Applications of Differential Calculus | Free AP Calculus AB Course

<!DOCTYPE html>

<head>

<title>AP Calculus AB: Applications of Differential Calculus | The School of Mathematics</title>

<style>

:root {

--blue: #4285f4;

--green: #34a853;

--red: #ea4335;

--gray-bg: #f8f9fa;

--border: #e0e0e0;

}

* { box-sizing: border-box; }

body {

font-family: -apple-system, BlinkMacSystemFont, "Segoe UI", Roboto, Helvetica, Arial, sans-serif;

background: #ffffff;

color: #202124;

line-height: 1.65;

margin: 0;

padding: 0;

}

.wrap { max-width: 820px; margin: 0 auto; padding: 32px 20px 80px; }

h1 { font-size: 2rem; line-height: 1.25; margin-bottom: 10px; }

h2 {

font-size: 1.5rem;

margin-top: 48px;

padding-top: 8px;

border-top: 2px solid var(--border);

}

h3 { font-size: 1.15rem; margin-bottom: 6px; }

.intro { font-size: 1.05rem; color: #3c4043; }

.cta-group {

display: flex;

gap: 12px;

flex-wrap: wrap;

margin: 22px 0 32px;

}

.cta-btn {

flex: 1 1 220px;

text-align: center;

text-decoration: none;

font-weight: 700;

font-size: 1rem;

padding: 15px 18px;

border-radius: 10px;

}

.cta-primary { background: var(--blue); color: #fff; }

.cta-primary:hover { background: #3367d6; }

.cta-secondary { background: #fff; color: var(--blue); border: 2px solid var(--blue); }

.cta-secondary:hover { background: #eef3fd; }

nav.toc {

background: var(--gray-bg);

border: 1px solid var(--border);

border-radius: 10px;

padding: 18px 22px;

margin-bottom: 36px;

}

nav.toc h2 { margin-top: 0; border-top: none; font-size: 1.1rem; }

nav.toc ol { margin: 0; padding-left: 22px; }

nav.toc a { color: var(--blue); text-decoration: none; }

nav.toc a:hover { text-decoration: underline; }

table.ref {

width: 100%;

border-collapse: collapse;

margin: 18px 0 26px;

font-size: 0.92rem;

}

table.ref th, table.ref td {

border: 1px solid var(--border);

padding: 9px 11px;

text-align: left;

vertical-align: top;

}

table.ref th { background: var(--gray-bg); }

.example {

background: var(--gray-bg);

border-left: 4px solid var(--blue);

border-radius: 6px;

padding: 16px 20px;

margin: 18px 0;

}

.example p { margin: 6px 0; }

.step-math { font-family: "Cambria Math", Georgia, serif; }

.frac {

display: inline-flex;

flex-direction: column;

vertical-align: middle;

text-align: center;

font-size: 0.85em;

line-height: 1;

margin: 0 2px;

position: relative;

top: 0.1em;

}

.frac .num { border-bottom: 1.3px solid #202124; padding: 0 4px 1px; }

.frac .den { padding: 1px 4px 0; }

.problem {

border: 1px solid var(--border);

border-radius: 10px;

padding: 18px 22px 20px;

margin: 18px 0;

}

.problem .prompt { font-weight: 600; margin-bottom: 12px; }

.options { display: flex; flex-direction: column; gap: 8px; margin: 0 0 4px; }

.option-btn {

text-align: left;

padding: 10px 14px;

border: 1px solid #ccc;

border-radius: 6px;

background: #fff;

cursor: pointer;

font-size: 0.98rem;

font-family: inherit;

}

.option-btn:hover:not(:disabled) { background: #f1f3f4; }

.option-btn.correct { background: #e6f4ea; border-color: var(--green); font-weight: 600; }

.option-btn.incorrect { background: #fce8e6; border-color: var(--red); }

.option-btn:disabled { cursor: default; }

.explanation {

margin-top: 0;

padding: 0 16px;

background: var(--gray-bg);

border-left: 4px solid var(--blue);

border-radius: 4px;

max-height: 0;

opacity: 0;

overflow: hidden;

transition: max-height 0.35s ease, opacity 0.35s ease, margin-top 0.35s ease, padding 0.35s ease;

