EduNinja

4.1 Interpreting the Meaning of the Derivative in Context

If $y=f(x)$ , then $f'(a)$ is the instantaneous rate at which the output $y$ is changing with respect to the input $x$ when $x=a$ . Its sign gives the direction of change, and its magnitude gives how quickly the output is changing at that instant.

\text{units of }f'(x)=\frac{\text{units of }f}{\text{units of }x}

A complete interpretation identifies: (1) the input value or instant, (2) the output quantity that is changing, (3) the independent variable it changes with respect to, (4) whether it is increasing or decreasing, and (5) the rate units.

Suppose $T(t)$ is the temperature of a drink in degrees Celsius, $t$ minutes after it is poured, and $T'(5)=-0.8$ . At $t=5$ minutes, the drink's temperature is decreasing at an instantaneous rate of $0.8$ degrees Celsius per minute. The negative sign indicates decreasing temperature; the units are $^\circ\mathrm{C}/\mathrm{min}$ .

$T'(5)=-0.8$ does not mean the temperature has changed by $-0.8^\circ\mathrm{C}$ in total, nor that it will keep changing at that rate for a long interval. It describes the local rate at the single instant $t=5$ .

4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration

For straight-line motion, let $s(t)$ be position at time $t$ . Velocity is the instantaneous rate of change of position, acceleration is the instantaneous rate of change of velocity, and speed is the magnitude of velocity.

v(t)=s'(t),\qquad \text{speed}=|v(t)|,\qquad a(t)=v'(t)=s''(t)

$v(t)>0$ : motion in the positive direction; $v(t)<0$ : motion in the negative direction.\n• Velocity and acceleration with the same sign: speed is increasing.\n• Velocity and acceleration with opposite signs: speed is decreasing.\nIf position is measured in meters and time in seconds, velocity uses $\mathrm{m/s}$ and acceleration uses $\mathrm{m/s^2}$ .

Suppose $s(t)=t^3-3t^2+2t$ meters. Then $v(t)=3t^2-6t+2$ and $a(t)=6t-6$ . At $t=2$ seconds, $s(2)=0\ \mathrm m,\qquad v(2)=2\ \mathrm{m/s},\qquad |v(2)|=2\ \mathrm{m/s},\qquad a(2)=6\ \mathrm{m/s^2}.$ The particle is moving in the positive direction and speeding up because $v(2)$ and $a(2)$ are both positive.

Velocity can be negative because it includes direction; speed cannot be negative. When $v(t)=0$ , the particle is momentarily at rest, and the same-sign/opposite-sign test alone does not classify speeding up at that instant—inspect velocity immediately around the time instead.

4.3 Rates of Change in Applied Contexts Other Than Motion

For a contextual model $Q(x)$ , the derivative $Q'(a)$ measures the instantaneous change in the output quantity $Q$ with respect to the input $x$ when $x=a$ . Its units are output units per input unit.

State what the input and output represent, including units.\n2. Differentiate the given model with respect to its input.\n3. Evaluate the derivative at the requested input.\n4. Interpret the sign, magnitude, instant, and quotient units in one contextual sentence.

As an illustrative model, let $P(t)=1000+50t-2t^2$ represent a population $t$ years after observation begins. Then $P'(t)=50-4t\quad\text{people per year},$ so $P'(6)=50-4(6)=26\quad\text{people per year}.$ At $t=6$ years, the model's population is increasing at an instantaneous rate of $26$ people per year.

A positive derivative means the modeled output is increasing at that input; a negative derivative means it is decreasing; a zero derivative means its instantaneous rate of change is zero. The sign belongs to the rate, not automatically to the output value itself.

$P'(6)=26$ is a local rate, not a claim that exactly 26 people are added every future year. The interpretation is also only as valid as the supplied model and the input interval for which that model is intended.

4.4 Introduction to Related Rates

In a related-rates problem, two or more quantities change while remaining linked by an equation. Treat each changing quantity as a function of the same independent variable—usually time $t$ —and differentiate the entire relation with respect to that variable.

Define each changing quantity and its rate.\n2. Write one equation relating the quantities.\n3. Differentiate both sides with respect to the common variable.\n4. Apply the chain rule to every changing quantity; use product or quotient rule when the equation's structure requires it.\n5. Substitute values from the specified instant, solve for the requested rate, and attach units.

A=\pi r^2\quad\Longrightarrow\quad \frac{dA}{dt}=2\pi r\frac{dr}{dt}

A circular region has radius $r=3$ meters at an instant when $dr/dt=0.5$ meter per second. Substituting into the differentiated relation gives $\frac{dA}{dt}=2\pi(3)(0.5)=3\pi\ \mathrm{m^2/s}.$ At that instant, the area is increasing at $3\pi$ square meters per second.

