IB AA: Differentiation Applications

In the study of calculus, learning to find derivatives is only half the battle. The true power of differentiation lies in its ability to solve tangible problems—from predicting the path of a rocket to designing a cost-efficient container. This unit on applications transforms abstract rules into indispensable tools for modeling and understanding the behavior of changing quantities in mathematics, physics, economics, and beyond.

Stationary Points and Curve Sketching

The first major application of the derivative is analyzing the shape of a graph. A stationary point (or critical point) occurs where the first derivative of a function is zero, $f^{\prime }(x)=0$. These points represent potential local maxima, local minima, or points of inflection. To classify them, we use the second derivative test. If $f^{\prime \prime }(x)>0$ at the stationary point, the curve is concave up, indicating a local minimum. If $f^{\prime \prime }(x)<0$, the curve is concave down, indicating a local maximum. If $f^{\prime \prime }(x)=0$, the test is inconclusive, and you must examine the sign change of $f^{\prime }(x)$ around the point.

This analysis is the cornerstone of systematic curve sketching. Beyond stationary points, you use the first derivative to determine intervals where the function is increasing ($f^{\prime }(x)>0$) or decreasing ($f^{\prime }(x)<0$). The second derivative reveals intervals of concavity (concave up: $f^{\prime \prime }(x)>0$; concave down: $f^{\prime \prime }(x)<0$). Points where the concavity changes are called points of inflection, found where $f^{\prime \prime }(x)=0$ and changes sign. Combining this information with intercepts and asymptotic behavior allows you to construct an accurate sketch that tells the story of the function's behavior.

Example: For $f(x)=x^3-3x+2$, find and classify stationary points.

1. Find $f^{\prime }(x)=3x^2-3$. Set to zero: $3x^2-3=0\Rightarrow x=\pm 1$.
2. Find $f^{\prime \prime }(x)=6x$. Classify: At $x=1$, $f^{\prime \prime }(1)=6>0$ ⇒ local minimum. At $x=-1$, $f^{\prime \prime }(-1)=-6<0$ ⇒ local maximum.
3. Find points of inflection: $f^{\prime \prime }(x)=6x=0\Rightarrow x=0$. Check sign change: $f^{\prime \prime }(x)$ changes from negative to positive at $x=0$, confirming a point of inflection.

Optimization Problems in Context

Optimization is the process of finding the best solution, often a maximum or minimum value, within a given context. These are word problems where you must first construct the function to be optimized. The general strategy is:

1. Identify the quantity to be optimized (e.g., area, volume, cost, profit) and express it as a function of one variable. This often requires using given constraints to relate variables.
2. Find the derivative of this function and locate stationary points by setting the derivative to zero.
3. Verify that the stationary point gives the desired maximum or minimum, typically using the second derivative test or analyzing endpoints of a logical domain.
4. Answer the question in context, with units.

Example: "A farmer has 100m of fencing to create a rectangular pen against a straight river (so fencing is only needed on three sides). Find the dimensions that maximize the enclosed area."

1. Let the side perpendicular to the river be $x$ m. The side parallel to the river is then $100-2x$ m. The area $A=x(100-2x)=100x-2x^2$.
2. $A^{\prime }(x)=100-4x$. Set to zero: $100-4x=0\Rightarrow x=25$.
3. $A^{\prime \prime }(x)=-4$, which is always negative, confirming a maximum at $x=25$.
4. Dimensions: The perpendicular sides are 25m, and the parallel side is $100-2(25)=50$m. The maximum area is $25\times 50=1250{\text{ m}}^2$.

Related Rates of Change

Related rates problems involve finding the rate at which one quantity changes with respect to time, given the rate at which a related quantity changes. The key is that the quantities are linked by an equation (often geometric, like the Pythagorean theorem or volume formulas). You differentiate this equation implicitly with respect to time, $t$, which introduces derivatives like $dx/dt$ (the rate of change of $x$). Then, you substitute known values and solve for the unknown rate.

The standard procedure is:

1. Draw a diagram (if applicable) and assign variables.
2. Write an equation relating the variables.
3. Differentiate implicitly with respect to time $t$.
4. Substitute all known values and rates at the specific instant in question.
5. Solve for the desired unknown rate.

Example: "A 5m ladder leans against a wall. The foot of the ladder slides away from the wall at 0.5 m/s. How fast is the top sliding down the wall when the foot is 3m from the wall?"

