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Applications of Derivatives

Optimization, related rates, and motion.

This module is where derivatives become powerful tools for solving real problems. You'll learn to find the maximum and minimum values of functions, sketch curves, analyze motion, and solve problems where multiple quantities change simultaneously.

The core idea: the derivative gives you information about how a function behaves. A positive derivative means the function is increasing. A zero derivative marks a potential extreme point. The second derivative tells you about curvature.

Mastering applications requires translating real-world scenarios into mathematical functions, then using derivative tools to extract answers. Practice the translation step as much as the calculus.

Critical points and the first derivative test

A critical point of ff occurs at x=cx=c where f(c)=0f'(c) = 0 or f(c)f'(c) is undefined (and f(c)f(c) exists).

The first derivative test: examine the sign of f(x)f'(x) on either side of the critical point.

If ff' changes from ++ to - at cc, then f(c)f(c) is a local maximum.

If ff' changes from - to ++ at cc, then f(c)f(c) is a local minimum.

If ff' does not change sign (e.g., f(x)=x3f(x) = x^3 at x=0x=0), the critical point is neither a max nor a min.

To apply: find all critical points, then build a sign chart for f(x)f'(x) using test values in each interval.

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The second derivative test

An alternative to the first derivative test that avoids building a sign chart. At a critical point where f(c)=0f'(c) = 0:

If f(c)>0f''(c) > 0: the curve is concave up (bowl-shaped) at cc, so f(c)f(c) is a local minimum. The function is curving upward — any nearby point is higher.

If f(c)<0f''(c) < 0: the curve is concave down (arch-shaped) at cc, so f(c)f(c) is a local maximum. The function is curving downward — any nearby point is lower.

If f(c)=0f''(c) = 0: the test is inconclusive. The critical point could be a min, max, or neither. Fall back to the first derivative test. Example: f(x)=x4f(x) = x^4 at x=0x=0 has f(0)=0f''(0)=0 but it's still a minimum. Meanwhile f(x)=x3f(x)=x^3 at x=0x=0 has f(0)=0f''(0)=0 and it's neither.

When to use which: the second derivative test is faster when ff'' is easy to compute. The first derivative test (sign chart) is more reliable — it never fails and also works when ff' is undefined.

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Concavity and inflection points

Concavity describes how the curve bends. f(x)>0f''(x) > 0 means concave up (opening upward), f(x)<0f''(x) < 0 means concave down (opening downward).

An inflection point is where the concavity changes: ff'' switches sign. At an inflection point, the curve transitions from bending one way to bending the other.

To find inflection points: set f(x)=0f''(x) = 0 (or find where ff'' is undefined), then verify ff'' changes sign across that point.

Example: f(x)=x3f(x) = x^3 has f(x)=6xf''(x) = 6x. This equals 00 at x=0x=0, and changes from negative to positive. So (0,0)(0, 0) is an inflection point.

Note: f(c)=0f''(c) = 0 alone is not enough. f(x)=x4f(x) = x^4 has f(0)=0f''(0) = 0 but no inflection point (concavity doesn't change).

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Absolute extrema on closed intervals (Extreme Value Theorem)

The Extreme Value Theorem (EVT): if ff is continuous on a closed interval [a,b][a,b], then ff attains an absolute maximum and an absolute minimum somewhere on [a,b][a,b]. Both 'continuous' and 'closed interval' are required — remove either condition and the theorem can fail.

The closed interval method to find them:

Step 1: Find all critical points of ff in the open interval (a,b)(a,b) — where f(x)=0f'(x)=0 or ff' is undefined.

Step 2: Evaluate ff at each critical point and at both endpoints aa and bb. Build a table of (x,f(x))(x, f(x)) values.

Step 3: The largest ff-value in the table is the absolute maximum, the smallest is the absolute minimum.

This is the go-to method for any 'find the max/min on an interval' problem. The absolute extrema must occur at a critical point or an endpoint — there's nowhere else for them to hide. Never forget the endpoints!

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Curve sketching (putting it all together)

A complete curve sketch synthesizes everything you've learned about derivatives into a picture. Follow this checklist systematically:

1. Domain and intercepts: where is ff defined? Set f(x)=0f(x)=0 for xx-intercepts, evaluate f(0)f(0) for the yy-intercept.

2. Symmetry: is ff even (f(x)=f(x)f(-x)=f(x), symmetric about yy-axis), odd (f(x)=f(x)f(-x)=-f(x), symmetric about origin), or neither? This can halve your work.

3. First derivative analysis: find critical points (where f=0f'=0 or undefined). Build a sign chart to determine intervals of increase (f>0f'>0) and decrease (f<0f'<0). Classify each critical point as local max, local min, or neither.

4. Second derivative analysis: find inflection point candidates (where f=0f''=0 or undefined). Build a sign chart for ff'' to determine concavity (up where f>0f''>0, down where f<0f''<0). Verify sign changes for actual inflection points.

