Applications of Derivatives

The Role of Derivatives

A derivative measures instantaneous change. Applications begin by translating a situation into a function or equation, differentiating carefully, and interpreting the result in context. The same core idea supports rate calculations, approximations, motion analysis, extreme-value problems, and optimization.

A useful first question is: what quantity is changing, what variable measures the change, and what relationship connects the quantities? Units and signs are part of the mathematical answer, not merely final annotations.

Takeaway: Derivatives connect a changing model to a meaningful rate, estimate, direction, or extremum.

Related Rates

A related-rates problem involves quantities that change with respect to a common variable, usually time. If a variable depends on time, write it as a function such as $r(t)$, and represent its rate by $\frac{dr}{dt}$.

A reliable procedure

1. Define every changing variable and assign units.

2. Write the given rates and the unknown rate, preserving positive and negative signs.

3. Find an equation relating the variables.

4. Differentiate the equation with respect to time, applying the chain rule to every changing quantity.

5. Substitute numerical values only after differentiating.

6. Solve and state whether the result represents an increase or decrease.

For a circular ripple with area $A=\pi r^2$, differentiation gives

If $r=5$ centimeters and $\frac{dr}{dt}=2$ centimeters per second, then

The area is therefore increasing at $20\pi\text{ cm}^2/\text{s}$. If a radius decreases at $3$ centimeters per second, use $\frac{dr}{dt}=-3\text{ cm}/\text{s}$; the negative sign must remain in the calculation and interpretation.

Common relationships include

and

For example, differentiating $V=\frac{4}{3}\pi r^3$ gives $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$, not merely $4\pi r^2$. The factor $\frac{dr}{dt}$ appears because $r$ changes with time.

Takeaway: Establish the relationship first, differentiate before substituting, and interpret the signed result with units.

Approximation and Differentials

Linearization uses the tangent line at a convenient point to approximate a function nearby. At $x=a$, the approximation is

so that $f(x)\approx L(x)$ when $x$ is close to $a$.

To estimate $\sqrt{4.1}$, choose $f(x)=\sqrt{x}$ and $a=4$. Then

Thus,

and

The actual error is $f(x)-L(x)$. Accuracy generally improves when $x$ is close to $a$, the function is differentiable nearby, and the graph has little curvature. For $f(x)=\sqrt{x}$, the graph is concave down, so the tangent-line approximation lies above the graph locally.

A differential expresses the same approximation in change notation. If $y=f(x)$, then

The actual change is $\Delta y=f(x+\Delta x)-f(x)$, and for a small change, $\Delta y\approx dy$. Consequently,

For a sphere with $V=\frac{4}{3}\pi r^3$,

At $r=10$ centimeters and $dr=0.02$ centimeters,

The relative-error relationship is

so a relative radius error of approximately $0.2\%$ produces a relative volume error of approximately $0.6\%$.

Takeaway: Linearization approximates function values, while differentials approximate small changes and propagated measurement errors.

Motion Along a Line

For motion along a line, let $s(t)$ denote position at time $t$. The related quantities are

Velocity has a sign: $v(t)>0$ means motion in the positive direction, $v(t)<0$ means motion in the negative direction, and $v(t)=0$ means the object is instantaneously at rest. A change of direction requires a sign change in velocity at a time when velocity is zero or undefined.

Consider

Its velocity is

The critical times are $t=1$ and $t=3$. The sign of velocity is positive on $(0,1)$, negative on $(1,3)$, and positive on $(3,4)$. Thus, the object changes direction at both $t=1$ and $t=3$.

Displacement is the net change in position,

whereas total distance must account for reversals:

Takeaway: Differentiate position to obtain velocity, use velocity signs to determine direction, and split distance calculations at direction changes.

Extreme Values and Derivative Tests

An absolute maximum is the greatest value on the entire domain, while an absolute minimum is the least value. A local maximum or minimum compares the function only with nearby values.

The Extreme Value Theorem guarantees both an absolute maximum and an absolute minimum when a function is continuous on a closed, bounded interval $[a,b]$. To find absolute extrema on such an interval:

1. Find all critical numbers in $(a,b)$, where $f'(x)=0$ or $f'(x)$ does not exist.

2. Evaluate the function at every interior critical number.

3. Evaluate the function at both endpoints.

4. Compare all resulting values.

For

on $[0,3]$,

The interior critical number is $x=2$. The relevant values are

Therefore, the absolute minimum is $-3$ at $x=2$, and the absolute maximum is $1$ at both $x=0$ and $x=3$.

The First Derivative Test uses the sign of $f'$ around a critical number. A change from positive to negative gives a local maximum; a change from negative to positive gives a local minimum. The Second Derivative Test provides a shortcut when $f'(c)=0$: $f''(c)>0$ indicates a local minimum, $f''(c)<0$ indicates a local maximum, and $f''(c)=0$ is inconclusive.

Takeaway: Critical numbers are candidates, not automatic extrema; absolute-extrema problems on closed intervals require endpoint comparisons.

The Mean Value Theorem

The Mean Value Theorem connects average and instantaneous rates. If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then at least one number $c\in(a,b)$ satisfies

The right side is the secant slope, or average rate of change, and $f'(c)$ is the tangent slope at an interior point.

For $f(x)=x^2$ on $[1,3]$, the average rate is

Since $f'(x)=2x$, the required point satisfies $2c=4$, so $c=2$.

Rolle's Theorem is the special case in which $f(a)=f(b)$. Under the same continuity and differentiability conditions, there is a point $c\in(a,b)$ where $f'(c)=0$.

Important consequences include:

• If $f'(x)>0$ throughout an interval, then $f$ is increasing there.

• If $f'(x)<0$ throughout an interval, then $f$ is decreasing there.

• If $f'(x)=0$ throughout an interval, then $f$ is constant there.

• If two functions have the same derivative on an interval, they differ by a constant on that interval.

Takeaway: The theorem turns derivative information into conclusions about average rates, monotonicity, and constancy, provided its hypotheses hold.

Optimization Problems

Optimization seeks the largest or smallest possible value of a quantity such as area, volume, cost, distance, time, or profit. A sound solution connects the real situation to a one-variable objective function.

General method

1. Identify the quantity to maximize or minimize.

2. Define variables and make a diagram when useful.

3. Write the constraint equation.

4. Use the constraint to express the objective in one variable.

5. Determine the feasible domain from the physical restrictions.

6. Find critical numbers by solving $f'(x)=0$ and including points where $f'$ is undefined.

7. Check critical numbers and endpoints.

8. State the result with units and in the original context.

For a rectangular garden with perimeter $40$ meters, let the dimensions be $x$ and $y$. The constraint is

so $y=20-x$. The area becomes the one-variable objective

Differentiate and solve:

Then $y=20-10=10$. Since the objective is a downward-opening parabola, the maximum occurs for a $10\text{ m}\times 10\text{ m}$ square, with area

A critical point is not automatically a maximum or minimum. Use a derivative sign chart, a derivative test, endpoint comparison, or direct comparison of objective values.

Takeaway: Translate the constraint into a feasible one-variable objective, then test every relevant candidate and interpret the result.