}

.explanation.show {

max-height: 600px;

opacity: 1;

margin-top: 14px;

padding: 14px 16px;

}

.explanation p { margin: 6px 0; }

.explanation .label { font-weight: 700; }

.mistake-list li { margin-bottom: 10px; }

.faq-item { margin-bottom: 18px; }

.faq-item h3 { margin-bottom: 4px; }

footer.cta-final { margin-top: 56px; }

</style>

</head>

<body>

<h1>AP Calculus AB: Applications of Differential Calculus</h1>

This is where derivatives start doing real work: describing how a function behaves, finding its extreme values, and modeling change in the physical world. This free, complete lesson covers slope and critical points, increasing and decreasing functions, concavity and inflection points, relating f, f', and f'' graphically, global extrema (the Closed Interval Test), optimization, motion along a line, local linear approximation, and related rates. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the full AP Calculus AB Question Bank linked below.

</p>

<a class="cta-btn cta-primary" href="https://theschoolofmathematics.com/quiz/ap-calculus-ab-applications-of-differential-calculus-quiz-1">Practice Applications of Differential Calculus Free</a>

<a class="cta-btn cta-secondary" href="https://theschoolofmathematics.com/quiz/course/AP-Calculus-AB-QBank">Explore the Full AP Calculus AB Qbank</a>

</div>

<h2>What's covered in this lesson</h2>

<ol>

<li><a href="#slope-critical">Slope, Critical Points, and Tangent Lines</a></li>

<li><a href="#incr-decr">Increasing and Decreasing Functions</a></li>

<li><a href="#concavity">Concavity, Inflection Points, and the Second Derivative Test</a></li>

<li><a href="#relating">Relating f, f', and f'' Graphically</a></li>

<li><a href="#global">Global Extrema: The Closed Interval Test</a></li>

<li><a href="#optimization">Optimization</a></li>

<li><a href="#motion">Motion Along a Line</a></li>

<li><a href="#linear-approx">Local Linear Approximation</a></li>

<li><a href="#related-rates">Related Rates</a></li>

<li><a href="#mistakes">Common Mistakes to Avoid</a></li>

<li><a href="#faq">Frequently Asked Questions</a></li>

</ol>

</nav>

<h2 id="slope-critical">1. Slope, Critical Points, and Tangent Lines</h2>

<p>The <strong>slope</strong> of a curve y = f(x) at a point is f'(x) there. A <strong>critical point</strong> is any c in the domain of f where either f'(c) = 0 or f'(c) is undefined. The equation of the tangent line at (x₁, y₁) is y − y₁ = f'(x₁)(x − x₁).</p>

<p><strong>Worked Example:</strong> For f(x) = 2x³ − 3x² − 12x + 5, find the critical points.</p>

<p>Setting f'(x) = 0: x = 2 and x = −1.</p>

</div>

<p class="prompt">1. For f(x) = x³ − 6x² + 9x, find the critical points.</p>

</div>

<p><span class="label">Explanation:</span> f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3), which is 0 at x = 1 and x = 3.</p>

</div>

<p class="prompt">2. Find an equation of the tangent to f(x) = x³ − 2x² at the point (2, 0).</p>

</div>

<p><span class="label">Explanation:</span> f'(x) = 3x² − 4x, so f'(2) = 12 − 8 = 4. Tangent: y − 0 = 4(x − 2), or y = 4x − 8.</p>

</div>

<p class="prompt">3. For g(x) = (x − 3)<sup>2/3</sup>, is x = 3 a critical point? Why?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p1-3',true)">A) Yes, because g'(3) is undefined</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-3',false)">B) No, because g'(3) = 0 nowhere</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-3',false)">C) No, since g is not defined at x = 3</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-3',false)">D) Yes, but only because g''(3) = 0</button>

</div>

<p><span class="label">Explanation:</span> g'(x) = <span class="frac"><span class="num">2</span><span class="den">3(x − 3)<sup>1/3</sup></span></span>, which is undefined at x = 3 even though g(3) = 0 is defined. A critical point can arise from an undefined derivative, not just a zero one.</p>

