Calculus and vector-valued functions

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With one input, and vector outputs, we work component-wise.

A question I’ve often asked myself is: “How do you know when you are doing a calculus problem?” The answer, I think, is that you are doing a calculus problem when you are computing: a limit, a derivative, or an integral. Now we are going to do calculus with vector-valued functions. To build a theory of calculus for vector-valued functions, we simply treat each component of a vector-valued function as a regular, single-variable function. Since we are currently thinking about vector-valued functions that only have a single input, we can work component-wise. Let’s see this in action.

Limits of vector-valued functions

With a vector-valued function, you have something like

$\begin{align*} \vec{f} &= \vector{x(t),y(t),z(t)}\\ &= \veci x(t) + \vecj y(t) + \veck z(t) \end{align*}$ where each component is completely independent of the other components. When computing a limit, we can write: $\begin{align*} \lim_{t\to a} \vec{f}(t) &= \lim_{t\to a}\left(\veci x(t) + \vecj y(t) + \veck z(t)\right)\\ &= \lim_{t\to a}\veci x(t) + \lim_{t\to a}\vecj y(t) + \lim_{t\to a}\veck z(t)\\ &= \veci \lim_{t\to a}\veci x(t) + \vecj\lim_{t\to a} y(t) + \veck\lim_{t\to a} z(t)\\ &= \vector{\lim_{t \to a}x(t),\lim_{t \to a}y(t),\lim_{t \to a}z(t)}. \end{align*}$

This is worth stating as a theorem.

Let $\vec {f}:\R \to \R ^3$ be a vector-valued function $\vec {f}(t) = \vector {x(t),y(t),z(t)}$ then the vector-valued limit as $t$ goes to $a$ is given by $\lim _{t \to a} \vec {f}(t) = \vector {\lim _{t \to a}x(t),\lim _{t \to a}y(t),\lim _{t \to a}z(t)}.$ This limit exists if and only if each of the limits of the components exist.

We evaluate limits by just taking the limit of each component separately.

Let $\vec {f}(t) = \vector {\sin (t),\cos (t),\frac {\sin (t)}{t}}$ . Compute: $\lim _{t \to 0} \vec {f}(t) \begin {prompt} =\vector {\answer {0},\answer {1},\answer {1}} \end {prompt}$

Take the limit of each component separately.

Now that we have the notion of limits, we may also define the concept of continuity of vector-valued functions:

A vector-valued function $\vec {f}$ is continuous at $t= a$ if and only if $\lim _{t \to a} \vec {f}(t) = \vec {f}(a)$

Because of the component-wise nature of limits, we can see that a function $\vec {f}$ is continuous if and only if each component function $x(t)$ , $y(t)$ , $z(t)$ is also continuous at $t=a$ .

Which of the following vector-valued functions are continuous for all real values of $t$ ?

$\vector {t,t^2,\tan (t)}$ $\vector {\frac {2}{t-5},\sin (t),0}$ $\vector {\cos (5 t), t^2-3t+1, e^t}$ $\vector {e^{\sqrt {t}}, 4\sin (t), t^7}$ $\vector {\ln (t^2), \ln (t), t^{3/2}}$ $\vector {0, 0, 0}$

Recall the following theorem:

Continuity of Famous Functions The following functions are continuous on their natural domains, for $k$ a real number and $b$ a positive real number:

Constant function: $f(x) =k$
Identity function: $f(x) = x$
Power function: $f(x)=x^b$
Exponential function: $f(x)=b^x$
Logarithmic function: $f(x)=\log _b(x)$
Sine and cosine: Both $\sin (x)$ and $\cos (x)$

In essence, we are saying that the functions listed above are continuous wherever they are defined.

Derivatives

Recall the limit definition of the derivative: $\ddx f(x) = \lim _{h\to 0} \frac {f(x+h)-f(x)}{h}$ We have a similar limit definition of vector-valued functions: $\dd {t} \vec {f}(t) = \lim _{h\to 0} \frac {\vec {f}(t+h)-\vec {f}(t)}{h}$ Since limits can be computed component-wise, this derivative can be computed component-wise.

