally defining integrals of vector-valued functions, consider the following equation that our calculus experience tells us should be true:

That is, the integral of a rate of change function should give total change. In the context of vector-valued functions, this total change is displacement. The above equation is true we now develop the theory to show why.

We can define antiderivatives and the indefinite integral of vector-valued functions in the same manner we defined indefinite integrals in Definition 5.1.2. However, we cannot define the definite integral of a vector-valued function as we did in Definition 5.2.6. That definition was based on the signed area between a function (y=f(x)) and the (x)-axis. An area-based definition will not be useful in the context of vector-valued functions. Instead, we define the definite integral of a vector-valued function in a manner similar to that of Theorem 5.3.26, utilizing Riemann sums.

Definition 12.2.25 . Antiderivatives, Indefinite and Definite Integrals of Vector-Valued Functions.

Let (vec r(t)) be a continuous vector-valued function on ([a,b] ext<.>) An of (vec r(t)) is a function (vec R(t)) such that (vec R'(t) = vec r(t) ext<.>)

The set of all antiderivatives of (vec r(t)) is the of (vec r(t) ext<,>) denoted by

where (Delta t_i) is the length of the (i)th subinterval of a partition of ([a,b] ext<,>) ( orm) is the length of the largest subinterval in the partition, and (c_i) is any value in the (i)th subinterval of the partition.

It is probably difficult to infer meaning from the definition of the definite integral. The important thing to realize from the definition is that it is built upon limits, which we can evaluate component-wise.

The following theorem simplifies the computation of definite integrals the rest of this section and the following section will give meaning and application to these integrals.

Theorem 12.2.26 . Indefinite and Definite Integrals of Vector-Valued Functions.

Let (vec r(t) = la f(t),g(t) a) be a vector-valued function in (mathbb^2) that is continuous on ([a,b] ext<.>)

(displaystyle ds int vec r(t), dt = la int f(t), dt, int g(t), dt a)

(displaystyle ds int_a^b vec r(t), dt = la int_a^b f(t), dt, int_a^b g(t), dt a)

A similar statement holds for vector-valued functions in (mathbb^3 ext<.>)

Example 12.2.27 . Evaluating a definite integral of a vector-valued function.

Let (vec r(t) = la e^<2t>,sin(t) a ext<.>) Evaluate (ds int_0^1 vec r(t) ,dt ext<.>)

Example 12.2.28 . Solving an initial value problem.

Let (vrp'(t) = la 2, cos(t) , 12t a ext<.>) Find (vec r(t)) where:

Knowing (vrp'(t) = la 2,cos(t) , 12t a ext<,>) we find (vrp(t)) by evaluating the indefinite integral.

Note how each indefinite integral creates its own constant which we collect as one constant vector (vec C ext<.>) Knowing (vrp(0) = la 5,3,0 a) allows us to solve for (vec C ext<:>)

So (vrp(t) = la 2t,sin(t) ,6t^2 a + la 5,3,0 a = la 2t+5, sin(t) + 3, 6t^2 a ext<.>) To find (vec r(t) ext<,>) we integrate once more.

With (vec r(0) = la -7,-1,2 a ext<,>) we solve for (vec C ext<:>)

What does the integration of a vector-valued function mean? There are many applications, but none as direct as “the area under the curve” that we used in understanding the integral of a real-valued function.

A key understanding for us comes from considering the integral of a derivative:

Integrating a rate of change function gives displacement.

Noting that vector-valued functions are closely related to parametric equations, we can describe the arc length of the graph of a vector-valued function as an integral. Given parametric equations (x=f(t) ext<,>) (y=g(t) ext<,>) the arc length on ([a,b]) of the graph is

as stated in Theorem 10.3.17 in Section 10.3. If (vrt = la f(t), g(t) a ext<,>) note that (sqrt = orm ext<.>) Therefore we can express the arc length of the graph of a vector-valued function as an integral of the magnitude of its derivative.

Theorem 12.2.29 . Arc Length of a Vector-Valued Function.

Let vrt be a vector-valued function where (vrp(t)) is continuous on ([a,b] ext<.>) The arc length (L) of the graph of vrt is

Note that we are actually integrating a scalar-function here, not a vector-valued function.

The next section takes what we have established thus far and applies it to objects in motion. We will let vrt describe the path of an object in the plane or in space and will discover the information provided by (vrp(t)) and (vrp'(t) ext<.>)

Exercises 12.2.5 Exercises

Limits, derivatives and integrals of vector-valued functions are all evaluated -wise.

The definite integral of a rate of change function gives .