</div>

<p class="prompt">4. If the tangent to a curve at x = 5 is horizontal, what must be true?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-4',false)">B) f'(5) is undefined</button>

</div>

<p><span class="label">Explanation:</span> A horizontal tangent line has slope 0, so f'(5) = 0.</p>

</div>

<h2 id="incr-decr">2. Increasing and Decreasing Functions</h2>

<p>f is <strong>increasing</strong> on an interval if f(a) < f(b) whenever a < b there; it's <strong>decreasing</strong> if f(a) > f(b). Where f'(x) ≥ 0, f is increasing; where f'(x) ≤ 0, f is decreasing. The <strong>First Derivative Test</strong>: if f' changes from + to − at a critical point c, f has a local maximum there; if it changes from − to +, a local minimum; if it doesn't change sign, neither.</p>

<p><strong>Worked Example:</strong> For f(x) = x³ − 3x², determine where f is increasing and decreasing.</p>

<p>f'(x) = 3x² − 6x = 3x(x − 2). Critical points: 0, 2.</p>

<p>Testing each interval: f is increasing for x ≤ 0 and x ≥ 2, decreasing for 0 ≤ x ≤ 2.</p>

</div>

<p class="prompt">1. For f(x) = x³ − 12x, on which interval is f decreasing?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-1',false)">D) All real numbers</button>

</div>

<p><span class="label">Explanation:</span> f'(x) = 3x² − 12 = 3(x − 2)(x + 2). Testing x = 0: f'(0) = −12 < 0, so f decreases on (−2, 2).</p>

</div>

<p class="prompt">2. A function f has f'(x) > 0 for x < 3 and f'(x) < 0 for x > 3, with f'(3) = 0. What does the First Derivative Test conclude at x = 3?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p2-2',true)">A) f has a local maximum at x = 3</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-2',false)">B) f has a local minimum at x = 3</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-2',false)">C) f has neither a max nor min at x = 3</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-2',false)">D) f has an inflection point at x = 3</button>

</div>

<p><span class="label">Explanation:</span> A sign change from + to − at a critical point signals a local maximum.</p>

</div>

<p class="prompt">3. For h(x) = <span class="frac"><span class="num">1</span><span class="den">x − 2</span></span>, determine the sign of h'(x) for all x ≠ 2.</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p2-3',true)">A) h'(x) is always negative</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-3',false)">B) h'(x) is always positive</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-3',false)">C) h'(x) changes sign at x = 2</button>

</div>

<p><span class="label">Explanation:</span> h'(x) = <span class="frac"><span class="num">−1</span><span class="den">(x − 2)²</span></span>. Since (x − 2)² is always positive for x ≠ 2, h'(x) is always negative.</p>

</div>

<p class="prompt">4. If f'(x) ≥ 0 on an interval and f'(x) = 0 only at an isolated point c within it, can c still be included in the interval where f is called increasing?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p2-4',true)">A) Yes, a single isolated zero of f' doesn't interrupt the classification</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-4',false)">B) No, f cannot be called increasing anywhere f' = 0</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p2-4',false)">D) Only if c is an endpoint of the domain</button>

</div>

<p><span class="label">Explanation:</span> The definition of increasing only requires f(a) < f(b) for a < b in the interval, a momentary flat tangent at one point doesn't violate this.</p>

</div>

<h2 id="concavity">3. Concavity, Inflection Points, and the Second Derivative Test</h2>

<p>A graph is <strong>concave upward</strong> where f''(x) > 0, and <strong>concave downward</strong> where f''(x) < 0. A <strong>point of inflection</strong> is where concavity changes. The <strong>Second Derivative Test</strong>: at a critical point c where f'(c) = 0, if f''(c) > 0, c yields a local minimum; if f''(c) < 0, a local maximum; if f''(c) = 0, the test fails and the First Derivative Test must be used instead.</p>

<p><strong>Worked Example:</strong> For f(x) = x³ − 6x² + 9x + 1, find any local extrema using the Second Derivative Test.</p>

<p>f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3). Critical points: 1, 3.</p>

<p class="step-math">f''(x) = 6x − 12. f''(1) = −6 < 0 (local max). f''(3) = 6 > 0 (local min).</p>

</div>

<p class="prompt">1. For f(x) = x³ − 3x² − 9x + 2, use the Second Derivative Test to classify the critical points.</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p3-1',true)">A) Local max at x = −1, local min at x = 3</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p3-1',false)">B) Local min at x = −1, local max at x = 3</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p3-1',false)">C) Local max at both</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p3-1',false)">D) Local min at both</button>

</div>

<p><span class="label">Explanation:</span> f'(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1); critical points −1, 3. f''(x) = 6x − 6. f''(−1) = −12 < 0 (max); f''(3) = 12 > 0 (min).</p>

</div>

<p class="prompt">2. For f(x) = x⁴ − 4x³, the Second Derivative Test fails at which critical point?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p3-2',false)">D) The test never fails for this function</button>

</div>

<p><span class="label">Explanation:</span> f'(x) = 4x³ − 12x² = 4x²(x − 3); critical points 0, 3. f''(x) = 12x² − 24x. f''(0) = 0 (test fails); f''(3) = 36 > 0 (test succeeds, local min).</p>

</div>

<p class="prompt">3. Continuing the previous problem, is x = 0 a local max, local min, or neither for f(x) = x⁴ − 4x³?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p3-3',true)">A) Neither</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p3-3',false)">B) Local max</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p3-3',false)">C) Local min</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p3-3',false)">D) Both a local max and a local min</button>

</div>

<p><span class="label">Explanation:</span> f'(x) = 4x²(x − 3). The factor x² is non-negative on both sides of 0, so the sign of f' is controlled entirely by (x − 3), which stays negative on both sides near x = 0. Since f' doesn't change sign, x = 0 yields neither a max nor a min.</p>

</div>

<p class="prompt">4. Find any inflection points of f(x) = x³ − 6x² + 9x + 1.</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p3-4',false)">D) There are no inflection points</button>

</div>

<p><span class="label">Explanation:</span> f''(x) = 6x − 12, which is 0 at x = 2 and changes sign (− to +) there, confirming an inflection point.</p>

</div>

<p class="prompt">5. A curve is concave upward on an interval. What does this tell us about f''(x) there?</p>

</div>

<p><span class="label">Explanation:</span> Concave upward is defined by a positive second derivative.</p>

</div>

<h2 id="relating">4. Relating f, f', and f'' Graphically</h2>

<p>Reading one graph off another is a core AP skill. This table summarizes the key relationships:</p>

<tr><th>At c or on an interval</th><th>What it tells you</th></tr>

<tr><td>f'(c) = 0, sign + to −</td><td>f has a local maximum at c</td></tr>

<tr><td>f'(c) = 0, sign − to +</td><td>f has a local minimum at c</td></tr>

<tr><td>f'(c) = 0, no sign change</td><td>Neither a max nor a min</td></tr>

<tr><td>f' is a local min (decreasing to increasing)</td><td>f has an inflection point at c</td></tr>

<tr><td>f' is a local max (increasing to decreasing)</td><td>f has an inflection point at c</td></tr>

<tr><td>f' increasing on an interval</td><td>f is concave upward there</td></tr>

<tr><td>f' decreasing on an interval</td><td>f is concave downward there</td></tr>

</table>

<p><strong>Worked Example:</strong> If f'(c) = 0 and f' changes from positive to negative as x increases through c, what does this indicate about f at c?</p>

<p>f has a local maximum at c.</p>

</div>

<p class="prompt">1. If f'(c) is a local minimum of f' (that is, f' changes from decreasing to increasing at c), what does this indicate about f at c?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p4-1',true)">A) f has a point of inflection at c</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-1',false)">B) f has a local minimum at c</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-1',false)">C) f has a local maximum at c</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-1',false)">D) f is undefined at c</button>

</div>

<p><span class="label">Explanation:</span> When f' itself has a local extremum, the concavity of f is switching, this is precisely a point of inflection of f, not an extremum of f.</p>