Let $\vec {f}:\R \to \R ^3$ be a vector-valued function $\vec {f}(t) = \vector {x(t),y(t),z(t)},$ then the vector-valued derivative is given by $\vec {f}'(t) = \vector {x'(t),y'(t),z'(t)}.$

The derivative of a vector-valued function gives a vector that points in the direction that the vector-valued function draws the curve.

Below we see the derivative of the vector-valued function along with an approximation of the limit for small values of $h$ :

Let $\vec {f}(t) = \vector {\arctan (t),7^t, \sec (t^2)}$ . Compute: $\dd {t} \vec {f}(t) \begin {prompt} =\vector {\answer {\frac {1}{1+t^2}},\answer {7^t \ln (7)},\answer { \sec (t^2)\tan (t^2)2t}} \end {prompt}$

We also have some (additional) derivative rules:

Let $\vec {f}$ and $\vec {g}$ be differentiable vector-valued functions, $s(t)$ be a differentiable scalar function, and $c$ be a constant. Then:

$\dd {t} \left (\vec {f}(t) + \vec {g}(t) \right ) = \vec {f}'(t) + \vec {g}'(t)$
$\dd {t} c\vec {f}(t) = c\vec {f}'(t)$
$\dd {t} s(t)\vec {f}(t) = s'(t)\vec {f}(t) + s(t)\vec {f}'(t)$
$\dd {t} \left (\vec {f}(t)\dotp \vec {g}(t)\right ) = \vec {f}'(t)\dotp \vec {g}(t) + \vec {f}(t)\dotp \vec {g}'(t)$
$\dd {t} \left (\vec {f}(t)\cross \vec {g}(t)\right ) = \vec {f}'(t)\cross \vec {g}(t) + \vec {f}(t)\cross \vec {g}'(t)$
$\dd {t} \vec {f}(s(t)) = \vec {f}'(s(t))\cdot s'(t)$

Suppose that $s:\R \to \R$ , $\vec {f}:\R \to \R ^3$ , and $\vec {g}:\R \to \R ^3$ where:

$s(2) = 3$ and $s'(2) =-1$ .
$\vec {f}(2) = \vector {1,-1,0}$ and $\vec {f}'(2) = \vector {-1,1,0}$ .
$\vec {g}(2) = \vector {0,1,2}$ and $\vec {g}'(2) = \vector {2,0,-1}$ .

Compute: $\eval {\dd {t} s(t)\vec {f}(t)}_{t=2}$ $\eval {\dd {t} s(t)\vec {f}(t)}_{t=2} = \vector {\answer {-4},\answer {4},\answer {0}}$

Compute: $\eval {\dd {t} \vec {f}(t)\dotp \vec {g}(t)}_{t=2}$ $\eval {\dd {t} \vec {f}(t)\dotp \vec {g}(t)}_{t=2} = \answer {3}$

Compute: $\eval {\dd {t} \vec {f}(t)\cross \vec {g}(t)}_{t=2}$ $\eval {\dd {t} \vec {f}(t)\cross \vec {g}(t)}_{t=2} = \vector {\answer {3},\answer {3},\answer {1}}$

The derivative of a vector-valued function gives a tangent vector. A tangent vector is a vector that points in the direction that the curve is drawn.

Below we see the derivative of the vector-valued function along with an approximation of the limit for small values of $h$ :

Let $\vec {f}(t)$ be a differentiable vector-valued function on an open interval $I$ containing $c$ , where $\vec {f}'(c)\neq \vec {0}$ .