Why is it generally not useful to graph both (vec r(t)) and (vrp(t)) on the same axes?

It is difficult to identify the points on the graphs of (vec r(t)) and (vrp(t)) that correspond to each other.

Theorem 12.2.20 contains three product rules. What are the three different types of products used in these rules?


Many vector-valued functions, like scalar-valued functions, can be differentiated by simply differentiating the components in the Cartesian coordinate system. Thus, if

is a vector-valued function, then

The vector derivative admits the following physical interpretation: if r(t) represents the position of a particle, then the derivative is the velocity of the particle

Likewise, the derivative of the velocity is the acceleration

Partial derivative

The partial derivative of a vector function a with respect to a scalar variable q is defined as [ 1 ]

where ai is the scalar component of a in the direction of ei. It is also called the direction cosine of a and ei or their dot product. The vectors e1,e2,e3 form an orthonormal basis fixed in the reference frame in which the derivative is being taken.

Ordinary derivative

If a is regarded as a vector function of a single scalar variable, such as time t, then the equation above reduces to the first ordinary time derivative of a with respect to t, [ 1 ]

Total derivative

If the vector a is a function of a number n of scalar variables qr (r = 1. n), and each qr is only a function of time t, then the ordinary derivative of a with respect to t can be expressed, in a form known as the total derivative, as [ 1 ]

Some authors prefer to use capital D to indicate the total derivative operator, as in D/Dt. The total derivative differs from the partial time derivative in that the total derivative accounts for changes in a due to the time variance of the variables qr.

Reference frames

Whereas for scalar-valued functions there is only a single possible reference frame, to take the derivative of a vector-valued function requires the choice of a reference frame (at least when a fixed Cartesian coordinate system is not implied as such). Once a reference frame has been chosen, the derivative of a vector-valued function can be computed using techniques similar to those for computing derivatives of scalar-valued functions. A different choice of reference frame will, in general, produce a different derivative function. The derivative functions in different reference frames have a specific kinematical relationship.

Derivative of a vector function with nonfixed bases

The above formulas for the derivative of a vector function rely on the assumption that the basis vectors e1,e2,e3 are constant, that is, fixed in the reference frame in which the derivative of a is being taken, and therefore the e1,e2,e3 each has a derivative of identically zero. This often holds true for problems dealing with vector fields in a fixed coordinate system, or for simple problems in physics. However, many complex problems involve the derivative of a vector function in multiple moving reference frames, which means that the basis vectors will not necessarily be constant. In such a case where the basis vectors e1,e2,e3 are fixed in reference frame E, but not in reference frame N, the more general formula for the ordinary time derivative of a vector in reference frame N is [ 1 ]

where the superscript N to the left of the derivative operator indicates the reference frame in which the derivative is taken. As shown previously, the first term on the right hand side is equal to the derivative of a in the reference frame where e1,e2,e3 are constant, reference frame E. It also can be shown that the second term on the right hand side is equal to the relative angular velocity of the two reference frames cross multiplied with the vector a itself. [ 1 ] Thus, after substitution, the formula relating the derivative of a vector function in two reference frames is [ 1 ]

where N ω E is the angular velocity of the reference frame E relative to the reference frame N.

One common example where this formula is used is to find the velocity of a space-borne object, such as a rocket, in the inertial reference frame using measurements of the rocket's velocity relative to the ground. The velocity N v R in inertial reference frame N of a rocket R located at position r R can be found using the formula

where N ω E is the angular velocity of the Earth relative to the inertial frame N. Since velocity is the derivative of position, N v R and E v R are the derivatives of r R in reference frames N and E, respectively. By substitution,

where E v R is the velocity vector of the rocket as measured from a reference frame E that is fixed to the Earth.

Derivative and vector multiplication

The derivative of the products of vector functions behaves similarly to the derivative of the products of scalar functions. [ 2 ] Specifically, in the case of scalar multiplication of a vector, if p is a scalar variable function of q, [ 1 ]

In the case of dot multiplication, for two vectors a and b that are both functions of q, [ 1 ]

Similarly, the derivative of the cross product of two vector functions is [ 1 ]


Level and Sub-level Sets

Level and sub-level sets correspond to the notion of contour of a function. Both are indexed on some scalar value .

Level sets

A level set is simply the set of points that achieve exactly some value for the function . For , the -level set of the function is defined as

x in mathbf^n, t = f(x) right> . " />

Sub-level sets

A related notion is that of sub-level set. This is now the set of points that achieve at most a certain value for , or below:

x in mathbf^n, t ge f(x) right> . " />

Level and sub-level sets of a function , with domain itself, and values on the domain given by


Infinite-dimensional vector functions

If the values of a function f lie in an infinite-dimensional vector space X, such as a Hilbert space, then f may be called an infinite-dimensional vector function.