5. Asymptotes: vertical (denominator → 0 with nonzero numerator), horizontal (limx±\lim_{x\to\pm\infty}), and oblique/slant (when degree of numerator = degree of denominator + 1).

6. Plot key points (intercepts, critical points, inflection points) and connect the dots, respecting the increase/decrease direction and concavity at every point.

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Optimization problems

The general strategy for optimization word problems:

Step 1: Draw a diagram and define variables. Label everything.

Step 2: Write the objective function (what you want to maximize or minimize).

Step 3: Write the constraint equation (the relationship that limits your variables).

Step 4: Use the constraint to eliminate one variable from the objective, getting a function of a single variable.

Step 5: Find the critical points of this function. Test them (and the endpoints, if the domain is closed) to identify the optimum.

Step 6: Answer the original question. Make sure you're solving for what was asked.

Common setups: maximize area given a perimeter constraint, minimize material given a volume constraint, maximize revenue.

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Related rates problems involve multiple quantities changing over time, connected by a geometric or physical equation.

Step 1: Draw a picture and identify all variables. Label which quantities are changing.

Step 2: Write an equation relating the variables (e.g., Pythagorean theorem, area formula, volume formula).

Step 3: Differentiate both sides with respect to time tt using implicit differentiation.

Step 4: Plug in all known values and rates at the specific instant in question.

Step 5: Solve for the unknown rate.

Critical: never plug in specific values before differentiating. The equation must remain general during differentiation because the variables are changing.

Common setups: expanding balloon (sphere volume), sliding ladder (Pythagorean theorem), filling cone (similar triangles + cone volume), spreading oil slick (circle area).

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Motion along a line

Given a position function s(t)s(t): velocity is v(t)=s(t)v(t) = s'(t) and acceleration is a(t)=v(t)=s(t)a(t) = v'(t) = s''(t). This is the most natural physical application of derivatives.

The particle is moving right (positive direction) when v(t)>0v(t) > 0 and moving left when v(t)<0v(t) < 0. The particle is at rest (momentarily stopped) when v(t)=0v(t) = 0.

Direction change: the particle changes direction when v(t)=0v(t) = 0 and the velocity changes sign. If vv doesn't change sign, the particle merely pauses.

Speed vs. velocity: velocity is signed (includes direction), speed is v(t)|v(t)| (always non-negative). The particle speeds up when velocity and acceleration have the same sign (both positive or both negative) and slows down when they have opposite signs.

Displacement vs. total distance: abv(t)dt\int_a^b v(t)\,dt gives displacement (net change in position, which can be zero if the particle returns). abv(t)dt\int_a^b |v(t)|\,dt gives total distance traveled (always ≥ 0). To compute total distance, find where v(t)=0v(t) = 0, split the integral, and flip signs on intervals where v<0v < 0.

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Linearization and differentials

The linearization of ff at aa is L(x)=f(a)+f(a)(xa)L(x) = f(a) + f'(a)(x-a). This is the equation of the tangent line at x=ax = a, repurposed as an approximation tool.

For values of xx near aa, f(x)L(x)f(x) \approx L(x). The closer xx is to aa, the better the approximation. This is the foundation of how engineers and scientists do quick estimates without a calculator.

Differentials formalize this: if y=f(x)y = f(x), then dy=f(x)dxdy = f'(x)\,dx is the approximate change in yy when xx changes by a small amount dxdx. The actual change is Δy=f(x+dx)f(x)\Delta y = f(x+dx) - f(x), and dyΔydy \approx \Delta y when dxdx is small.

Why this works: the tangent line is the best linear approximation to the curve at that point. For tiny changes, a curve behaves almost exactly like its tangent line. The error is proportional to (xa)2(x-a)^2, so it vanishes quickly as xax \to a.

Applications: estimating values like 4.02\sqrt{4.02} or sin(0.1)\sin(0.1) by hand, error propagation in measurements, and building the intuition that leads to Taylor series (which extend this idea to quadratic, cubic, and higher-degree approximations).

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The Mean Value Theorem

If ff is continuous on [a,b][a,b] and differentiable on (a,b)(a,b), then there exists at least one cc in (a,b)(a,b) such that f(c)=f(b)f(a)baf'(c) = \frac{f(b)-f(a)}{b-a}.

In words: somewhere between aa and bb, the instantaneous rate of change equals the average rate of change.

Geometric interpretation: there's a point where the tangent line is parallel to the secant line connecting (a,f(a))(a, f(a)) and (b,f(b))(b, f(b)).

Applications: proving that a function must have a certain derivative value, establishing speed limits (if you drove 120 miles in 2 hours, you must have been going 60 mph at some instant).

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Newton's method

Newton's method finds approximate roots of f(x)=0f(x) = 0 using tangent lines.

Start with an initial guess x0x_0. The iteration formula is: xn+1=xnf(xn)f(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.

Each step draws the tangent line at (xn,f(xn))(x_n, f(x_n)) and finds where it crosses the xx-axis. That crossing point is xn+1x_{n+1}.