A vector $\vec v$ is a tangent vector to the graph of $\vec {f}(t)$ at $t=c$ , if $\vec v$ is parallel to $\vec {f}'(c)$ .
The tangent line to the graph of $\vec f(t)$ at $t=c$ is the line through $\vec f(c)$ in a direction parallel to $\vec {f}'(c)$ . An equation of the tangent line is $\vec { \boldsymbol {\l }}(t) = \vec f(c) + t\vec {f}'(c).$
The direction of $\vec {f}$ is given by the unit tangent vector of $\vec {f}$ is given by $\utan (t)=\frac {\vec {f}'(t)}{|\vec {f}'(t)|} = \frac {\vec {f}'(t)}{\sqrt {\vec {f}'(t)\dotp \vec {f}'(t)}}.$

Let $\vec {f}(t) = \vector {t,t^2,t^3}$ on $[-1.5,1.5]$ . Find the vector equation of the line tangent to the graph of $\vec f$ at $t=-1$ . $\vec {\boldsymbol {\l }}(t) = \vector {\answer {-1},\answer {1},-1} + t\vector {\answer {1},-2,\answer {3}}$

We want to be able to predict how a curve drawn by a vector-valued function behaves based on the function’s derivative. However, if the derivative is ever the zero vector, we loose all such information. When this happens we cannot use the tools of calculus. Because of this, we have a special name for functions where the tangent vector is never the zero vector:

Let $\vec {f}(t)$ be a differentiable vector-valued function on an open interval $I$ . $\vec f(t)$ is smooth on $I$ if $\vec {f}'(t)\neq \vec {0}$ for $t$ in $I$ .

Is the vector-valued function $\vec {f}(t) = \vector {t^2,t^2}$ smooth on $\R$ ?

yes no

This function looks like a “ray.” Conceptually it isn’t smooth because it goes to the origin and turns around “abruptly.” Check it out: $\graph {(t^2,t^2)}$

Finally, we point out that if a vector-valued function has constant length, then there is a special relationship between the function and its tangent vectors.

Let $\vec f(t)$ be a differentiable vector-valued function on an open interval $I$ . $\vec {f}(t)$ is of constant length if and only if the function and its tangent vector are orthogonal.

Suppose that $|\vec f(t)| = c$ for all $t$ in $I$ . In particular, we see $\vec {f}(t) \dotp \vec {f}(t) = c^2.$ Now take the derivative with respect to $t$ of both sides and apply the derivative rules from above. Write with me $\begin{align*} \vec{f}'(t) \dotp \vec{f}(t) + \vec{f}(t)\dotp \vec{f}'(t) &= 0\\ 2\vec{f}'(t) \dotp \vec{f}(t) &= 0\\ \vec{f}'(t) \dotp \vec{f}(t) &= 0 \end{align*}$ and this means that the vectors $\vec {f}'(t)$ and $\vec {f}(t)$ are parallelperpendicular .

Now suppose that for all $t$ in $I$ , $\vec {f}'(t)$ and $\vec {f}(t)$ are orthogonal. This means

$\begin{align*} \vec{f}'(t) \dotp \vec{f}(t) &= 0\\ 2\vec{f}'(t) \dotp \vec{f}(t) &= 0\\ \vec{f}'(t) \dotp \vec{f}(t) + \vec{f}(t)\dotp \vec{f}'(t) &= 0, \end{align*}$ but we know where the derivative on the left came from! So we may write $\vec {f}(t) \dotp \vec {f}(t) = c^2$ for some constant $c$ . Now we see that $\vec {f}$ is a vector-valued function of constant length on an open interval $I$ .

The theorem above says a lot in a very little. To start, it says that if the magnitude of a vector-valued function is constant, then the vector-value of the function and its tangent vector are orthogonal. Moreover, this means that the curve looks locally like a circle. However, it does not mean the curve looks like a circle globally. For example, consider this curve:

By moving the slider around you can see the vector of constant length that draws the curve and the tangent vector. Note that these vectors are orthogonal.

Integrals

Since we took the derivative of vector-valued functions by differentiating each component, we will also compute indefinite and definite integrals by computing antiderivatives of each component.

