Derivatives from Difference Quotients

From Average Change to Instantaneous Change

A vector-valued function describes a quantity that has both a parameter and a vector value. For a parametrized curve, the parameter may represent time and the vector value may represent position. Over a small parameter interval, the change in position divided by the length of that interval is an average rate of change. To describe the rate at one particular parameter value, we take a limit as the interval shrinks.

This is the central idea behind the derivative of a vector-valued function. The definition uses the same difference-quotient structure as an ordinary derivative, but the quotient is a vector. Its limit, when it exists, is also a vector.

Definition: Let $\mathbf{r}:I\to\mathbb{R}^n$, where $I$ is an interval, and let $a$ be an interior point of $I$. The derivative of $\mathbf{r}$ at $a$ is $$ \mathbf{r}'(a) = \lim_{h\to 0} \frac{\mathbf{r}(a+h)-\mathbf{r}(a)}{h}, $$ provided this vector limit exists. If the limit does not exist as a finite vector, then $\mathbf{r}$ is not differentiable at $a$.

The scalar $h$ is a change in the parameter. The numerator is the change in the function value over that change. Since dividing a vector by a nonzero scalar means multiplying each vector by the reciprocal scalar, the difference quotient is a vector for every permitted $h\ne 0$.

For this two-sided definition, $a+h$ must belong to the domain for all sufficiently small positive and negative $h$. At an endpoint of the parameter interval, a derivative can instead be defined using the appropriate one-sided limit, if a one-sided derivative is being considered.

Worked Example: A Difference Quotient for a Polynomial Curve

Let $\mathbf{r}(t)=\langle t^2,3t+1,5\rangle$. Find the derivative at $t=2$ from the definition.

First calculate the function value at the parameter of interest:

\mathbf{r}(2)=\langle 2^2,3(2)+1,5\rangle =\langle 4,7,5\rangle.

For $h\ne 0$, evaluate the function at $2+h$:

\begin{aligned} \mathbf{r}(2+h) &=\langle (2+h)^2,3(2+h)+1,5\rangle\\ &=\langle 4+4h+h^2,7+3h,5\rangle. \end{aligned}

Subtract the original value, then divide the whole vector by $h$:

\begin{aligned} \frac{\mathbf{r}(2+h)-\mathbf{r}(2)}{h} &=\frac{\langle 4+4h+h^2,7+3h,5\rangle-\langle 4,7,5\rangle}{h}\\ &=\frac{\langle 4h+h^2,3h,0\rangle}{h}\\ &=\langle 4+h,3,0\rangle. \end{aligned}

Now take the limit as $h\to 0$:

\mathbf{r}'(2) = \lim_{h\to 0}\langle 4+h,3,0\rangle = \langle 4,3,0\rangle.

As a check, the exact change in position is $\mathbf{r}(2+h)-\mathbf{r}(2)=\langle 4h+h^2,3h,0\rangle$. Dividing this change by $h$ gives the quotient above, whose difference from $\langle 4,3,0\rangle$ is $\langle h,0,0\rangle$, tending to $\mathbf{0}$. Thus the proposed derivative agrees with the defining limit.

Why This Limit Is an Instantaneous Rate

For a curve $\mathbf{r}(t)$ describing position, the quotient $\frac{\mathbf{r}(a+h)-\mathbf{r}(a)}{h}$ compares the displacement from time $a$ to time $a+h$ with the time elapsed. It is therefore an average velocity over that interval. When its limit exists as $h\to 0$, it gives the instantaneous velocity at $a$.

The definition also gives a precise way to describe the local change in the function. Rearranging the difference quotient, for $h\ne 0$, gives

\mathbf{r}(a+h)-\mathbf{r}(a) = h\left( \frac{\mathbf{r}(a+h)-\mathbf{r}(a)}{h} \right).

If the derivative exists, the vector in parentheses approaches $\mathbf{r}'(a)$. In other words, for small $h$, the displacement is approximately $h\mathbf{r}'(a)$. More precisely, define the error vector

\boldsymbol{\varepsilon}(h) = \frac{\mathbf{r}(a+h)-\mathbf{r}(a)}{h} -\mathbf{r}'(a).

By the definition of the limit, $\boldsymbol{\varepsilon}(h)\to\mathbf{0}$ as $h\to 0$. Multiplying its defining equation by $h$ yields

\mathbf{r}(a+h)-\mathbf{r}(a) = h\mathbf{r}'(a)+h\boldsymbol{\varepsilon}(h), \qquad \boldsymbol{\varepsilon}(h)\to\mathbf{0}.

This shows why the derivative captures the leading-order change: the displacement is $h\mathbf{r}'(a)$ plus a remainder that becomes negligible compared with $h$. If the parameter measures time, the units of $\mathbf{r}'(a)$ are the units of position divided by the units of time.

Worked Example: A Derivative on a Circular Path

Let $\mathbf{r}(t)=\langle 3\cos t,3\sin t\rangle$. Find $\mathbf{r}'(0)$ using the difference quotient.

The point at $t=0$ is

\mathbf{r}(0)=\langle 3\cos 0,3\sin 0\rangle=\langle 3,0\rangle.

For $h\ne 0$, use the definition and the values $\cos 0=1$, $\sin 0=0$:

\begin{aligned} \frac{\mathbf{r}(h)-\mathbf{r}(0)}{h} &=\frac{\langle 3\cos h,3\sin h\rangle-\langle 3,0\rangle}{h}\\ &=\left\langle 3\frac{\cos h-1}{h},3\frac{\sin h}{h}\right\rangle. \end{aligned}

Using the standard trigonometric limits $\lim_{h\to0}\frac{\sin h}{h}=1$ and $\lim_{h\to0}\frac{\cos h-1}{h}=0$, we obtain

\mathbf{r}'(0) = \langle 3(0),3(1)\rangle = \langle 0,3\rangle.

