Related Rates

Finding how rates of change are related to each other

⏱️ Related Rates

What are Related Rates?

Related rates problems involve finding how fast one quantity is changing based on how fast another related quantity is changing.

Real-World Examples

  1. Balloon: If you're inflating a balloon at 2 cubic inches per second, how fast is the radius increasing?

  2. Ladder: A ladder is sliding down a wall. If the bottom is moving away at 3 ft/sec, how fast is the top sliding down?

  3. Shadow: You're walking toward a lamppost. How fast is your shadow shrinking?

  4. Water tank: Water is draining from a cone-shaped tank. How fast is the water level dropping?

💡 Key Idea: Use derivatives to connect rates of change of related quantities!


The General Strategy

Step-by-Step Process

  1. Draw a picture (if applicable)
  2. Identify known and unknown information
    • What rates are given? (these are derivatives)
    • What rate are we finding? (this is the unknown derivative)
  3. Write an equation relating the variables
  4. Differentiate both sides with respect to time tt (implicit differentiation!)
  5. Substitute known values
  6. Solve for the unknown rate

⚠️ IMPORTANT: Don't substitute values until AFTER differentiating!


Common Rate Notation

In related rates, we use derivatives with respect to time:

  • drdt\frac{dr}{dt} = rate of change of radius
  • dVdt\frac{dV}{dt} = rate of change of volume
  • dAdt\frac{dA}{dt} = rate of change of area
  • dhdt\frac{dh}{dt} = rate of change of height
  • dxdt\frac{dx}{dt} = rate of change of position

Units matter! If dxdt=5\frac{dx}{dt} = 5 ft/sec, that means the position is increasing at 5 feet per second.


Example 1: Expanding Circle

Problem: The radius of a circle is increasing at 3 cm/sec. How fast is the area increasing when the radius is 10 cm?


Step 1: Identify information

Given: drdt=3\frac{dr}{dt} = 3 cm/sec (radius increasing)

Find: dAdt\frac{dA}{dt} when r=10r = 10 cm


Step 2: Write equation relating variables

A=πr2A = \pi r^2


Step 3: Differentiate with respect to time

dAdt=ddt[πr2]\frac{dA}{dt} = \frac{d}{dt}[\pi r^2]

dAdt=π2rdrdt\frac{dA}{dt} = \pi \cdot 2r \cdot \frac{dr}{dt}

dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r \frac{dr}{dt}


Step 4: Substitute known values

When r=10r = 10 and drdt=3\frac{dr}{dt} = 3:

dAdt=2π(10)(3)=60π cm2/sec\frac{dA}{dt} = 2\pi(10)(3) = 60\pi \text{ cm}^2/\text{sec}

Answer: The area is increasing at 60π188.560\pi \approx 188.5 square cm per second.


Example 2: Ladder Sliding Down Wall

Problem: A 13-foot ladder is leaning against a wall. The bottom is being pulled away from the wall at 2 ft/sec. How fast is the top sliding down when the bottom is 5 feet from the wall?


Step 1: Draw and identify

Let xx = distance from wall to bottom of ladder

Let yy = height of top of ladder on wall

Given: dxdt=2\frac{dx}{dt} = 2 ft/sec (bottom moving away)

Find: dydt\frac{dy}{dt} when x=5x = 5 ft


Step 2: Write equation

By Pythagorean theorem:

x2+y2=132=169x^2 + y^2 = 13^2 = 169


Step 3: Find yy when x=5x = 5

52+y2=1695^2 + y^2 = 169 y2=144y^2 = 144 y=12y = 12 feet


Step 4: Differentiate with respect to time

ddt[x2+y2]=ddt[169]\frac{d}{dt}[x^2 + y^2] = \frac{d}{dt}[169]

2xdxdt+2ydydt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0


Step 5: Substitute and solve

When x=5x = 5, y=12y = 12, and dxdt=2\frac{dx}{dt} = 2:

2(5)(2)+2(12)dydt=02(5)(2) + 2(12)\frac{dy}{dt} = 0

20+24dydt=020 + 24\frac{dy}{dt} = 0

dydt=2024=56\frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6} ft/sec

Answer: The top is sliding down at 56\frac{5}{6} ft/sec (negative means decreasing).


