Unit 4 Related Rates: Connecting Changing Quantities in Time

Related rates

Problems where two or more quantities change over time and are connected by a known relationship, so their rates of change are related.

Rate of change (with respect to time)

A derivative like d(variable)/dt that measures how fast a quantity changes per unit time, including units (e.g., cm/s).

Implicit differentiation (in related rates)

Differentiating an equation relating variables with respect to t, even if t is not shown explicitly, because the variables depend on time.

Chain Rule in related rates

When differentiating a quantity like $r^2$ with respect to $t$, you must multiply by $\frac{dr}{dt}$ (e.g., $\frac{d}{dt}(r^2)=2r \bullet \frac{dr}{dt}$).

Constraint equation

A single equation (often from geometry) that relates the changing variables in a related rates problem (e.g., $x^2+y^2=L^2$).

“Differentiate with respect to time”

The core move in related rates: apply d/dt to both sides of the constraint equation to connect the variables’ rates.

Instantaneous rate

A rate evaluated at a specific moment (“at the moment when…”), using the variable values at that instant.

Variable vs. rate

Distinguishing a quantity (like r or y) from its time derivative (dr/dt or dy/dt); they are not interchangeable.

Units check

Using units to verify correctness (e.g., $(cm)(\frac{cm}{s})=\frac{cm^2}{s}$ for $\frac{dA}{dt}$).

Sign of a rate

Positive means increasing in the chosen positive direction; negative means decreasing or moving in the opposite direction.

“Equation first, derivative second, numbers last”

Workflow memory aid: write the relationship, differentiate, then substitute numerical values to avoid breaking the variable relationship.

Standard related rates workflow

Define variables, write one constraint equation, differentiate w.r.t. $t$, substitute values at the instant, solve for the unknown rate (with sign/units).

Diagram (related rates)

A labeled picture of the situation that helps choose correct variables, identify constants, and write the correct geometric relationship.

Constant (in related rates)

A quantity that does not change over time (e.g., ladder length), so its derivative with respect to t is 0.

Pythagorean theorem constraint

For a right-triangle situation: $x^2 + y^2 = L^2$ (often used for ladders, distances, and perpendicular motion).

Differentiated Pythagorean equation

From $x^2+y^2=L^2$ (L constant): $2x \bullet \frac{dx}{dt} + 2y \bullet \frac{dy}{dt}=0.$

Ladder sliding interpretation

In the ladder problem, as the bottom moves away (x increases), the top moves down (y decreases), so dy/dt is negative.

Circle area formula

$A = \text{π}r^2$, relating a circle’s area $A$ to its radius $r$.

Area related-rates equation (circle)

Differentiating $A=\frac{\pi}{r^2}$ gives \frac{dA}{dt} = 2\pir \bullet \frac{dr}{dt}.

“Plugging in too early” mistake

Substituting numerical values before differentiating (e.g., r=12) can turn a changing variable into a constant and destroy the rate relationship.

Cone volume formula

$V = \frac{1}{3}\text{π}r^2h$, relating the volume of a cone to radius $r$ and height $h$.

Similar triangles (cone problems)

A proportionality relationship (like r/h = constant) used to eliminate one variable so the volume can be written in terms of a single changing variable.

Cone setup example ratio

If the full cone has radius 4 and height 12, similar triangles give r/h = 4/12 = 1/3, so r = h/3 for the water cone.

Perpendicular motion distance relationship

If one object is $x$ east and another is $y$ north, the distance between them satisfies $s^2 = x^2 + y^2$.

Distance-between-objects rate equation

From $s^2=x^2+y^2$: $2s \bullet \frac{ds}{dt} = 2x \bullet \frac{dx}{dt} + 2y \bullet \frac{dy}{dt}$, so $\frac{ds}{dt} = \frac{x \bullet \frac{dx}{dt} + y \bullet \frac{dy}{dt}}{s}.$