</div>

<p class="prompt">2. If f' does not change sign as x increases through a critical point c, what can we conclude about f at c?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p4-2',true)">A) f has neither a local maximum nor a local minimum at c</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-2',false)">B) f has a local maximum at c</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-2',false)">C) f has a local minimum at c</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-2',false)">D) f has an inflection point at c</button>

</div>

<p><span class="label">Explanation:</span> Without a sign change, f keeps moving in the same direction through c, ruling out both types of local extremum.</p>

</div>

<p class="prompt">3. On an interval where f' is increasing, what must be true of the graph of f?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p4-3',true)">A) The graph of f is concave upward</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-3',false)">B) The graph of f is concave downward</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-3',false)">C) f is decreasing</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-3',false)">D) f has a local maximum</button>

</div>

<p><span class="label">Explanation:</span> f' increasing means f'' > 0, which is exactly the definition of concave upward.</p>

</div>

<p class="prompt">4. At a point of inflection where f'(c) is a local maximum of f', how does f'' behave as x increases through c?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p4-4',true)">A) f'' changes from positive to negative</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-4',false)">B) f'' changes from negative to positive</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-4',false)">C) f'' remains positive throughout</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p4-4',false)">D) f'' equals zero everywhere near c</button>

</div>

<p><span class="label">Explanation:</span> Since f' switches from increasing to decreasing at its own local maximum, f'' (the derivative of f') switches from positive to negative there.</p>

</div>

<h2 id="global">5. Global Extrema: The Closed Interval Test</h2>

<p>If f is continuous on [a, b], the <strong>Closed Interval Test</strong> (Candidates Test) finds global extrema by evaluating f at every critical point in [a, b] and at both endpoints, then comparing the results. The largest value found is the global maximum; the smallest is the global minimum.</p>

<p><strong>Worked Example:</strong> Find the global max and global min of f(x) = x³ − 3x on [−2, 3].</p>

<p>f'(x) = 3x² − 3 = 3(x − 1)(x + 1); critical points −1 and 1 (both in range).</p>

<p>Global max: 18 at x = 3. Global min: −2 (tied at x = −2 and x = 1).</p>

</div>

<p class="prompt">1. Find the global maximum of f(x) = x³ − 12x on [−3, 3].</p>

</div>

<p><span class="label">Explanation:</span> Critical points at ±2. f(−3) = 9, f(−2) = 16, f(2) = −16, f(3) = −9. The largest is 16, at x = −2.</p>

</div>

<p class="prompt">2. Using the same function on the same interval, find the global minimum.</p>

</div>

<p><span class="label">Explanation:</span> From the same four candidate values, −16 (at x = 2) is the smallest.</p>

</div>

<p class="prompt">3. For f(x) = x² − 4x + 1 on [0, 5], the Closed Interval Test requires evaluating f at which x-values?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-3',false)">D) Every integer from 0 to 5</button>

</div>

<p><span class="label">Explanation:</span> f'(x) = 2x − 4 = 0 gives the critical point x = 2, which falls in [0, 5]. Combined with both endpoints: x = 0, 2, 5.</p>

</div>

<p class="prompt">4. Can a global maximum occur at an endpoint of a closed interval, even if it isn't a critical point?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p5-4',true)">A) Yes, endpoints must always be checked alongside critical points</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-4',false)">B) No, only critical points can yield global extrema</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-4',false)">C) Only if the function is a polynomial</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-4',false)">D) Only if the endpoint is also where f' = 0</button>

</div>

<p><span class="label">Explanation:</span> This is exactly why the Closed Interval Test includes endpoints as candidates, a global extremum can occur there even without a horizontal tangent.</p>

</div>

<h2 id="optimization">6. Optimization</h2>

<p>To maximize or minimize a quantity y in terms of x: express y as a function of a single variable, find critical points using y', classify them, and check endpoints if the domain is restricted.</p>

<p><strong>Worked Example:</strong> A farmer wants to fence a rectangular field using 200 feet of fencing, with one side of a barn serving as one side of the field (no fencing needed there). If x is the width and y is the length, find the value of x that maximizes the area, given 2x + y = 200.</p>