Convergence is typically quadratic: the number of correct digits roughly doubles each step. For example, if x3x_3 has 3 correct digits, x4x_4 may have 6.

Error estimation: xn+1rf(r)2f(r)xnr2|x_{n+1} - r| \approx \frac{|f''(r)|}{2|f'(r)|} \cdot |x_n - r|^2 where rr is the true root. This quantifies the quadratic convergence.

Pitfalls: the method can fail if f(xn)=0f'(x_n) = 0 (horizontal tangent), if the initial guess is too far from the root, or if the function oscillates. Cycling is possible — Newton's method applied to x32x+2x^3 - 2x + 2 starting at x0=0x_0 = 0 enters an infinite loop between 11 and 00.

Stopping criterion: iterate until xn+1xn<ε|x_{n+1} - x_n| < \varepsilon for a desired tolerance ε\varepsilon, or until f(xn)<ε|f(x_n)| < \varepsilon.

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Rates of change in the sciences

The derivative is the universal language for rates. Here are key applications across disciplines.

Physics — velocity and acceleration: if s(t)s(t) is position, then v(t)=s(t)v(t) = s'(t) is velocity and a(t)=v(t)=s(t)a(t) = v'(t) = s''(t) is acceleration. Force equals mass times acceleration: F=ma=ms(t)F = ma = ms''(t).

Biology — population growth: if P(t)P(t) is population, P(t)P'(t) is the growth rate. The per-capita growth rate is P(t)/P(t)P'(t)/P(t). This leads to the logistic equation P=rP(1P/K)P' = rP(1 - P/K).

Chemistry — reaction rates: the rate of a chemical reaction d[A]dt\frac{d[A]}{dt} measures how fast a reactant AA is consumed. The rate law relates this to concentrations: d[A]dt=k[A]n-\frac{d[A]}{dt} = k[A]^n.

Economics — marginal analysis: the marginal cost C(x)C'(x) is the cost of producing one additional unit. The marginal revenue R(x)R'(x) is the revenue from one more sale. Profit is maximized when R(x)=C(x)R'(x) = C'(x).

Engineering — sensitivity analysis: if an output yy depends on a parameter pp, the derivative dy/dpdy/dp measures how sensitive the output is to small changes in pp. This is fundamental to error propagation: Δyy(p)Δp\Delta y \approx |y'(p)| \cdot \Delta p.

The mathematical tools (chain rule, implicit differentiation, related rates) are identical across all these fields. Only the variable names change.

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Worked Practice Problems (5 examples)

Study these solved examples to understand the techniques. Then head to Practice to try problems on your own.

Practice on your own →
1

Problem 1

A particle moves with v(t)=4tv(t) = 4t. Find the distance from t=0t=0 to t=3t=3.

Step-by-Step Solution

1

Distance =034tdt=2t203= \int_0^3 4t\,dt = 2t^2\Big|_0^3.

2

2(9)0=182(9) - 0 = 18.

Final Answer: 1818 units

2

Problem 2

Find the critical point of f(x)=x26x+5f(x) = x^2 - 6x + 5.

Step-by-Step Solution

1

f(x)=2x6f'(x) = 2x - 6.

2

Set f(x)=0f'(x)=0: 2x=62x=6, so x=3x=3.

Final Answer: x=3x=3

3

Problem 3

A ball is thrown up with h(t)=16t2+64th(t)=-16t^2+64t. Find the max height.

Step-by-Step Solution

1

h(t)=32t+64=0t=2h'(t) = -32t + 64 = 0 \Rightarrow t = 2.

2

h(2)=16(4)+64(2)=64+128=64h(2) = -16(4)+64(2) = -64+128 = 64.

Final Answer: 6464 ft

4

Problem 4

If f(x)=x312xf(x) = x^3 - 12x, find the local minimum value.

Step-by-Step Solution

1

f(x)=3x212=0x=±2f'(x)=3x^2-12=0 \Rightarrow x=\pm 2.

2

f(x)=6xf''(x)=6x. At x=2x=2: f=12>0f''=12>0 (min).

3

f(2)=824=16f(2)=8-24=-16.

Final Answer: 16-16

5

Problem 5

Find the slope of the tangent to y=x2y=x^2 at x=3x=3.

Step-by-Step Solution

1

y=2xy' = 2x.

2

At x=3x=3: slope =2(3)=6= 2(3) = 6.

Final Answer: 66

Frequently asked questions about applications of derivatives

What is optimization in calculus?
Optimization uses derivatives to find the maximum or minimum value of a function. Common applications include maximizing area, minimizing cost, and finding shortest paths.
What are related rates problems?
Related rates problems involve finding how fast one quantity is changing given the rate of change of another related quantity, using implicit differentiation with respect to time.
How do you find critical points?
Set the first derivative equal to zero and solve. Critical points are where the function may have a local maximum, minimum, or inflection point.