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
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Balloon: If you're inflating a balloon at 2 cubic inches per second, how fast is the radius increasing?
-
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?
-
Shadow: You're walking toward a lamppost. How fast is your shadow shrinking?
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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
- Draw a picture (if applicable)
- Identify known and unknown information
- What rates are given? (these are derivatives)
- What rate are we finding? (this is the unknown derivative)
- Write an equation relating the variables
- Differentiate both sides with respect to time (implicit differentiation!)
- Substitute known values
- 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:
- = rate of change of radius
- = rate of change of volume
- = rate of change of area
- = rate of change of height
- = rate of change of position
Units matter! If 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: cm/sec (radius increasing)
Find: when cm
Step 2: Write equation relating variables
Step 3: Differentiate with respect to time
Step 4: Substitute known values
When and :
Answer: The area is increasing at 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 = distance from wall to bottom of ladder
Let = height of top of ladder on wall
Given: ft/sec (bottom moving away)
Find: when ft
Step 2: Write equation
By Pythagorean theorem:
Step 3: Find when
feet
Step 4: Differentiate with respect to time
Step 5: Substitute and solve
When , , and :
ft/sec
Answer: The top is sliding down at ft/sec (negative means decreasing).
Common Formulas for Related Rates
Geometry Formulas
Circle:
- Area:
- Circumference:
Sphere:
- Volume:
- Surface Area:
Cone:
- Volume:
Cylinder:
- Volume:
Triangle:
- Area:
Pythagorean Theorem:
Similar Triangles:
⚠️ 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 with respect to :
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
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:
- Draw a clear diagram
- Label variables that change with time
- Label constants (like ladder length)
- Write the equation relating variables
When Differentiating:
- Differentiate the entire equation with respect to
- Use Chain Rule for every variable
- Don't substitute until after differentiating
After Differentiating:
- Substitute known values
- Solve for the unknown rate
- Check the sign (increasing or decreasing?)
- 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
- Read carefully and identify what's given and what's asked
- Draw and label everything
- Find the equation relating your variables
- Differentiate with respect to time (use implicit differentiation)
- Substitute values at the specific instant
- Solve for the unknown rate
- 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: in³/min (volume increasing)
Find: when inches
Step 2: Write volume formula for a sphere
Step 3: Differentiate with respect to time
Step 4: Substitute known values
When and :
Step 5: Solve for
Answer: The radius is increasing at 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: , )
💡 Show Solution
Solution:
Given: cm/s, cm
Part (a): Find when .
Differentiate with respect to time:
Substitute and :
cm³/s
Part (b): Find when .
Substitute:
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: , )
💡 Show Solution
Solution:
Given: cm/s, cm
Part (a): Find when .
Differentiate with respect to time:
Substitute and :
cm³/s
Part (b): Find when .
Substitute:
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: m³/min (volume increasing)
Find: when meters
Tank dimensions: radius = 2 m, height = 4 m
Step 2: Set up volume formula
Volume of cone:
But and both vary! We need to eliminate one variable.
Step 3: Use similar triangles
The cone and the water always have the same shape, so:
Therefore:
Step 4: Substitute into volume formula
Step 5: Differentiate with respect to time
Step 6: Substitute and solve
When and :
Answer: The water level is rising at 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 hours:
- Car A is miles north
- Car B is miles east
- Let = distance between them
Given: mph (Car B going east) mph (Car A going north)
Find: when hours
Step 2: Set up relationship (Pythagorean theorem)
Let = distance Car B has traveled (east)
Let = distance Car A has traveled (north)
Step 3: Find positions at hours
miles
miles
miles
Step 4: Differentiate with respect to time
Divide by 2:
Step 5: Substitute values
Step 6: Solve for
Answer: The distance between the cars is increasing at mph.
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