Contextual Applications of Differentiation

If $y=f(x)$ describes a real quantity, then $f'(x)=\dfrac{dy}{dx}$ is the instantaneous rate of change of $y$ with respect to $x$.

• Value at a point: $f'(a)$ is the rate of change of $y$ at the instant $x=a$.
• Units: the units of $f'$ are always (units of $y$) per (units of $x$). If $V$ is in liters and $t$ in seconds, then $\dfrac{dV}{dt}$ is in liters per second.
• Sign: $f'(a)>0$ means $y$ is increasing at $x=a$; $f'(a)<0$ means decreasing.
• Second derivative: $f''(a)$ tells whether the rate itself is increasing or decreasing (concavity).

For a particle moving along a line with position $s(t)$:

• Velocity $v(t)=s'(t)$ (has a sign: direction of motion).
• Speed $=|v(t)|$ (always $\ge 0$; “how fast”).
• Acceleration $a(t)=v'(t)=s''(t)$.
• At rest when $v(t)=0$; the particle may change direction there.
• Speeding up when $v$ and $a$ have the same sign; slowing down when they have opposite signs.
• Displacement on $[a,b]$ is $s(b)-s(a)$ (net change). Total distance is found by splitting at every time $v=0$ and adding the absolute changes.

Related-rates problems link the rates of change of two or more quantities that vary with time.

• Relate: write an equation connecting the variables (geometry, similar triangles, Pythagorean theorem, area/volume formulas). Eliminate variables you have no rate for.
• Differentiate both sides implicitly with respect to $t$ (chain rule on every variable).
• Substitute the known instantaneous values last, then solve for the unknown rate.

Never plug in the moving numbers before differentiating.

Near $x=a$ a differentiable function is close to its tangent line. The linearization of $f$ at $a$ is

Differentials: for a small change $dx$, the estimated change in $y$ is $dy=f'(x)\,dx$, so $\Delta y\approx dy$.

Over/under-estimate (use concavity):

• If $f''>0$ on the interval (concave up), the tangent line lies below the curve $\Rightarrow$ $L(x)$ is an underestimate.
• If $f''<0$ (concave down), the tangent line lies above the curve $\Rightarrow$ $L(x)$ is an overestimate.

If $\displaystyle\lim_{x\to c}\dfrac{f(x)}{g(x)}$ gives the indeterminate form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, then

provided the right-hand limit exists (differentiate top and bottom separately — not the quotient rule). Repeat if still indeterminate.

Other forms must be rewritten first:

• $0\cdot\infty$: rewrite as $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ by moving a factor to the denominator.
• $\infty-\infty$: combine into a single fraction.
• $1^\infty,\ 0^0,\ \infty^0$: set $y=(\text{expression})$, take $\ln$, find $\lim \ln y$, then exponentiate.

Warning: only apply the rule to a genuine $\tfrac00$ or $\tfrac\infty\infty$. Check the form each time.

To approximate a root of $f(x)=0$, start with a guess $x_0$ and iterate

Each step follows the tangent line at $x_n$ down to the $x$-axis.

Example ($\sqrt2$ as a root of $f(x)=x^2-2$): $f'(x)=2x$. With $x_0=1.5$,

This converges rapidly ($\sqrt2=1.41421\ldots$). It can fail if $f'(x_n)=0$ or the guess is poor.

The error $E(x)=f(x)-L(x)$ of a tangent-line estimate is controlled by concavity: its sign is opposite the estimate's bias (concave up $\Rightarrow L$ under $\Rightarrow E>0$). The farther $x$ is from $a$ and the larger $|f''|$, the bigger the error — $E$ grows roughly like $\tfrac12 f''(a)(x-a)^2$. So a good estimate needs $x$ close to $a$ and small curvature. Always state “over” or “under” by checking the sign of $f''$ between $a$ and $x$.

For $1^\infty,\ 0^0,\ \infty^0$, let $y$ equal the expression and take $\ln$.

Example ($1^\infty$): $\displaystyle\lim_{x\to\infty}\Big(1+\tfrac{3}{x}\Big)^{x}$. Let $y=\big(1+\tfrac3x\big)^x$, so

Therefore $\displaystyle\lim_{x\to\infty}\big(1+\tfrac3x\big)^x=e^{3}$.

Water drains from an inverted cone (height $12$ ft, top radius $6$ ft) at $\dfrac{dV}{dt}=-4$ ft$^3$/min. Find $\dfrac{dh}{dt}$ when $h=6$ ft.

Relate & reduce: similar triangles give $\dfrac{r}{h}=\dfrac{6}{12}=\dfrac12$, so $r=\dfrac h2$ and

Differentiate: $\dfrac{dV}{dt}=\dfrac{\pi h^2}{4}\dfrac{dh}{dt}$. Substitute $h=6$: $-4=\dfrac{\pi(36)}{4}\dfrac{dh}{dt}=9\pi\dfrac{dh}{dt}$, so $\dfrac{dh}{dt}=-\dfrac{4}{9\pi}\approx-0.141$ ft/min (falling).

A balloon rises straight up at $3$ m/s. An observer stands $50$ m away. How fast is the angle of elevation $\theta$ increasing when the balloon is $50$ m high?

Relate: $\tan\theta=\dfrac{h}{50}$. Differentiate: $\sec^2\theta\,\dfrac{d\theta}{dt}=\dfrac{1}{50}\dfrac{dh}{dt}$. At $h=50$, $\theta=45^\circ$ so $\sec^2\theta=2$: $2\dfrac{d\theta}{dt}=\dfrac{1}{50}(3)$, giving $\dfrac{d\theta}{dt}=\dfrac{3}{100}=0.03$ rad/s.

For a full motion analysis on $[a,b]$:

• Find $v=s'$ and $a=s''$; solve $v=0$ and $a=0$.
• Build a number line; mark the sign of $v$ (direction) and of $a$ on each subinterval.
• Same signs $\Rightarrow$ speeding up; opposite signs $\Rightarrow$ slowing down.
• For total distance, split $[a,b]$ at every time $v=0$ and add $|s(t_{i+1})-s(t_i)|$; for displacement, just compute $s(b)-s(a)$.