AP Calculus AB

Applications of Derivatives

Calculus AB

✓By the end of this lesson students will be able to solve related rates problems involving various geometric and physical scenarios.
✓By the end of this lesson students will be able to solve optimization problems by finding the absolute or local maximum/minimum values of functions.
✓By the end of this lesson students will be able to analyze the behavior of functions, including intervals of increasing/decreasing, local extrema, concavity, and points of inflection.
✓By the end of this lesson students will be able to use derivative analysis to accurately sketch the graph of a function.

Key concepts

Related Rates

Related rates problems involve finding the rate at which a quantity changes by relating it to other quantities whose rates of change are known. The core idea is to establish an equation relating the variables, then differentiate both sides of the equation with respect to time (t) using the Chain Rule. This process is known as implicit differentiation. It is crucial to identify all given rates and the rate to be found, and to only substitute constant values after differentiation.

Optimization

Optimization problems involve finding the maximum or minimum value of a function, often subject to certain constraints. This typically involves setting up a function to be optimized (the objective function), finding its critical points (where the first derivative is zero or undefined), and then using the First Derivative Test or the Second Derivative Test to determine whether these critical points correspond to a local maximum or minimum. For absolute extrema on a closed interval, the Extreme Value Theorem requires checking critical points and endpoints.

To find critical points, set f'(x) = 0 or find where f'(x) is undefined.

First Derivative Test

The First Derivative Test is used to determine the intervals where a function is increasing or decreasing and to classify local extrema (maxima or minima). If f'(x) > 0 on an interval, the function f(x) is increasing on that interval. If f'(x) < 0 on an interval, the function f(x) is decreasing on that interval. A local maximum occurs at a critical point where f'(x) changes from positive to negative. A local minimum occurs at a critical point where f'(x) changes from negative to positive. If f'(x) does not change sign, there is no local extremum.

If f'(x) > 0, f is increasing. If f'(x) < 0, f is decreasing.

Second Derivative Test

The Second Derivative Test provides an alternative method to classify local extrema at critical points. If c is a critical point (f'(c) = 0) and f''(c) > 0, then f has a local minimum at x = c. If f''(c) < 0, then f has a local maximum at x = c. If f''(c) = 0, the test is inconclusive, and the First Derivative Test must be used.

If f'(c)=0 and f''(c) > 0, local minimum at c. If f'(c)=0 and f''(c) < 0, local maximum at c.

Concavity and Points of Inflection

Concavity describes the direction in which the graph of a function opens. A function is concave up on an interval if its graph lies above its tangent lines on that interval, which occurs when f''(x) > 0. A function is concave down on an interval if its graph lies below its tangent lines on that interval, which occurs when f''(x) < 0. A point of inflection is a point on the graph where the concavity changes (from concave up to concave down or vice versa). This typically occurs where f''(x) = 0 or f''(x) is undefined, provided f''(x) changes sign at that point.

If f''(x) > 0, concave up. If f''(x) < 0, concave down.

Key facts to remember

1Related rates problems require implicit differentiation with respect to time (t) after establishing a relationship between variables.
2Optimization problems involve finding critical points (where f'(x) = 0 or f'(x) is undefined) and classifying them as maxima or minima.
3The First Derivative Test uses the sign change of f'(x) to identify local maxima (f' changes from + to -) and local minima (f' changes from - to +).
4If f'(x) > 0, the function is increasing; if f'(x) < 0, the function is decreasing.
5The Second Derivative Test uses the sign of f''(c) at a critical point c (where f'(c)=0) to classify local extrema: f''(c) > 0 implies a local minimum, f''(c) < 0 implies a local maximum.
6If f''(x) > 0, the function is concave up; if f''(x) < 0, the function is concave down.
7Points of inflection occur where the concavity changes, typically where f''(x) = 0 or is undefined and f''(x) changes sign.
8For absolute extrema on a closed interval, always check the function values at critical points and at the endpoints of the interval.

Worked examples

Example 1

A ladder 13 feet long is leaning against a vertical wall. If the bottom of the ladder is pulled away from the wall at a rate of 2 ft/s, how fast is the top of the ladder sliding down the wall when the bottom is 5 feet from the wall?

IDraw a diagram. Let x be the distance of the bottom of the ladder from the wall, and y be the height of the top of the ladder on the wall. The ladder itself is the hypotenuse, 13 feet.

IIIdentify knowns and unknowns:\n - Length of ladder = 13 ft (constant)\n - dx/dt = 2 ft/s (rate at which the bottom is pulled away)\n - We want to find dy/dt when x = 5 ft.

IIIRelate the variables using the Pythagorean theorem: x^2 + y^2 = 13^2.

IVDifferentiate both sides with respect to time (t) using implicit differentiation:\n d/dt (x^2 + y^2) = d/dt (169)\n 2x (dx/dt) + 2y (dy/dt) = 0

VFind the value of y when x = 5 ft:\n 5^2 + y^2 = 13^2\n 25 + y^2 = 169\n y^2 = 144\n y = 12 ft (since height must be positive)

VISubstitute the known values (x=5, y=12, dx/dt=2) into the differentiated equation:\n 2(5)(2) + 2(12)(dy/dt) = 0\n 20 + 24(dy/dt) = 0\n 24(dy/dt) = -20\n dy/dt = -20/24 = -5/6

VIIState the final answer with units.

Answer

The top of the ladder is sliding down the wall at a rate of -5/6 ft/s.

The negative sign indicates that the height (y) is decreasing.

Example 2

A rectangular page is to contain 30 square inches of print. The margins at the top and bottom of the page are 1 inch, and the margins on each side are 0.5 inches. What are the dimensions of the page that will minimize the amount of paper used?

