Related Rates, Differentials, and Error Propagation in Calculus

Related Rates

Problem Solving Strategy

Related rates problems involve finding the rate at which one quantity changes with respect to another, often time, when the quantities are related by an equation. The following steps outline a systematic approach:

• Read the problem carefully to understand what is being asked.

• Draw a diagram if possible to visualize the situation.

• Introduce notation for all quantities that are functions of time.

• Express the given information and required rate in terms of derivatives.

• Write an equation relating the variables. Use geometry if necessary.

• Differentiate both sides of the equation with respect to time using the Chain Rule.

• Substitute known values and solve for the unknown rate.

Example 1: The Ladder Problem

A 10 ft ladder rests against a vertical wall. If the bottom slides away at a rate of $4\frac{ft}{s}$, how fast is the top sliding down when the bottom is 6 ft from the wall?

• Let $x(t)$ be the distance from the wall to the bottom, $y(t)$ the height of the top.

• Given: $\frac{dx}{dt} = 4$, $x = 6$, ladder length $= 10$.

• Relationship: $x^2 + y^2 = 100$

• Differentiating: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

• Solve for $\frac{dy}{dt}$: $\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}$

• At $x=6$: $y = 8$, $\frac{dy}{dt} = -\frac{6}{8} \times 4 = -3 \frac{ft}{s}$

Interpretation: The top of the ladder is sliding down at $3\frac{ft}{s}$ when the bottom is 6 ft from the wall.

Example 2: Sphere Surface Area

The radius of a sphere is increasing at $2\frac{cm}{min}$. At what rate is the surface area increasing when the radius is 8 cm?

• Let $r(t)$ be the radius, $S$ the surface area.

• Given: $\frac{dr}{dt} = 2$, $r = 8$

• Surface area: $S = 4\pi r^2$

• Differentiating: $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$

• Substitute: $\frac{dS}{dt} = 8\pi \times 8 \times 2 = 128\pi \frac{cm^2}{min}$

Example 3: Triangle Area

The altitude of a triangle increases at $1\frac{cm}{min}$, and the area increases at $2\frac{cm^2}{min}$. At what rate is the base changing when the altitude is 10 cm and the area is $100cm^2$?

• Let $h(t)$ be altitude, $b(t)$ base, $A$ area.

• Given: $\frac{dh}{dt} = 1$, $\frac{dA}{dt} = 2$, $h = 10$, $A = 100$

• Area: $A = \frac{1}{2} b h$

• Differentiating: $\frac{dA}{dt} = \frac{1}{2}(b \frac{dh}{dt} + h \frac{db}{dt})$

• Substitute values: $2 = \frac{1}{2}(b \times 1 + 10 \frac{db}{dt})$

• Find $b$: $100 = \frac{1}{2} b \times 10 \Rightarrow b = 20$

• Solve: $2 = \frac{1}{2}(20 + 10 \frac{db}{dt}) \Rightarrow 4 = 20 + 10 \frac{db}{dt} \Rightarrow \frac{db}{dt} = -1.6 \frac{cm}{min}$

Interpretation: The base is decreasing at $1.6\frac{cm}{min}$.

Related Rates in Volume Problems

• Cylinder: $V = \pi r^2 h$, $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$ (if $r$ is constant)

• Cone: $V = \frac{1}{3} \pi r^2 h$, $\frac{dV}{dt} = \frac{1}{3} \pi (2 r h \frac{dr}{dt} + r^2 \frac{dh}{dt})$

Application: These formulas are used to relate the rates of change of volume, radius, and height in tank problems.

Differentials

Definition and Use

Differentials provide a way to estimate how much a function changes as a result of a small change in its independent variable. They are closely related to the tangent line approximation.

• If $y = f(x)$, then a small change $dx$ in $x$ produces a change $dy$ in $y$.

• Formula: $dy = f'(x) dx$

• Interpretation: $dy$ approximates the actual change $\Delta y$ for small $dx$.

Example: Tangent Line Approximation

• If $x$ changes from $a$ to $a + dx$, then $\Delta y = f(a + dx) - f(a)$

• Linear approximation: $\Delta y \approx f'(a) dx$

Graphical Interpretation: The tangent line at $x = a$ gives a good estimate of the function's change for small $dx$.

Error Propagation

Estimating Measurement Error

Measurement errors are inevitable and can compound when used in further calculations. Differentials can be used to estimate the error in a computed quantity based on the error in the measured variable.

• If $x$ is measured with error $dx$, and $y = f(x)$, then the error in $y$ is approximately $dy = f'(x) dx$.

• Application: Useful for estimating errors in computed areas, volumes, etc.

Example: Error in Cube Volume

• Suppose the side length $a$ of a cube is measured as $6$cm with error $0.2$ cm.

• Volume: $V = a^3$

• Differential: $dV = 3a^2 da$

• Estimate error: $dV = 3 \times 6^2 \times 0.2 = 3 \times 36 \times 0.2 = 21.6 \text{ cm}^3$

Interpretation: The error in the computed volume is approximately $21.6 \text{ cm}^3$.

Summary Table: Related Rates Formulas

Shape	Volume Formula	Related Rate Formula
Cylinder	$V = \pi r^2 h$	$\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$
Sphere	$V = \frac{4}{3} \pi r^3$	$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$
Cone	$V = \frac{1}{3} \pi r^2 h$	$\frac{dV}{dt} = \frac{1}{3} \pi (2 r h \frac{dr}{dt} + r^2 \frac{dh}{dt})$

Additional info: These notes cover key applications of differentiation, including related rates, differentials, and error propagation, which are essential topics in a college Calculus course (Ch. 5).