Functions with values in a Hilbert space

If the argument of f is a real number and X is a Hilbert space, then the derivative of f at a point t can be defined as in the finite-dimensional case:

Most results of the finite-dimensional case also hold in the infinite-dimensional case too, mutatis mutandis. Differentiation can also be defined to functions of several variables (e.g., or even , where Y is an infinite-dimensional vector space).

N.B. If X is a Hilbert space, then one can easily show that any derivative (and any other limit) can be computed componentwise: if

(i.e., , where is an orthonormal basis of the space X), and exists, then

.

However, the existence of a componentwise derivative does not guarantee the existence of a derivative, as componentwise convergence in a Hilbert space does not guarantee convergence with respect to the actual topology of the Hilbert space.


Plotly.subplots .make_subplots¶

Return an instance of plotly.graph_objects.Figure with predefined subplots configured in ‘layout’.

rows (int (default 1)) – Number of rows in the subplot grid. Must be greater than zero.

cols (int (default 1)) – Number of columns in the subplot grid. Must be greater than zero.

Assign shared (linked) x-axes for 2D cartesian subplots

  • True or ‘columns’: Share axes among subplots in the same column

  • ’rows’: Share axes among subplots in the same row

  • ’all’: Share axes across all subplots in the grid.

Assign shared (linked) y-axes for 2D cartesian subplots

  • ’columns’: Share axes among subplots in the same column

  • True or ‘rows’: Share axes among subplots in the same row

  • ’all’: Share axes across all subplots in the grid.

Choose the starting cell in the subplot grid used to set the domains_grid of the subplots.

  • ’top-left’: Subplots are numbered with (1, 1) in the top

    left corner

  • ’bottom-left’: Subplots are numbererd with (1, 1) in the bottom

    left corner

print_grid (boolean (default True):) – If True, prints a string representation of the plot grid. Grid may also be printed using the Figure.print_grid() method on the resulting figure.

Space between subplot columns in normalized plot coordinates. Must be a float between 0 and 1.

Applies to all columns (use ‘specs’ subplot-dependents spacing)

Space between subplot rows in normalized plot coordinates. Must be a float between 0 and 1.

Applies to all rows (use ‘specs’ subplot-dependents spacing)

Title of each subplot as a list in row-major ordering.

Empty strings (“”) can be included in the list if no subplot title is desired in that space so that the titles are properly indexed.

Per subplot specifications of subplot type, row/column spanning, and spacing.

Indices of the outer list correspond to subplot grid rows starting from the top, if start_cell=’top-left’, or bottom, if start_cell=’bottom-left’. The number of rows in ‘specs’ must be equal to ‘rows’.

Indices of the inner lists correspond to subplot grid columns starting from the left. The number of columns in ‘specs’ must be equal to ‘cols’.

Each item in the ‘specs’ list corresponds to one subplot in a subplot grid. (N.B. The subplot grid has exactly ‘rows’ times ‘cols’ cells.)

Use None for a blank a subplot cell (or to move past a col/row span).

Note that specs[0][0] has the specs of the ‘start_cell’ subplot.

The available keys are: * type (string, default ‘xy’): Subplot type. One of

  • ’xy’: 2D Cartesian subplot type for scatter, bar, etc.

  • ’scene’: 3D Cartesian subplot for scatter3d, cone, etc.

  • ’polar’: Polar subplot for scatterpolar, barpolar, etc.

  • ’ternary’: Ternary subplot for scatterternary

  • ’mapbox’: Mapbox subplot for scattermapbox

  • ’domain’: Subplot type for traces that are individually

    positioned. pie, parcoords, parcats, etc.

  • trace type: A trace type which will be used to determine

    the appropriate subplot type for that trace

y-axis positioned on the right side of the subplot. Only valid if type=’xy’.

l (float, default 0.0): padding left of cell

r (float, default 0.0): padding right of cell

t (float, default 0.0): padding right of cell

b (float, default 0.0): padding bottom of cell

Note: Use ‘horizontal_spacing’ and ‘vertical_spacing’ to adjust the spacing in between the subplots.