Common Formulas for Related Rates

Geometry Formulas

Circle:

  • Area: A=πr2A = \pi r^2
  • Circumference: C=2πrC = 2\pi r

Sphere:

  • Volume: V=43πr3V = \frac{4}{3}\pi r^3
  • Surface Area: S=4πr2S = 4\pi r^2

Cone:

  • Volume: V=13πr2hV = \frac{1}{3}\pi r^2 h

Cylinder:

  • Volume: V=πr2hV = \pi r^2 h

Triangle:

  • Area: A=12bhA = \frac{1}{2}bh

Pythagorean Theorem: a2+b2=c2a^2 + b^2 = c^2

Similar Triangles: a1a2=b1b2\frac{a_1}{a_2} = \frac{b_1}{b_2}


⚠️ Common Mistakes

Mistake 1: Substituting Too Early

❌ Substitute values before differentiating ✅ Differentiate first, then substitute

Mistake 2: Wrong Signs

If a quantity is decreasing, its rate is negative!

Mistake 3: Forgetting Chain Rule

When differentiating r2r^2 with respect to tt: ddt[r2]=2rdrdt\frac{d}{dt}[r^2] = 2r\frac{dr}{dt}

Mistake 4: Units

Always include units in your answer! (ft/sec, cm³/min, etc.)

Mistake 5: Missing the Relationship

Make sure your equation actually relates the variables in the problem.


Problem Types

Type 1: Area and Volume

Expanding/shrinking circles, spheres, cones, etc.

Key: Use geometry formulas and differentiate with respect to time.

Type 2: Pythagorean Relationships

Ladders, shadows, moving vehicles, etc.

Key: Set up right triangle and use a2+b2=c2a^2 + b^2 = c^2

Type 3: Similar Triangles

Shadows, conical tanks with changing water levels, etc.

Key: Set up proportion using similar triangles, then differentiate.

Type 4: Distance and Position

Two objects moving, finding rate of change of distance between them.

Key: Use distance formula or Pythagorean theorem.


Tips for Success

Before Differentiating:

  1. Draw a clear diagram
  2. Label variables that change with time
  3. Label constants (like ladder length)
  4. Write the equation relating variables

When Differentiating:

  1. Differentiate the entire equation with respect to tt
  2. Use Chain Rule for every variable
  3. Don't substitute until after differentiating

After Differentiating:

  1. Substitute known values
  2. Solve for the unknown rate
  3. Check the sign (increasing or decreasing?)
  4. Include proper units

Word Problem Clues

Certain phrases tell you what rate is given or asked:

  • "increasing at..." → positive rate
  • "decreasing at..." → negative rate
  • "how fast is..." → find a derivative
  • "at the instant when..." → specific value to substitute
  • "at what rate..." → find a derivative
  • "constant rate" → the rate doesn't change

📝 Practice Strategy

  1. Read carefully and identify what's given and what's asked
  2. Draw and label everything
  3. Find the equation relating your variables
  4. Differentiate with respect to time (use implicit differentiation)
  5. Substitute values at the specific instant
  6. Solve for the unknown rate
  7. Check sign and units

📚 Practice Problems

1Problem 1medium

Question:

A spherical balloon is being inflated at a rate of 100 cubic inches per minute. How fast is the radius increasing when the radius is 5 inches?