Let $\vec {f}:\R \to \R ^3$ be a continuous vector-valued function $\vec {f}(t) = \vector {x(t),y(t),z(t)},$ then the vector-valued integral is given by $\int \vec {f}(t) \d t= \vector {\int x(t) \d t,\int y(t)\d t,\int z(t)\d t}.$

Let $\vec f(t) = \vector {e^{2t},\sin t}$ . Compute: $\int _0^1 \vec f(t) \d t \begin {prompt} = \vector {\answer {\frac {1}{2}(e^2-1)},\answer {-\cos (1)+1}} \end {prompt}$

We can also solve initial value problems, check out our next example:

Let $\vec {f}''(t) = \vector {2, \cos (t), 12t}$ . Find $\vec f(t)$ where:

$\vec f(0) = \vector {-7,-1,2}$ and
$\vec {f}'(0) = \vector {5,3,0}.$

Knowing $\vec {f}''(t) = \vector {2,\cos (t), 12t}$ , we find $\vec {f}'(t)$ by evaluating the indefinite integral. $\begin{align*} \int \vec{f}''(t)\d t &= \vector{\int \answer[given]{2}\d t, \int \answer[given]{\cos(t)}\d t , \int \answer[given]{12t}\d t} \\ &= \vector{\answer[given]{2t}+C_1, \answer[given]{\sin(t)}+ C_2, \answer[given]{6t^2} + C_3} \\ &= \vector{2t,\sin(t),6t^2} + \vector{C_1,C_2,C_3} \\ &= \vector{2t,\sin(t),6t^2} + \vec{C}. \end{align*}$ Note how each indefinite integral creates its own constant which we collect as one constant vector $\vec C$ . Knowing $\vec {f}'(0) = \vector {5,3,0}$ allows us to solve for $\vec C$ : $\begin{align*} \vec{f}'(t) & = \vector{2t,\sin(t),6t^2} + \vec C\\ \vec{f}'(0) &= \vector{\answer[given]{0},\answer[given]{0},\answer[given]{0}} + \vec C\\ \vector{5,3,0} &= \vec C. \end{align*}$ So $\begin{align*} \vec{f}'(t) &= \vector{2t,\sin(t),6t^2} + \vector{\answer[given]{5},\answer[given]{3},\answer[given]{0}}\\ &=\vector{2t+5, \sin(t) + 3, 6t^2}. \end{align*}$ To find $\vec f(t)$ , we integrate once more. $\begin{align*} \int \vec{f}'(t)\d t &= \vector{\int\answer[given]{2t+5}\d t, \int \answer[given]{\sin(t) + 3}\d t, \int \answer[given]{6t^2}\d t }\\ &= \vector{\answer[given]{t^2+5t}, \answer[given]{-\cos(t) + 3t}, \answer[given]{2t^3}} + \vec C \end{align*}$ With $\vec f(0) = \vector {-7,-1,2}$ , we solve for $\vec C$ : $\begin{align*} \vec f(t) &= \vector{\answer[given]{t^2+5t}, \answer[given]{-\cos(t) + 3t}, \answer[given]{2t^3}} + \vec C\\ \vec f(0) &= \vector{\answer[given]{0},\answer[given]{-1},\answer[given]{0}} + \vec C\\ \vector{-7,-1,2} &= \vector{0,-1,0} + \vec C. \end{align*}$ Hence: $\vec {C}=\vector {\answer [given]{-7},\answer [given]{0},\answer [given]{2}}$ So $\begin{align*} \vec f(t) &= \vector{t^2+5t, -\cos(t) + 3t, 2t^3} + \vector{-7,0,2}\\ &= \vector{\answer[given]{t^2+5t-7},\answer[given]{-\cos(t)+3t},\answer[given]{2t^3+2}}. \end{align*}$

We now leave you with a question: What does 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. We will explore this later in our study of calculus.