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Related Rates: 4 Steps to Solve Any Word Problem (Visual Guide)

Finding the rate at which one quantity changes by relating it to other quantities whose rates of change are known. This is a practical application of Implicit Differentiation with respect to time ($t$).

Core Theorem
1. General Chain Rule
If y=f(x)y = f(x), then dydt=f′(x)⋅dxdt\frac{dy}{dt} = f'(x) \cdot \frac{dx}{dt}.

 2. Common Geometric Models
 The Sliding Ladder (Pythagorean Theorem): For a ladder of constant length LL, x2+y2=L2x^2 + y^2 = L^2. Differentiating gives 2xdxdt+2ydydt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0. The speed at which the base slides (dx/dtdx/dt) determines the speed at which the top slides down (dy/dtdy/dt).

 The Conical Tank (Volume): V=13πr2hV = \frac{1}{3}\pi r^2 h. To solve, use Similar Triangles (e.g., r/h=R/Hr/h = R/H) to substitute rr in terms of hh (or vice versa) before differentiating to reduce variables.

 The Shadow Problem (Similar Triangles): Relates the height of a street lamp and a person to the distances. The speed of the person walking away from the lamp (dx/dtdx/dt) determines the rate at which the shadow length (ds/dtds/dt) increases.
Step-by-Step SOP
  1. 1

    1. Label & List

    Draw a diagram. Label constants and variables. List given rates and the 'snapshot' value.
  2. 2

    2. Establish Equation

    Find a formula relating the variables (Area, Volume, Pythagorean Theorem, or Similar Triangles).
  3. 3

    3. Differentiate w.r.t. $t$

    Perform implicit differentiation on both sides with respect to time (tt).
  4. 4

    4. Solve

    Plug in the known values at that specific instant and isolate the unknown rate.

Practice Exercises

Example 01Easy
A circle's radius increases at 3 cm/s. How fast is the area increasing when r=10r = 10 cm?
NEED A HINT?
Equation: A=πr2A = \pi r^2. We know dr/dt=3dr/dt = 3. Find dA/dtdA/dt.
SHOW DETAILED EXPLANATION

Step 1: Identify Given Information

We are given drdt=3\frac{dr}{dt} = 3 cm/s and need to find dAdt\frac{dA}{dt} when r=10r = 10 cm.

Step 2: Differentiate Equation

Start with A=πr2A = \pi r^2. Differentiate both sides with respect to tt: dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r \frac{dr}{dt}.

Step 3: Substitute and Solve

Substitute the values: dAdt=2π(10)(3)=60π\frac{dA}{dt} = 2\pi(10)(3) = 60\pi. The area is increasing at 60π60\pi cm2^2/s.
Example 02Medium
A 13-foot ladder is leaning against a wall. If the bottom of the ladder is pulled away from the wall at 2 ft/s, how fast is the top sliding down when the bottom is 5ft from the wall?
NEED A HINT?
Use x2+y2=L2x^2 + y^2 = L^2. Note that LL is constant, so dL/dt=0dL/dt = 0.
SHOW DETAILED EXPLANATION

Step 1: Set up the Geometry

Let xx be the distance from the wall and yy be the height. By Pythagorean theorem: x2+y2=132x^2 + y^2 = 13^2. When x=5x=5, y=169−25=12y = \sqrt{169-25} = 12.

Step 2: Implicit Differentiation

Differentiate x2+y2=169x^2 + y^2 = 169 with respect to tt: 2xdxdt+2ydydt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0.

Step 3: Solve for Rate

Substitute x=5,y=12,dxdt=2x=5, y=12, \frac{dx}{dt}=2: 2(5)(2)+2(12)dydt=0⇒dydt=−2024=−562(5)(2) + 2(12)\frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6} ft/s. The top is sliding down at 5/65/6 ft/s.
Example 03Hard
Water is poured into a conical tank (height 10m, radius 4m) at 2m3/min2 m^3/min. How fast is the water level rising when the water is 5m deep?
NEED A HINT?
The radius rr and height hh change together. Use similar triangles to eliminate rr.
SHOW DETAILED EXPLANATION

Step 1: Relate r and h

By similar triangles, rh=410⇒r=0.4h\frac{r}{h} = \frac{4}{10} \Rightarrow r = 0.4h. Substitute this into V=13πr2hV = \frac{1}{3}\pi r^2 h to get V=13π(0.4h)2h=0.163πh3V = \frac{1}{3}\pi (0.4h)^2 h = \frac{0.16}{3}\pi h^3.

Step 2: Differentiate w.r.t. Time

dVdt=0.163π(3h2)dhdt=0.16πh2dhdt\frac{dV}{dt} = \frac{0.16}{3}\pi (3h^2) \frac{dh}{dt} = 0.16\pi h^2 \frac{dh}{dt}.

Step 3: Plug in Values

Given dVdt=2\frac{dV}{dt} = 2 and h=5h = 5: 2=0.16π(25)dhdt⇒2=4πdhdt2 = 0.16\pi (25) \frac{dh}{dt} \Rightarrow 2 = 4\pi \frac{dh}{dt}. Thus, dhdt=12π≈0.159\frac{dh}{dt} = \frac{1}{2\pi} \approx 0.159 m/min.
Example 04Medium
A 6ft tall man walks away from a 15ft lamppost at 5 ft/s. How fast is his shadow lengthening when he is 10ft from the post?
NEED A HINT?
Draw two similar triangles: one for the lamp and one for the man.
SHOW DETAILED EXPLANATION

Step 1: Set up Proportions

Let xx be man's distance from post and ss be shadow length. By similar triangles: 15x+s=6s\frac{15}{x+s} = \frac{6}{s}.

Step 2: Simplify and Differentiate

Cross-multiply: 15s=6x+6s⇒9s=6x⇒s=23x15s = 6x + 6s \Rightarrow 9s = 6x \Rightarrow s = \frac{2}{3}x. Differentiating gives dsdt=23dxdt\frac{ds}{dt} = \frac{2}{3}\frac{dx}{dt}.

Step 3: Conclusion

Since dxdt=5\frac{dx}{dt} = 5, dsdt=23(5)=3.33\frac{ds}{dt} = \frac{2}{3}(5) = 3.33 ft/s. The shadow length is independent of the distance xx.
Common Pitfalls
  • ⚠
    Early SubstitutionSubstituting constant 'snapshot' values (like r=10r=10 or h=5h=5) before differentiating. This turns variables into constants, making their derivatives zero incorrectly.
  • ⚠
    Sign ErrorsForgetting that rates of decreasing quantities must be negative (e.g., a ladder sliding 'down' means dy/dt<0dy/dt < 0).
  • ⚠
    Chain Rule NeglectForgetting to multiply by d(variable)/dtd(\text{variable})/dt after differentiating each term.
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