Inset specifications. Insets are subplots that overlay grid subplots

    Each item in ‘insets’ is a dictionary.

    cell (tuple, default=(1,1)): (row, col) index of the

subplot cell to overlay inset axes onto.

type (string, default ‘xy’): Subplot type

in fraction of cell width

in fraction of cell width (‘to_end’: to cell right edge)

in fraction of cell height

in fraction of cell height (‘to_end’: to cell top edge)

list of length cols of the relative widths of each column of suplots. Values are normalized internally and used to distribute overall width of the figure (excluding padding) among the columns.

For backward compatibility, may also be specified using the column_width keyword argument.

list of length rows of the relative heights of each row of subplots. If start_cell=’top-left’ then row heights are applied top to bottom. Otherwise, if start_cell=’bottom-left’ then row heights are applied bottom to top.

For backward compatibility, may also be specified using the row_width kwarg. If specified as row_width , then the width values are applied from bottom to top regardless of the value of start_cell. This matches the legacy behavior of the row_width argument.

column_titles (list of str or None (default None)) – list of length cols of titles to place above the top subplot in each column.

row_titles (list of str or None (default None)) – list of length rows of titles to place on the right side of each row of subplots. If start_cell=’top-left’ then row titles are applied top to bottom. Otherwise, if start_cell=’bottom-left’ then row titles are applied bottom to top.

x_title (str or None (default None)) – Title to place below the bottom row of subplots, centered horizontally

y_title (str or None (default None)) – Title to place to the left of the left column of subplots, centered vertically

figure (go.Figure or None (default None)) – If None, a new go.Figure instance will be created and its axes will be populated with those corresponding to the requested subplot geometry and this new figure will be returned. If a go.Figure instance, the axes will be added to the layout of this figure and this figure will be returned. If the figure already contains axes, they will be overwritten.

This is the format of your plot grid: [ (1,1) xaxis1,yaxis1 ] [ (2,1) xaxis2,yaxis2 ]


5.1: Vector-Valued Functions and Space Curves

    Recall that a parameterized curve in the plane is the image of a straight line segment "bent" by some mapping r(t) = (x(t),y(t)). For example, the vector-valued function r(t) = (t,t 2 ) describes a parameterized parabola in the plane. Write down a vector-valued function that describes a circle in the plane. Plot your parametric curve in your worksheet to confirm that it represents a circle.

Often the two parameters are labeled u and v. The parameterized surface is a vector valued function r(u,v) of two variables, whether written in ijk vector notation or as an ordered triple of functions of u and v. Since each of the variables u and v ranges over an interval, the domain for r(u,v) is a coordinate rectangle, say [a,b] x [c,d], in the uv-plane. (Either or both intervals may be infinitely long.)

  1. Every surface that is the graph of a function f(x,y) can also be described parametrically by letting the parameters be x and y. In you r worksheet, confirm that the saddle surface which is the graph of f(x,y) = xy is the same as the graph of r(u,v) = ui + vj + uvk.

Just as we can use Cartesian coordinates as parameters, we can use other coordinate systems as well. The next two steps illustrate this.

    Use the cylindrical coordinates u = and v = z to construct a parametric representation of a circular cylinder of radius 2 and height 3. Plot your parametric surface in your worksheet.

Now we consider a parameterization of the torus pictured above before step 1. We can visualize this surface by first imagining a circle of radius a in the xy-plane that runs through the center of the "tube". From each point on this circle, we can reach a circle of points on the surface by making it the center of a circle of radius b, where b < a. This second circle is drawn in a vertical plane that includes the z-axis. In the figure at the right, we show the horizontal circle of radius a in blue and a typical circle of radius b in red. As the red circle travels around the blue one, it sweeps out the entire torus. We take as parameters u and v, respectively, the central angles for the blue and red circles. Then we can construct the parameterization for each point r(u,v) on the surface by adding a vector from the origin to a point on the blue circle and a vector from that point to a point on the corresponding red circle.

    Explain why the construction just described leads to the parameterization


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Examples

The examples below show how OFFSET can be configured to return different kinds of ranges. These screens were taken with Excel 365, so OFFSET returns a dynamic array when the result is more than one cell. In older versions of Excel, you can use the F9 key to check results returned from OFFSET.

Example #1

In the screen below, we use OFFSET to return the third value (March) in the second column (West). The formula in H4 is:

Example #2

In the screen below, we use OFFSET to return the last value (June) in the third column (North). The formula in H4 is:

Example #3

Below, we use OFFSET to return all values in the third column (North). The formula in H4 is:

Example #4

Below, we use OFFSET to return all values for May (fifth row). The formula in H4 is:

Example #5

Below, we use OFFSET to return April, May, and June value for the West region. The formula in H4 is:

Example #6

Below, we use OFFSET to return April, May, and June value for West and North. The formula in H4 is:


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