1. Let $x$ be the distance from the wall to the ladder's foot, and $y$ be the height of the ladder on the wall. By Pythagoras: $x^2+y^2=5^2=25$.
2. Differentiate w.r.t. $t$: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$.
3. Known: $x=3$ m, $\frac{dx}{dt}=0.5$ m/s. Find $y$ when $x=3$: $3^2+y^2=25\Rightarrow y=4$ m.
4. Substitute: $2(3)(0.5)+2(4)\frac{dy}{dt}=0\Rightarrow 3+8\frac{dy}{dt}=0$.
5. Solve: $\frac{dy}{dt}=-\frac{3}{8}$ m/s. The negative signifies the height $y$ is decreasing.

Tangents, Normals, and Kinematics Applications

The derivative provides precise tools for analyzing lines and motion. The tangent line to a curve at point $P(x_1,y_1)$ has a slope equal to $f^{\prime }(x_1)$. Its equation is found using $y-y_1=f^{\prime }(x_1)(x-x_1)$. The normal line is perpendicular to the tangent at the same point; its slope is the negative reciprocal, $m_{normal}=-\frac{1}{f^{\prime }(x_1)}$, provided $f^{\prime }(x_1)\ne 0$.

In kinematics (the study of motion), these concepts take on physical meaning for an object moving along a straight line. If $s(t)$ represents the displacement (position) from a fixed origin, then:

• The velocity is the rate of change of displacement: $v(t)=s^{\prime }(t)$.
• The acceleration is the rate of change of velocity: $a(t)=v^{\prime }(t)=s^{\prime \prime }(t)$.

You can analyze motion by applying derivative rules: velocity is zero at an instantaneous rest, and acceleration indicates changes in speed. Integration (the reverse process) would be used to find displacement from velocity, or velocity from acceleration.

Example: A particle moves such that its displacement is $s(t)=t^3-6t^2+9t$ meters.

• Velocity: $v(t)=s^{\prime }(t)=3t^2-12t+9$. Set $v(t)=0$ to find when at rest: $3(t^2-4t+3)=0\Rightarrow t=1$s and $t=3$s.
• Acceleration: $a(t)=v^{\prime }(t)=6t-12$.
• At $t=1$, $a(1)=-6{\text{ m/s}}^2$ (negative acceleration while at rest, indicating it's about to move backward).

Common Pitfalls

1. Misapplying the Second Derivative Test: A zero second derivative ($f^{\prime \prime }(x)=0$) does not guarantee a point of inflection; the concavity must change on either side of the point. Similarly, if $f^{\prime \prime }(x)=0$, you must resort to the first derivative test to classify the stationary point.
2. Forgetting the Domain in Optimization: Always consider the practical domain of your variable. The optimal value from your derivative work might be a local extremum, but the absolute maximum or minimum could occur at an endpoint of the domain (e.g., a negative length is mathematically possible from an equation but physically nonsense).
3. Substituting Too Early in Related Rates: A frequent error is substituting numerical values for variables before differentiating. You must differentiate the general equation first to get a relationship between the rates, then substitute the specific values for that instant. Substituting early "freezes" the variables and loses the relationship between their rates.
4. Confusing Velocity and Speed: In kinematics, velocity, $v(t)$, is a signed quantity (can be positive or negative). Its sign indicates direction. Speed is the magnitude of velocity, $|v(t)|$, and is always non-negative. An object can have zero velocity (momentarily at rest) but non-zero acceleration.

Summary

• The first derivative, $f^{\prime }(x)$, identifies stationary points and intervals of increase/decrease, while the second derivative, $f^{\prime \prime }(x)$, determines concavity and helps classify those points for comprehensive curve sketching.
• Optimization problems require translating a real-world scenario into a single-variable function, using derivatives to find its optimum, and justifying that it is a maximum or minimum.
• Related rates problems are solved by implicitly differentiating an equation linking two or more variables with respect to time, then substituting known instantaneous values.
• The derivative at a point gives the slope of the tangent line; the normal is perpendicular to it. In motion, the first and second derivatives of displacement $s(t)$ yield velocity $v(t)$ and acceleration $a(t)$, respectively.
• Always be mindful of the context, domain, and the correct order of operations (differentiate before substituting) to avoid common analytical mistakes.