</div>

<p class="prompt">1. A rectangular box with a square base has volume 32 cubic feet, with an open top. If x is the side of the square base and h is the height (so x²h = 32), express the surface area S in terms of x alone.</p>

</div>

<p><span class="label">Explanation:</span> h = 32/x². S = x² + 4xh = x² + 4x(32/x²) = x² + 128/x.</p>

</div>

<p class="prompt">2. Using S(x) = x² + 128/x, find the value of x that minimizes the surface area.</p>

</div>

<p><span class="label">Explanation:</span> S'(x) = 2x − 128/x² = 0, so 2x³ = 128, giving x³ = 64, so x = 4.</p>

</div>

<p class="prompt">3. Using the farmer's fencing problem (A(x) = 200x − 2x²) from the worked example, what is the maximum area?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p6-3',true)">A) 5,000 square feet</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-3',false)">B) 10,000 square feet</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-3',false)">C) 2,500 square feet</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-3',false)">D) 200 square feet</button>

</div>

<p><span class="label">Explanation:</span> A(50) = 200(50) − 2(50)² = 10,000 − 5,000 = 5,000 square feet.</p>

</div>

<p class="prompt">4. In an optimization problem where the domain of the quantity being optimized is a restricted closed interval, what must be checked in addition to critical points?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p6-4',true)">A) The endpoints of the interval</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-4',false)">B) Only the second derivative</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-4',false)">C) Nothing else is needed</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-4',false)">D) Only inflection points</button>

</div>

<p><span class="label">Explanation:</span> Just as with the Closed Interval Test, a restricted domain means the true optimum could occur at a boundary rather than at an interior critical point.</p>

</div>

<h2 id="motion">7. Motion Along a Line</h2>

<p>If a particle's position is s(t), its <strong>velocity</strong> is v(t) = s'(t) and its <strong>acceleration</strong> is a(t) = v'(t) = s''(t). The particle's <strong>speed</strong> is |v(t)|. Speed is increasing when v and a share the same sign, and decreasing when they have opposite signs.</p>

<p><strong>Worked Example:</strong> A particle moves with position s(t) = t³ − 6t² + 9t, t ≥ 0. Find the velocity and determine when the particle moves right.</p>

<p class="step-math">v > 0 (moving right) for t < 1 and t > 3; v < 0 (moving left) for 1 < t < 3</p>

</div>

<p class="prompt">1. A particle's position is s(t) = t³ − 9t² + 24t. Find the velocity function v(t).</p>

</div>

<p><span class="label">Explanation:</span> v(t) = s'(t) = 3t² − 18t + 24.</p>

</div>

<p class="prompt">2. Using v(t) = 3t² − 18t + 24 = 3(t − 2)(t − 4), for what t ≥ 0 is the particle moving to the right?</p>

</div>

<p><span class="label">Explanation:</span> v > 0 outside the roots 2 and 4, so the particle moves right for 0 ≤ t < 2 and for t > 4.</p>

</div>

<p class="prompt">3. A particle has position s(t) = t² − 6t + 5, so v(t) = 2t − 6 and a(t) = 2 (constant). For what t is the particle's speed increasing?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-3',false)">D) Never</button>

</div>

<p><span class="label">Explanation:</span> Speed increases when v and a share the same sign. Since a is always positive, speed increases wherever v > 0, that is, t > 3.</p>

</div>

<p class="prompt">4. A particle moves with position s(t) = t³ − 3t. What is its acceleration at t = 2?</p>

</div>

<p><span class="label">Explanation:</span> v(t) = 3t² − 3, so a(t) = 6t. a(2) = 12.</p>

</div>

<p class="prompt">5. If a particle's velocity and acceleration have opposite signs at some instant, what is happening to its speed?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p7-5',true)">A) The speed is decreasing</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-5',false)">B) The speed is increasing</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-5',false)">C) The particle has stopped</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-5',false)">D) The particle has reversed direction at that exact instant</button>