💡 Show Solution

Step 1: Identify information

Given: dVdt=100\frac{dV}{dt} = 100 in³/min (volume increasing)

Find: drdt\frac{dr}{dt} when r=5r = 5 inches


Step 2: Write volume formula for a sphere

V=43πr3V = \frac{4}{3}\pi r^3


Step 3: Differentiate with respect to time

dVdt=ddt[43πr3]\frac{dV}{dt} = \frac{d}{dt}\left[\frac{4}{3}\pi r^3\right]

dVdt=43π3r2drdt\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt}

dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}


Step 4: Substitute known values

When r=5r = 5 and dVdt=100\frac{dV}{dt} = 100:

100=4π(5)2drdt100 = 4\pi(5)^2\frac{dr}{dt}

100=4π(25)drdt100 = 4\pi(25)\frac{dr}{dt}

100=100πdrdt100 = 100\pi\frac{dr}{dt}


Step 5: Solve for drdt\frac{dr}{dt}

drdt=100100π=1π\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi}

Answer: The radius is increasing at 1π0.318\frac{1}{\pi} \approx 0.318 inches per minute.

2Problem 2hard

Question:

A spherical balloon is being inflated so that its radius is increasing at a rate of 2 cm/s.

a) How fast is the volume increasing when the radius is 10 cm? b) How fast is the surface area increasing at that moment?

(Formulas: V=43πr3V = \frac{4}{3}\pi r^3, S=4πr2S = 4\pi r^2)

💡 Show Solution

Solution:

Given: drdt=2\frac{dr}{dt} = 2 cm/s, r=10r = 10 cm

Part (a): Find dVdt\frac{dV}{dt} when r=10r = 10.

V=43πr3V = \frac{4}{3}\pi r^3

Differentiate with respect to time:

dVdt=43π3r2drdt=4πr2drdt\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}

Substitute r=10r = 10 and drdt=2\frac{dr}{dt} = 2:

dVdt=4π(10)2(2)=4π(100)(2)=800π\frac{dV}{dt} = 4\pi (10)^2 (2) = 4\pi (100)(2) = 800\pi cm³/s

Part (b): Find dSdt\frac{dS}{dt} when r=10r = 10.

S=4πr2S = 4\pi r^2

dSdt=4π2rdrdt=8πrdrdt\frac{dS}{dt} = 4\pi \cdot 2r \cdot \frac{dr}{dt} = 8\pi r \frac{dr}{dt}

Substitute:

dSdt=8π(10)(2)=160π\frac{dS}{dt} = 8\pi (10)(2) = 160\pi cm²/s

3Problem 3hard

Question:

A spherical balloon is being inflated so that its radius is increasing at a rate of 2 cm/s.

a) How fast is the volume increasing when the radius is 10 cm? b) How fast is the surface area increasing at that moment?

(Formulas: V=43πr3V = \frac{4}{3}\pi r^3, S=4πr2S = 4\pi r^2)

💡 Show Solution

Solution:

Given: drdt=2\frac{dr}{dt} = 2 cm/s, r=10r = 10 cm

Part (a): Find dVdt\frac{dV}{dt} when r=10r = 10.

V=43πr3V = \frac{4}{3}\pi r^3

Differentiate with respect to time:

dVdt=43π3r2drdt=4πr2drdt\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}

Substitute r=10r = 10 and drdt=2\frac{dr}{dt} = 2:

dVdt=4π(10)2(2)=4π(100)(2)=800π\frac{dV}{dt} = 4\pi (10)^2 (2) = 4\pi (100)(2) = 800\pi cm³/s

Part (b): Find dSdt\frac{dS}{dt} when r=10r = 10.

S=4πr2S = 4\pi r^2

dSdt=4π2rdrdt=8πrdrdt\frac{dS}{dt} = 4\pi \cdot 2r \cdot \frac{dr}{dt} = 8\pi r \frac{dr}{dt}

Substitute:

dSdt=8π(10)(2)=160π\frac{dS}{dt} = 8\pi (10)(2) = 160\pi cm²/s

4Problem 4hard

Question:

A water tank has the shape of an inverted cone with base radius 2 meters and height 4 meters. If water is being pumped into the tank at 2 cubic meters per minute, find the rate at which the water level is rising when the water is 3 meters deep.

