Related rates and extreme values

This week the derivative goes into the “real world” for the first time. There are two different question types, both word problems, and both come up very often. In a related rates question, connected quantities change with time; they give you the rate of one and ask for the rate of another. In an extrema question, you find the largest and smallest value of a function on an interval. Both have an algorithm; whoever knows the algorithm does not miss the question.

1. In related rates, substitute the numbers last. First differentiate the general equation.
2. The time variable is hidden. Every quantity is a function of $t$, and when you differentiate the chain rule is automatically in play.
3. The rate of a decreasing quantity is written as negative.
4. Critical point: $f'=0$ or $f'$ undefined.
5. On a closed interval, test the endpoints too. The largest value often occurs at an endpoint.

Related rates

If the radius of a balloon grows, its volume grows too. The relation $V=\frac43\pi r^3$ also gives a relation between the rates of change of the two quantities. To see this, differentiate both sides with respect to $t$:

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

Since $r$ is a function of $t$, the derivative of $r^3$ is $3r^2\dfrac{dr}{dt}$ (chain rule). This formula is true at every moment. When you substitute the numbers at a specific moment, the unknown rate comes out.

The five-step method

1. Draw a figure. Give a letter to every changing quantity (x, y, r, h, θ).
2. Write the given and the asked as derivatives:  dx/dt = 0.5,  dy/dt = ?
3. Set up the equation linking the variables (Pythagoras, volume, similar triangles, tan).
4. Differentiate the equation with respect to t.
5. Substitute the values at that moment (computing them from the equation if needed), and solve.

Example 1 (area). The radius of a circular oil slick grows at $2$ m/s. How fast does the area grow when the radius is $30$ m?

$A=\pi r^2 \Rightarrow \dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}=2\pi(30)(2)=120\pi\approx377$ m²/s.

Example 2 (Pythagoras). A plane flies horizontally at an altitude of $3$ km at $500$ km/h and has just passed directly over a radar station. How fast is the distance between the radar and the plane increasing when it is $5$ km?

x: horizontal distance along the ground, D: radar-plane distance, altitude constant 3
D^2 = x^2 + 9                  (the constant 3 can be written before differentiating, it doesn't change)
2D D' = 2x x'   ->   D' = x x' / D
at that moment: D = 5  ->  x = sqrt(25 - 9) = 4
D' = 4 · 500 / 5 = 400 km/h

Example 3 (angle). A camera is $30$ m from a launch pad. A rocket rises vertically at $10$ m/s. How fast is the camera’s angle of elevation changing when the rocket is at a height of $40$ m?

$\tan\theta=\dfrac{h}{30}$. Derivative: $\sec^2\theta\cdot\theta'=\dfrac{h'}{30}$. At that moment $\tan\theta=\tfrac43$, $\sec^2\theta=1+\tfrac{16}{9}=\tfrac{25}{9}$. $\theta'=\dfrac{10}{30}\cdot\dfrac{9}{25}=\dfrac{3}{25}=0.12$ rad/s.

PITFALL · substituting the number before differentiating. In the ladder problem, if you put $x=5$ before differentiating, you get $25+y^2=169$, the derivative is $2yy'=0$, and the result is nonsense. Substitute the instantaneous value of a changing quantity only after differentiating. Constants that do not change (the ladder length $13$, the cone’s $r/h$ ratio, the plane’s altitude) can be substituted beforehand.

Cones and similar triangles

In cone problems $V=\frac13\pi r^2h$ has two variables. The water depth and the surface radius are linked by similar triangles: $\dfrac rh=\dfrac{R}{H}$ (the tank’s dimensions). Write $r$ in terms of $h$, reduce $V$ to a single variable, then differentiate. That way you don’t have to deal with a second unknown rate like $\dfrac{dr}{dt}$.

Extreme values

• If $f(c)\ge f(x)$ for every $x$ in the domain, $f(c)$ is the absolute maximum.
• If $f(c)\ge f(x)$ for every $x$ near $c$, it is a local maximum. For a minimum the inequality is reversed.

Extreme Value Theorem: if $f$ is continuous on the closed interval $[a,b]$, an absolute maximum and minimum definitely exist. On an open interval or for a discontinuous function there is no guarantee: $\tfrac1x$ attains neither a max nor a min on $(0,1)$.

Fermat’s theorem and critical points

If $f$ has a local extremum at an interior point $c$ and $f'(c)$ exists, then $f'(c)=0$. At a peak or a valley the tangent is horizontal.

So the candidates for extrema are the critical numbers: the values $c$ where $f$ is defined and $f'(c)=0$ or $f'(c)$ is undefined.

The converse is not true: for $f(x)=x^3$, $f'(0)=0$ but there is no extremum at $0$, the graph passes straight through and keeps increasing. A critical point is a candidate, not a guarantee.

Example 4 (a critical point at a corner). $f(x)=x^{1/3}(4-x)=4x^{1/3}-x^{4/3}$.

$f'(x)=\frac43x^{-2/3}-\frac43x^{1/3}=\frac43x^{-2/3}(1-x)=\frac{4(1-x)}{3\sqrt[3]{x^2}}$

$f'=0$: $x=1$. $f'$ undefined: $x=0$, and $f(0)=0$ is defined. The critical points are $0$ and $1$. A root left in the denominator with fractional powers always produces a critical point; this is the candidate most often forgotten on exams.

The closed interval method

To find the absolute extrema of $f$ continuous on $[a,b]$:

1. Compute f'(x), find the critical points inside (a,b).
2. Evaluate f at the critical points and at the endpoints a, b.
3. The largest value is the absolute max, the smallest value is the absolute min.

No need for the second derivative test or a sign table. Just a table of values.

Example 5. $f(x)=2x^3-3x^2-12x+1$, $[-2,3]$.

$f'(x)=6x^2-6x-12=6(x-2)(x+1)$. The critical points are $-1$ and $2$, both in the interval.

The absolute maximum is $8$ (at $x=-1$), the absolute minimum is $-19$ (at $x=2$). In the answer write both the value and the location; if the question says “values” give the value, if it says “where” give the location.

PITFALL · a critical point outside the interval. When working on the interval $[0,1]$, if the critical point $x=2$ comes out, do not put it in the table. A value outside the interval does not determine the max/min on the interval.

PITFALL · forgetting an endpoint. If Example 5 had been on $[-2,4]$, $f(4)=128-48-48+1=33$ would be the largest value. The max is not at a critical point just because the tangent is horizontal there.

Common mistakes