</div>

<p><span class="label">Explanation:</span> Opposite signs of v and a mean the particle is decelerating, its speed is decreasing.</p>

</div>

<h2 id="linear-approx">8. Local Linear Approximation</h2>

<p>For values of x close to a, the <strong>tangent-line (local linear) approximation</strong> is:</p>

<p><strong>Worked Example:</strong> Find the tangent-line approximation for f(x) = √x at a = 9, and use it to estimate √9.3.</p>

</div>

<p class="prompt">1. Find the tangent-line approximation for f(x) = x³ at a = 2.</p>

</div>

<p><span class="label">Explanation:</span> f(2) = 8. f'(x) = 3x², so f'(2) = 12. Approximation: 8 + 12(x − 2).</p>

</div>

<p class="prompt">2. Using f(x) ≈ 8 + 12(x − 2) for f(x) = x³, estimate the value of (2.1)³.</p>

</div>

<p><span class="label">Explanation:</span> 8 + 12(0.1) = 8 + 1.2 = 9.2. (The true value, 9.261, is close, confirming the approximation is reasonable for x near a.)</p>

</div>

<p class="prompt">3. Using the local linearization (1 + x)<sup>k</sup> ≈ 1 + kx for x near 0, estimate (1.02)⁵.</p>

</div>

<p><span class="label">Explanation:</span> With k = 5 and x = 0.02: 1 + 5(0.02) = 1.1.</p>

</div>

<p class="prompt">4. If a function f is concave up on an interval containing a, is its tangent-line approximation at a an overestimate or underestimate of f(x) for x near a?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p8-4',true)">A) An underestimate</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p8-4',false)">B) An overestimate</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p8-4',false)">C) Exactly equal to f(x)</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p8-4',false)">D) Cannot be determined</button>

</div>

<p><span class="label">Explanation:</span> A concave-up curve always lies above its tangent lines, so the tangent-line approximation underestimates the true value.</p>

</div>

<h2 id="related-rates">9. Related Rates</h2>

<p>When several quantities that are functions of time t are related by an equation, differentiating both sides with respect to t (using the Chain Rule) relates their rates of change.</p>

<p><strong>Worked Example:</strong> A ladder 10 feet long leans against a wall; the bottom slides away at 2 ft/sec. Find how fast the top is sliding down when the bottom is 6 feet from the wall.</p>

<p>x² + y² = 100. Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0.</p>

<p>At x = 6, y = 8 (since 6² + 8² = 100), and dx/dt = 2:</p>

</div>

<p class="prompt">1. A ladder 13 feet long leans against a wall, with the bottom sliding away at 3 ft/sec. Using x² + y² = 169, find dy/dt when x = 5 (so y = 12).</p>

</div>

<p><span class="label">Explanation:</span> 2(5)(3) + 2(12)(dy/dt) = 0, so 30 + 24(dy/dt) = 0, giving dy/dt = −30/24 = −1.25 ft/sec.</p>

</div>

<p class="prompt">2. A spherical balloon's volume is V = <span class="frac"><span class="num">4</span><span class="den">3</span></span>πr³. Find dV/dr.</p>

</div>

<p><span class="label">Explanation:</span> Using the Power Rule: <span class="frac"><span class="num">4</span><span class="den">3</span></span>π · 3r² = 4πr².</p>

</div>

<p class="prompt">3. Using dV/dt = (dV/dr)(dr/dt) and dV/dr = 4πr², find dr/dt when r = 5 inches and dV/dt = 100 cubic inches/second.</p>

</div>

<p><span class="label">Explanation:</span> 100 = 4π(25)(dr/dt) = 100π(dr/dt), so dr/dt = 100/(100π) = 1/π.</p>

</div>

<p class="prompt">4. Two cars leave an intersection at the same time, one heading north at 40 mph and the other east at 30 mph. Using z² = x² + y², find dz/dt one hour later (when x = 30, y = 40, z = 50).</p>

</div>

<p><span class="label">Explanation:</span> z(dz/dt) = x(dx/dt) + y(dy/dt) = 30(30) + 40(40) = 900 + 1,600 = 2,500. dz/dt = 2,500/50 = 50 mph.</p>