IDefine variables:\n - Let w be the width of the printed area and h be the height of the printed area.\n - Area of print = wh = 30 => h = 30/w.\n - Let W be the total width of the page and H be the total height of the page.

IIExpress page dimensions in terms of print dimensions and margins:\n - W = w + 0.5 + 0.5 = w + 1\n - H = h + 1 + 1 = h + 2

IIIFormulate the objective function (Area of the page to minimize):\n - A = W * H = (w + 1)(h + 2)

IVSubstitute h = 30/w into the objective function to get A in terms of a single variable, w:\n - A(w) = (w + 1)(30/w + 2)\n - A(w) = 30 + 2w + 30/w + 2\n - A(w) = 32 + 2w + 30w^(-1)

VFind the derivative of A(w) with respect to w:\n - A'(w) = 2 - 30w^(-2) = 2 - 30/w^2

VIFind critical points by setting A'(w) = 0:\n - 2 - 30/w^2 = 0\n - 2 = 30/w^2\n - 2w^2 = 30\n - w^2 = 15\n - w = sqrt(15) (since width must be positive)

VIIUse the Second Derivative Test to confirm it's a minimum:\n - A''(w) = d/dw (2 - 30w^(-2)) = 60w^(-3) = 60/w^3\n - A''(sqrt(15)) = 60/(sqrt(15))^3. Since sqrt(15) > 0, A''(sqrt(15)) > 0, which confirms a local minimum.

VIIICalculate h and the page dimensions (W and H):\n - w = sqrt(15) inches\n - h = 30/w = 30/sqrt(15) = 30*sqrt(15)/15 = 2*sqrt(15) inches\n - W = w + 1 = sqrt(15) + 1 inches\n - H = h + 2 = 2*sqrt(15) + 2 inches

9State the final answer with units.

Answer

The dimensions of the page that minimize the amount of paper used are (sqrt(15) + 1) inches by (2*sqrt(15) + 2) inches.

Always check the domain of your objective function. Here, w must be positive.

Example 3

Analyze the function f(x) = x^4 - 4x^3 for intervals of increasing/decreasing, local extrema, concavity, and points of inflection.

IFind the first derivative, f'(x):\n - f'(x) = 4x^3 - 12x^2

IIFind critical points by setting f'(x) = 0:\n - 4x^3 - 12x^2 = 0\n - 4x^2(x - 3) = 0\n - Critical points at x = 0 and x = 3.

IIIUse the First Derivative Test to determine increasing/decreasing intervals and local extrema:\n - Test points in intervals: x < 0 (e.g., x=-1), 0 < x < 3 (e.g., x=1), x > 3 (e.g., x=4).\n - f'(-1) = -16 < 0 (decreasing)\n - f'(1) = -8 < 0 (decreasing)\n - f'(4) = 64 > 0 (increasing)\n - Intervals: Decreasing on (-infinity, 3); Increasing on (3, infinity).\n - Local Extrema: At x = 0, f'(x) does not change sign, so no local extremum. At x = 3, f'(x) changes from negative to positive, so there is a local minimum. f(3) = (3)^4 - 4(3)^3 = 81 - 108 = -27. Local minimum at (3, -27).

IVFind the second derivative, f''(x):\n - f''(x) = d/dx (4x^3 - 12x^2) = 12x^2 - 24x

VFind possible points of inflection by setting f''(x) = 0:\n - 12x^2 - 24x = 0\n - 12x(x - 2) = 0\n - Possible points of inflection at x = 0 and x = 2.

VIUse concavity analysis to determine concavity and points of inflection:\n - Test points in intervals: x < 0 (e.g., x=-1), 0 < x < 2 (e.g., x=1), x > 2 (e.g., x=3).\n - f''(-1) = 36 > 0 (concave up)\n - f''(1) = -12 < 0 (concave down)\n - f''(3) = 36 > 0 (concave up)\n - Concavity: Concave up on (-infinity, 0) and (2, infinity); Concave down on (0, 2).\n - Points of Inflection: At x = 0, f''(x) changes from positive to negative. f(0) = 0. Point of inflection at (0, 0). At x = 2, f''(x) changes from negative to positive. f(2) = (2)^4 - 4(2)^3 = 16 - 32 = -16. Point of inflection at (2, -16).

VIISummarize all findings.

Answer

Increasing: (3, infinity)\nDecreasing: (-infinity, 3)\nLocal Minimum: (3, -27)\nConcave Up: (-infinity, 0) and (2, infinity)\nConcave Down: (0, 2)\nPoints of Inflection: (0, 0) and (2, -16)

Remember to evaluate the original function f(x) to find the y-coordinates of local extrema and points of inflection.

Common mistakes

✗In related rates, substituting a variable's specific value before differentiating with respect to time, especially if that value is changing.
✗For optimization problems, failing to define the domain of the objective function, particularly for physical constraints, which can lead to incorrect absolute extrema.
✗Not justifying the nature of extrema (max/min) or points of inflection with a clear statement (e.g., 'by the First Derivative Test,' 'because f'' changes sign').
✗Confusing the conditions for increasing/decreasing (based on f'(x)) with concavity (based on f''(x)).
✗Assuming that f''(x) = 0 automatically means a point of inflection without checking for a sign change in f''(x) around that point.

Exam tips

★For related rates and optimization problems, always start by drawing a clear, labeled diagram to visualize the problem and define variables.
★Clearly define all variables and state their units, especially for rates of change, to avoid errors and ensure clarity.
★Show all steps in your derivative calculations and tests (First Derivative Test, Second Derivative Test, concavity analysis) to earn full credit.
★When asked to justify a conclusion (e.g., a local maximum or a point of inflection), explicitly state the theorem or test you are using and explain how it applies (e.g., 'By the First Derivative Test, f has a local minimum at x=c because f'(x) changes from negative to positive').

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