The quotient approaches $\langle 0,3\rangle$, so the derivative exists. The position at $t=0$ is $\langle 3,0\rangle$, while the instantaneous rate of change is $\langle 0,3\rangle$. They are different kinds of vectors: one gives the location, and the other describes the limiting change in location per unit parameter.

Checking Whether the Derivative Exists

A derivative exists only if the difference quotient approaches one finite vector as $h$ approaches zero from both sides. If the limits from positive and negative $h$ disagree, there is no two-sided derivative. This is the vector version of the usual limit-existence test. The earlier material on vector limits establishes that a vector limit exists exactly when the limits of all its components exist; here, that fact can help identify a failure, but the definition itself is a limit of vectors.

Worked Example: A Curve with No Derivative at a Corner

Let $\mathbf{r}(t)=\langle |t|,t\rangle$. Test whether $\mathbf{r}'(0)$ exists.

Since $\mathbf{r}(0)=\langle 0,0\rangle$, the difference quotient is

\frac{\mathbf{r}(h)-\mathbf{r}(0)}{h} = \left\langle\frac{|h|}{h},1\right\rangle.

If $h>0$, then $|h|=h$, so the quotient is $\langle 1,1\rangle$. Therefore, the limit from the right is

\lim_{h\to0^+}\frac{\mathbf{r}(h)-\mathbf{r}(0)}{h} = \langle 1,1\rangle.

If $h<0$, then $|h|=-h$, so the quotient is $\langle -1,1\rangle$. Therefore, the limit from the left is

\lim_{h\to0^-}\frac{\mathbf{r}(h)-\mathbf{r}(0)}{h} = \langle -1,1\rangle.

These one-sided limits differ, so the two-sided limit does not exist. Hence $\mathbf{r}'(0)$ does not exist. The calculation checks both sides of the parameter value rather than relying on a single direction of approach.

A Consequence: Differentiability Implies Continuity

The difference-quotient definition also shows that differentiability at a point implies continuity there. Suppose $\mathbf{r}'(a)$ exists. For nonzero $h$ sufficiently close to zero, write

\mathbf{r}(a+h)-\mathbf{r}(a) = h\frac{\mathbf{r}(a+h)-\mathbf{r}(a)}{h}.

The quotient approaches the finite vector $\mathbf{r}'(a)$, so it remains bounded for $h$ close enough to zero. The scalar $h$ tends to zero; therefore the product tends to $\mathbf{0}$. It follows that $\mathbf{r}(a+h)\to\mathbf{r}(a)$, which is continuity at $a$. The converse need not hold: continuity alone does not ensure that the difference quotient has a limit, as the corner example illustrates.

Common Pitfall: Forgetting to Subtract the Starting Value

The numerator must measure the change from the value at $a$, not simply the value at $a+h$. Consider $\mathbf{r}(t)=\langle t+2,1\rangle$ at $t=0$. An incorrect calculation would divide $\mathbf{r}(h)$ by $h$:

\frac{\mathbf{r}(h)}{h} = \left\langle\frac{h+2}{h},\frac{1}{h}\right\rangle.

This expression does not approach a finite vector, but it is not the derivative quotient. The required subtraction is $\mathbf{r}(h)-\mathbf{r}(0)$. Since $\mathbf{r}(0)=\langle 2,1\rangle$, the correct calculation is

\frac{\mathbf{r}(h)-\mathbf{r}(0)}{h} = \frac{\langle h+2,1\rangle-\langle 2,1\rangle}{h} = \frac{\langle h,0\rangle}{h} = \langle 1,0\rangle.

The quotient is the same for every $h\ne0$, so its limit is $\mathbf{r}'(0)=\langle 1,0\rangle$. Subtracting $\mathbf{r}(0)$ removes the starting position and leaves only the change.

Summary

To define or test a vector derivative at an interior parameter $a$, form $\frac{\mathbf{r}(a+h)-\mathbf{r}(a)}{h}$ and examine its two-sided limit as $h\to0$. If that limit exists as a finite vector, it is the instantaneous rate of change. If the one-sided limits disagree, the derivative does not exist. Keep the roles distinct: $\mathbf{r}(a)$ is a position or value, while $\mathbf{r}'(a)$ describes the limiting change per unit parameter.

Check Your Understanding

Use the difference-quotient definition to think through these questions.

For $\mathbf{r}:I\to\mathbb{R}^n$, write the difference quotient used to define $\mathbf{r}'(a)$.
What does the quotient represent when $\mathbf{r}(t)$ gives position as a function of time?
Why must the numerator subtract $\mathbf{r}(a)$?
What must be true of the one-sided limits for a two-sided vector derivative to exist?
Explain why existence of $\mathbf{r}'(a)$ implies continuity at $a$.

Derivatives from Difference Quotients

What You'll Learn

From Average Change to Instantaneous Change

Worked Example: A Difference Quotient for a Polynomial Curve

Worked Example: A Difference Quotient for a Polynomial Curve

Why This Limit Is an Instantaneous Rate

Worked Example: A Derivative on a Circular Path

Worked Example: A Derivative on a Circular Path

Checking Whether the Derivative Exists

Worked Example: A Curve with No Derivative at a Corner

A Consequence: Differentiability Implies Continuity

Common Pitfall: Forgetting to Subtract the Starting Value

Summary

Check Your Understanding

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