</div>

<p class="prompt">5. Why is it essential to differentiate the relating equation BEFORE substituting specific numerical values in a related rates problem?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p9-5',true)">A) Substituting numbers first would incorrectly treat variables as constants during differentiation</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p9-5',false)">B) It doesn't actually matter; either order works fine</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p9-5',false)">C) Substituting first makes the arithmetic simpler with no downside</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p9-5',false)">D) The Chain Rule requires numerical values to be present first</button>

</div>

<p><span class="label">Explanation:</span> Plugging in a specific numerical value for a variable before differentiating locks it in as a constant, erasing the very relationship between the rates that the problem is asking about.</p>

</div>

<h2 id="mistakes">Common Mistakes to Avoid</h2>

<li><strong>Forgetting that a critical point can come from an undefined derivative, not just f'(x) = 0.</strong> Always check for x-values where f' fails to exist, not only where it equals zero.</li>

<li><strong>Confusing a local extremum with a global extremum.</strong> A local max or min only needs to beat its immediate neighbors; a global extremum must beat every value in the entire interval, which is why the Closed Interval Test also checks the endpoints.</li>

<li><strong>Mixing up which sign change of f' signals a max versus a min.</strong> Plus-to-minus is a local maximum; minus-to-plus is a local minimum, it's easy to flip these under time pressure.</li>

<li><strong>Treating "speed is increasing" and "velocity is increasing" as the same thing.</strong> Speed increases only when velocity and acceleration share the same sign; velocity alone increasing whenever acceleration is positive says nothing about speed if velocity is negative.</li>

<li><strong>Substituting known numerical values before differentiating in a related rates problem.</strong> Every variable that changes with time must stay a variable until after you differentiate, only then do you plug in the specific values for that instant.</li>

</ul>

<h2 id="faq">Frequently Asked Questions</h2>

<h3>When should I use the First Derivative Test instead of the Second Derivative Test?</h3>

<p>Use the Second Derivative Test as a quick first check, it's usually faster. But whenever f''(c) = 0 at a critical point, the test is inconclusive, and you must fall back on the First Derivative Test's sign analysis instead.</p>

</div>

<h3>How is an inflection point different from a local extremum?</h3>

<p>A local extremum is where f itself changes from increasing to decreasing (or vice versa), found from the sign of f'. An inflection point is where the concavity of f changes, found from the sign of f''. They describe different features of the same curve.</p>

</div>

<h3>What's the general strategy for setting up an optimization or related rates word problem?</h3>

<p>Draw a diagram if possible, identify which quantity is changing and which equation relates the variables involved, then differentiate (with respect to the input variable for optimization, or with respect to time for related rates) before substituting any given numerical values.</p>

</div>

<h3>Where can I practice more problems like these?</h3>

<p>The <a href="https://theschoolofmathematics.com/quiz/ap-calculus-ab-applications-of-differential-calculus-quiz-1">Applications of Differential Calculus quizzes</a> in the AP Calculus AB Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested throughout the course.</p>

</div>

<a class="cta-btn cta-primary" href="https://theschoolofmathematics.com/quiz/ap-calculus-ab-applications-of-differential-calculus-quiz-1">Practice Applications of Differential Calculus Free</a>

<a class="cta-btn cta-secondary" href="https://theschoolofmathematics.com/quiz/course/AP-Calculus-AB-QBank">Explore the Full AP Calculus AB Qbank</a>

</div>

</footer>

</div>

function checkAnswer(btn, problemId, isCorrect) {

var problem = document.getElementById(problemId);

var buttons = problem.querySelectorAll('.option-btn');

buttons.forEach(function(b) { b.disabled = true; });

if (isCorrect) {

btn.classList.add('correct');

} else {

btn.classList.add('incorrect');

var correctBtn = problem.querySelector('.option-btn[data-correct="true"]');

if (correctBtn) { correctBtn.classList.add('correct'); }

}

var exp = problem.querySelector('.explanation');

if (exp) { exp.classList.add('show'); }

}

</script>

</body>

</html>