💡 Show Solution

Step 1: Identify information

Given: dVdt=2\frac{dV}{dt} = 2 m³/min (volume increasing)

Find: dhdt\frac{dh}{dt} when h=3h = 3 meters

Tank dimensions: radius = 2 m, height = 4 m


Step 2: Set up volume formula

Volume of cone: V=13πr2hV = \frac{1}{3}\pi r^2 h

But rr and hh both vary! We need to eliminate one variable.


Step 3: Use similar triangles

The cone and the water always have the same shape, so:

rh=24=12\frac{r}{h} = \frac{2}{4} = \frac{1}{2}

Therefore: r=h2r = \frac{h}{2}


Step 4: Substitute into volume formula

V=13πr2h=13π(h2)2hV = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{h}{2}\right)^2 h

V=13πh24h=πh312V = \frac{1}{3}\pi \cdot \frac{h^2}{4} \cdot h = \frac{\pi h^3}{12}


Step 5: Differentiate with respect to time

dVdt=π123h2dhdt\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt}

dVdt=πh24dhdt\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}


Step 6: Substitute and solve

When h=3h = 3 and dVdt=2\frac{dV}{dt} = 2:

2=π(3)24dhdt2 = \frac{\pi(3)^2}{4}\frac{dh}{dt}

2=9π4dhdt2 = \frac{9\pi}{4}\frac{dh}{dt}

dhdt=249π=89π\frac{dh}{dt} = \frac{2 \cdot 4}{9\pi} = \frac{8}{9\pi}

Answer: The water level is rising at 89π0.283\frac{8}{9\pi} \approx 0.283 meters per minute.

5Problem 5hard

Question:

Two cars start from the same point. Car A travels north at 50 mph and Car B travels east at 40 mph. How fast is the distance between them increasing 2 hours later?

💡 Show Solution

Step 1: Draw and identify

After tt hours:

  • Car A is 50t50t miles north
  • Car B is 40t40t miles east
  • Let zz = distance between them

Given: dxdt=40\frac{dx}{dt} = 40 mph (Car B going east) dydt=50\frac{dy}{dt} = 50 mph (Car A going north)

Find: dzdt\frac{dz}{dt} when t=2t = 2 hours


Step 2: Set up relationship (Pythagorean theorem)

Let xx = distance Car B has traveled (east)

Let yy = distance Car A has traveled (north)

x2+y2=z2x^2 + y^2 = z^2


Step 3: Find positions at t=2t = 2 hours

x=40(2)=80x = 40(2) = 80 miles

y=50(2)=100y = 50(2) = 100 miles

z=802+1002=6400+10000=16400=2041z = \sqrt{80^2 + 100^2} = \sqrt{6400 + 10000} = \sqrt{16400} = 20\sqrt{41} miles


Step 4: Differentiate with respect to time

ddt[x2+y2]=ddt[z2]\frac{d}{dt}[x^2 + y^2] = \frac{d}{dt}[z^2]

2xdxdt+2ydydt=2zdzdt2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}

Divide by 2:

xdxdt+ydydt=zdzdtx\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}


Step 5: Substitute values

80(40)+100(50)=2041dzdt80(40) + 100(50) = 20\sqrt{41}\frac{dz}{dt}

3200+5000=2041dzdt3200 + 5000 = 20\sqrt{41}\frac{dz}{dt}

8200=2041dzdt8200 = 20\sqrt{41}\frac{dz}{dt}


Step 6: Solve for dzdt\frac{dz}{dt}

dzdt=82002041=41041=4104141=1041\frac{dz}{dt} = \frac{8200}{20\sqrt{41}} = \frac{410}{\sqrt{41}} = \frac{410\sqrt{41}}{41} = 10\sqrt{41}

Answer: The distance between the cars is increasing at 104164.010\sqrt{41} \approx 64.0 mph.