
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

where each component is completely independent of the other components. When computing a limit, we can write:

This is worth stating as a theorem.

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:
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:

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:

### Derivatives

Recall the limit definition of the derivative: We have a similar limit definition of vector-valued functions: Since limits can be computed component-wise, this derivative can be computed component-wise.

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:

We also have some (additional) derivative rules:

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: $\eval {\dd {t} s(t)\vec {f}(t)}_{t=2}$

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

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

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:

Is the vector-valued function $\vec {f}(t) = \vector {t^2,t^2}$ smooth on $\R$?
yes no

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.

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 $\vec f(t) = \vector {e^{2t},\sin t}$. Compute:

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

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.