Do not replace a changing quantity with its instantaneous number before differentiating; doing so can incorrectly turn it into a constant. Also, $d(r^2)/dt=2r\,dr/dt$ , not merely $2r$ : the rate factor is the chain-rule record that $r$ changes with time.

4.5 Solving Related Rates Problems

To solve a related-rates problem, connect the changing quantities with one equation, differentiate with respect to their common variable, and use known values from the same instant to determine the requested rate.

A 10-foot ladder rests against a wall. Let $x$ be the bottom's distance from the wall and $y$ the top's height, both in feet. The fixed length gives $x^2+y^2=100$ . Suppose the bottom moves away at $dx/dt=2$ feet per second. Find the top's rate when $x=6$ .

2x\frac{dx}{dt}+2y\frac{dy}{dt}=0

At the specified instant, $6^2+y^2=100$ , so $y=8$ feet. Substitute $x=6$ , $y=8$ , and $dx/dt=2$ : $2(6)(2)+2(8)\frac{dy}{dt}=0,$ hence $\frac{dy}{dt}=-\frac{24}{16}=-1.5\ \mathrm{ft/s}.$ When the bottom is 6 feet from the wall, the top is moving downward at 1.5 feet per second.

The negative result is not an error: with upward chosen as positive, it means the height is decreasing. Use positions and rates from the same instant, and state both direction and units; writing only $-1.5$ leaves the contextual answer incomplete.

4.6 Approximating Values of a Function Using Local Linearity and Linearization

Near $x=a$ , a differentiable function is approximately linear, so its tangent line can stand in for the curve. The linearization uses the known value $f(a)$ and the tangent slope $f'(a)$ .

L(x)=f(a)+f'(a)(x-a),\qquad f(x)\approx L(x)\text{ for }x\text{ near }a

Choose a nearby base point $a$ where $f(a)$ and $f'(a)$ are easy to evaluate.\n2. Build $L(x)=f(a)+f'(a)(x-a)$ .\n3. Substitute the target input into $L$ .\n4. Use local concavity to classify the estimate.

To approximate $\sqrt{4.1}$ , let $f(x)=\sqrt{x}$ and use $a=4$ . Since $f(4)=2$ and $f'(4)=1/4$ , $L(x)=2+\frac14(x-4).$ Therefore $\sqrt{4.1}\approx L(4.1)=2+\frac14(0.1)=2.025.$

Where the curve is concave up, it lies locally above its tangent line, so $L(x)$ is an underestimate. Where the curve is concave down, it lies locally below its tangent line, so $L(x)$ is an overestimate. Because $\sqrt{x}$ is concave down near $4$ , $2.025$ is an overestimate.

The approximation is not an identity: $f(x)$ and $L(x)$ agree in value and slope at $x=a$ , but their difference can grow as $x$ moves farther from the point of tangency.

4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms

A quotient approaching $0/0$ or $\infty/\infty$ is indeterminate: those symbols do not give the limit. L’Hospital’s Rule may replace the quotient with the quotient of its derivatives when the relevant derivatives exist and the new quotient has a limit.

\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}

Substitute or analyze the numerator and denominator limits.\n2. Continue only if the quotient has form $0/0$ or $\infty/\infty$ .\n3. Differentiate the numerator and denominator separately.\n4. Evaluate the new quotient limit; if it is still one of the allowed indeterminate forms, the rule may be applied again when its conditions still hold.

For $\lim_{x\to0}\frac{e^x-1}{x},$ direct substitution gives $0/0$ , so L’Hospital’s Rule applies: $\lim_{x\to0}\frac{e^x}{1}=1.$

This is not the quotient rule: use $f'/g'$ , not $(f/g)'$ . Do not apply L’Hospital’s Rule unless the original limit is an allowed quotient form. Other indeterminate forms such as $\infty-\infty$ are outside the assessed AP Calculus AB/BC scope stated here.

Unit 4: Contextual Applications of Differentiation

Exam analysis

Most tested topics

4.1 Interpreting the Meaning of the Derivative in Context

Read a Derivative as a Contextual Rate

4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration

Differentiate Position to Describe Motion

4.3 Rates of Change in Applied Contexts Other Than Motion

Turn a Contextual Model into a Rate

4.4 Introduction to Related Rates

Differentiate the Relationship Between Changing Quantities

4.5 Solving Related Rates Problems

Use a Known Rate to Find an Unknown Rate

4.6 Approximating Values of a Function Using Local Linearity and Linearization

Approximate a Curve with Its Tangent Line

4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms

Check the Form Before Using L’Hospital’s Rule