Notes
Derivatives in context
$f'(a)$ is the rate of change of $f$ at $a$, in output units per input unit. Example: if $V(t)$ is gallons in a tank after $t$ minutes, $V'(5) = -3$ means at 5 minutes the volume is decreasing at 3 gallons per minute.
Motion along a line
- Position $s(t)$, velocity $v(t) = s'(t)$, acceleration $a(t) = v'(t)$.
- The particle moves right when $v > 0$, left when $v < 0$, and changes direction when $v$ changes sign.
- Speed $= |v(t)|$. Speed increases when $v$ and $a$ have the same sign.
Related rates
1. Draw a picture and name the changing quantities. 2. Write an equation that relates them (Pythagorean theorem, area, volume, similar triangles). 3. Differentiate both sides with respect to $t$. 4. Substitute known values after differentiating, then solve.
Linear approximation
Near $x = a$: $f(x) \approx f(a) + f'(a)(x - a)$. The tangent line overestimates when $f$ is concave down and underestimates when concave up.
L'Hôpital's rule
If $\displaystyle \lim \frac{f(x)}{g(x)}$ gives $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\pm\infty}{\pm\infty}$, then $\displaystyle \lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}$ (if the new limit exists).
Topics in this unit
Each topic has its own short notes and practice.
- 4.1Interpreting the Meaning of the Derivative in Context
- 4.2Straight-Line Motion: Connecting Position, Velocity, and Acceleration
- 4.3Rates of Change in Applied Contexts Other Than Motion
- 4.4Introduction to Related Rates
- 4.5Solving Related Rates Problems
- 4.6Approximating Values of a Function Using Local Linearity and Linearization
- 4.7Using L'Hôpital's Rule for Determining Limits of Indeterminate Forms
Practice
A derivative is an instantaneous rate with units of gallons per minute.
$v = (t - 1)(t - 3)$ changes sign at $t = 1$ and $t = 3$.
Velocity and acceleration have the same sign, so the speed is increasing.
$\displaystyle A = \pi r^2 \Rightarrow \frac{dA}{dt} = 2\pi r\frac{dr}{dt} = 2\pi(5)(2)$.
$f(9) = 3$, $\displaystyle f'(9) = \frac{1}{6}$. $\displaystyle 3 + \frac{1}{6}(0.6) = 3.1$.
It's $\displaystyle \frac{0}{0}$. L'Hôpital twice: $\displaystyle \frac{e^x - 1}{2x} \to \frac{e^x}{2} \to \frac{1}{2}$.
$x^2 + y^2 = 169$, with $x = 5$, $y = 12$. $\displaystyle 2x x' + 2y y' = 0 \Rightarrow y' = -\frac{5 \cdot 2}{12} = -\frac{5}{6}$.
A concave-up graph lies above its tangent lines.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A spherical balloon is inflated so that its volume increases at 100 cm³ per second. ($V = \frac{4}{3}\pi r^3$)
(a) How fast is the radius increasing when $r = 5$ cm?
Show the answer and scoring guide
$\displaystyle \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$, so $\displaystyle 100 = 4\pi(25)\frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{\pi} \approx 0.318$ cm/s.
Check off the points you earned:
(b) How fast is the surface area $S = 4\pi r^2$ increasing at that moment?
Show the answer and scoring guide
$\displaystyle \frac{dS}{dt} = 8\pi r\frac{dr}{dt} = 8\pi(5)\left(\frac{1}{\pi}\right) = 40$ cm²/s.
Check off the points you earned:
(c) Is $\frac{dr}{dt}$ increasing or decreasing as the balloon grows? Explain.
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Decreasing. $\displaystyle \frac{dr}{dt} = \frac{100}{4\pi r^2}$, and as $r$ increases the denominator grows.
Check off the points you earned: