The Central Idea
Suppose two quantities \(x\) and \(y\) both change with time and satisfy
\[ F(x,y)=0. \]Because \(x=x(t)\) and \(y=y(t)\), differentiating with respect to time gives a relationship involving their rates:
\[ \frac{d}{dt}F(x(t),y(t))=0. \]The derivatives
\[ \frac{dx}{dt} \qquad\text{and}\qquad \frac{dy}{dt} \]tell us how the quantities are changing at the same instant.
A related-rates problem uses the equation connecting the quantities to connect their rates.
Rates Are Derivatives with Respect to Time
A rate such as
\[ \frac{dx}{dt} \]is the instantaneous rate at which \(x\) changes as time changes.
If \(x\) is measured in meters and \(t\) in seconds, then
\[ \frac{dx}{dt} \]has units of meters per second.
Related-rates problems therefore combine derivative information with the units and geometry of the quantities being modeled.
The Quantities Are Functions of Time
A problem may write quantities simply as \(x\), \(y\), \(r\), \(A\), or \(V\), but when those quantities are changing, we should think of them as functions of time:
\[ x=x(t),\qquad y=y(t),\qquad r=r(t),\qquad A=A(t),\qquad V=V(t). \]This viewpoint explains why the chain rule appears automatically when we differentiate.
Related Rates Are Not a New Differentiation Rule
Related rates do not require a special derivative rule.
The calculus usually comes from rules already developed:
- the power rule;
- the product rule;
- the quotient rule when needed;
- the chain rule;
- implicit differentiation.
What makes related-rates problems distinctive is the modeling workflow: we must first determine which quantities are changing and how they are connected.
A Simple Rate Relationship
Suppose
\[ y=x^2 \]and both \(x\) and \(y\) change with time.
Differentiate with respect to \(t\):
\[ \frac{dy}{dt} = 2x\frac{dx}{dt}. \]This equation says that the rate of change of \(y\) depends on both the current value of \(x\) and the current rate of change of \(x\).
Example: Using a Known Rate
Suppose
\[ y=x^2, \]and at a particular instant,
\[ x=3 \qquad\text{and}\qquad \frac{dx}{dt}=4. \]Since
\[ \frac{dy}{dt} = 2x\frac{dx}{dt}, \]substitute the instantaneous values:
\[ \frac{dy}{dt} = 2(3)(4)=24. \]Therefore,
\[ \boxed{\frac{dy}{dt}=24}. \]The Snapshot Principle
A related-rates equation describes quantities that vary over time. The problem usually asks about one specific instant.
The workflow therefore has two levels:
- derive a general relationship among the rates;
- evaluate that relationship using the values at the requested instant.
The changing equation comes first. The numerical snapshot comes second.
Why We Usually Differentiate Before Substituting Instantaneous Values
Suppose a circle has radius \(r\) that changes with time and we want to connect the radius rate to the area rate.
Start with
\[ A=\pi r^2. \]Differentiate while \(r\) is still understood as a variable:
\[ \frac{dA}{dt} = 2\pi r\frac{dr}{dt}. \]Only then should we substitute the radius at the instant of interest.
Replacing \(r\) by a numerical value too early can destroy the information that \(r\) is changing.
Constants May Be Substituted Immediately
There is an important distinction between a quantity that is fixed for the entire situation and a changing quantity whose value is known only at one instant.
A ladder of fixed length \(13\) feet may be represented immediately by
\[ x^2+y^2=13^2. \]The number \(13\) is constant throughout the motion.
By contrast, if \(x=5\) only at the instant being studied, we normally keep \(x\) as a variable until after differentiation.
A Reliable Related-Rates Workflow
- Read the problem and identify the changing quantities.
- Assign variables and include units.
- Record which rates are known and which rate is requested.
- Draw or interpret the geometry when useful.
- Write one equation relating the relevant changing quantities.
- Eliminate unnecessary variables if an additional geometric relationship is available.
- Differentiate the relationship with respect to time.
- Substitute the values that hold at the requested instant.
- Solve for the unknown rate.
- Interpret the sign and state the units.
Known Values and Known Rates Are Different
Suppose a problem tells us
\[ r=5\text{ cm} \]and
\[ \frac{dr}{dt}=2\text{ cm/s}. \]These statements provide different kinds of information.
- \(r=5\) is a value of the radius at one instant.
- \(\frac{dr}{dt}=2\) is the rate at which the radius is changing at that instant.
Confusing a value with a rate is one of the most common related-rates errors.
Signs Carry Direction Information
A positive rate means the quantity is increasing:
\[ \frac{dx}{dt}>0. \]A negative rate means the quantity is decreasing:
\[ \frac{dx}{dt}<0. \]Words such as increasing, expanding, and rising usually indicate positive rates. Words such as decreasing, shrinking, falling, and approaching often indicate negative rates relative to the chosen coordinate.
Units Provide a Built-In Check
Suppose \(A\) is measured in square centimeters and \(t\) in seconds. Then
\[ \frac{dA}{dt} \]must have units
\[ \text{cm}^2/\text{s}. \]If a final answer for an area rate is given only in centimeters per second, the units reveal that something is wrong.
Example: Expanding Circle
A circle's radius increases at
\[ \frac{dr}{dt}=3\text{ cm/s}. \]How fast is its area increasing when \(r=5\) cm?
Start with
\[ A=\pi r^2. \]Differentiate with respect to time:
\[ \frac{dA}{dt} = 2\pi r\frac{dr}{dt}. \]At the requested instant,
\[ r=5, \qquad \frac{dr}{dt}=3. \]Therefore,
\[ \frac{dA}{dt} = 2\pi(5)(3) = 30\pi. \]Thus,
\[ \boxed{\frac{dA}{dt}=30\pi\text{ cm}^2/\text{s}}. \]Why the Area Rate Depends on the Current Radius
For a circle,
\[ \frac{dA}{dt} = 2\pi r\frac{dr}{dt}. \]Even if \(\frac{dr}{dt}\) stays constant, the area rate changes as the radius changes.
A one-centimeter increase in radius adds much more area to a large circle than to a small circle. The factor
\[ 2\pi r \]captures that dependence.
Example: Shrinking Sphere
A spherical object has radius decreasing at
\[ \frac{dr}{dt}=-0.2\text{ cm/s}. \]Find the rate of change of its volume when \(r=10\) cm.
The volume is
\[ V=\frac43\pi r^3. \]Differentiate:
\[ \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}. \]Substitute:
\[ \frac{dV}{dt} = 4\pi(10)^2(-0.2) = -80\pi. \]Therefore,
\[ \boxed{\frac{dV}{dt}=-80\pi\text{ cm}^3/\text{s}}. \]The negative sign confirms that the volume is decreasing.
Geometry Often Supplies the Governing Equation
Many related-rates problems are built from familiar geometric relationships.
- Circle: \(A=\pi r^2\)
- Sphere: \(V=\frac43\pi r^3\)
- Cylinder: \(V=\pi r^2h\)
- Rectangle: \(A=LW\)
- Right triangle: \(x^2+y^2=L^2\)
- Similar triangles: ratios of corresponding sides are equal
The calculus begins only after the appropriate mathematical relationship has been identified.
Example: A Sliding Ladder
A \(13\)-foot ladder rests against a vertical wall. Let
\[ x \]be the distance from the foot of the ladder to the wall and
\[ y \]be the height of the top of the ladder above the ground.
Because the ladder length is constant,
\[ x^2+y^2=13^2. \]Differentiating the Ladder Relationship
Differentiate
\[ x^2+y^2=169 \]with respect to time:
\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0. \]Divide by \(2\):
\[ \boxed{ x\frac{dx}{dt} + y\frac{dy}{dt} = 0 }. \]This equation connects the horizontal and vertical speeds.
Example: Ladder Rate at a Particular Instant
Suppose the foot of the \(13\)-foot ladder moves away from the wall at
\[ \frac{dx}{dt}=2\text{ ft/s}. \]Find how fast the top is moving when \(x=5\) ft.
First find \(y\):
\[ 5^2+y^2=13^2. \] \[ y^2=144, \qquad y=12. \]Use the rate equation:
\[ 5(2)+12\frac{dy}{dt}=0. \]Therefore,
\[ 12\frac{dy}{dt}=-10, \] \[ \boxed{\frac{dy}{dt}=-\frac56\text{ ft/s}}. \]The negative sign means the top of the ladder is moving downward.
The Ladder Example Is Implicit Differentiation in Time
The equation
\[ x^2+y^2=169 \]does not solve one changing quantity explicitly in terms of the other. We differentiate the relationship directly.
This is the same basic reasoning used in implicit differentiation, except that both variables are now functions of time.
Example: Changing Rectangle
A rectangle has length \(L\) and width \(W\), both changing with time. Its area is
\[ A=LW. \]Differentiate using the product rule:
\[ \boxed{ \frac{dA}{dt} = L\frac{dW}{dt} + W\frac{dL}{dt} }. \]Both dimensions contribute to the area rate.
Example: Rectangle with One Side Increasing and One Decreasing
Suppose at a particular instant,
\[ L=10\text{ m}, \qquad W=6\text{ m}, \] \[ \frac{dL}{dt}=3\text{ m/s}, \qquad \frac{dW}{dt}=-1\text{ m/s}. \]Then
\[ \frac{dA}{dt} = 10(-1)+6(3) = -10+18 = 8. \]Therefore,
\[ \boxed{\frac{dA}{dt}=8\text{ m}^2/\text{s}}. \]The width is shrinking, but the length is increasing fast enough that the area is still increasing.
Several Changing Factors May Require the Product Rule
Consider a cylinder:
\[ V=\pi r^2h. \]If both \(r\) and \(h\) vary with time, then
\[ \frac{dV}{dt} = \pi \left( 2rh\frac{dr}{dt} + r^2\frac{dh}{dt} \right). \]We cannot treat \(h\) as constant unless the physical situation actually says that the height remains fixed.
Choose Variables That Match the Geometry
In applied problems, good variable choices simplify the mathematics.
For a right-triangle configuration, it is often natural to use horizontal and vertical distances such as
\[ x(t) \qquad\text{and}\qquad y(t). \]For a tank, radius \(r(t)\), depth \(h(t)\), and volume \(V(t)\) may be natural.
For an angle-of-elevation problem, an angle \(\theta(t)\) and a distance \(x(t)\) may be enough.
Some Problems Need an Auxiliary Relationship
The first equation we write may contain too many changing variables.
A second geometric relationship can often eliminate one of them before differentiation.
Similar triangles are especially useful in cone, shadow, and projection problems because they can connect two changing lengths through a fixed ratio.
Example: Water in a Conical Tank
Consider an inverted conical tank with fixed height \(12\) m and fixed top radius \(4\) m. Water inside forms a smaller similar cone with water depth \(h\) and surface radius \(r\).
Similar triangles give
\[ \frac{r}{h} = \frac{4}{12} = \frac13. \]Therefore,
\[ r=\frac{h}{3}. \]Eliminating a Variable in the Conical-Tank Model
The water volume is
\[ V=\frac13\pi r^2h. \]Substitute
\[ r=\frac{h}{3}. \]Then
\[ V = \frac13\pi \left(\frac{h}{3}\right)^2h = \frac{\pi}{27}h^3. \]Now the volume is expressed using only the changing depth \(h\).
Example: Rising Water Level
For the conical tank,
\[ V=\frac{\pi}{27}h^3. \]Differentiate:
\[ \frac{dV}{dt} = \frac{\pi}{9}h^2\frac{dh}{dt}. \]Suppose water enters at
\[ \frac{dV}{dt}=2\text{ m}^3/\text{min} \]and the water is \(h=3\) m deep.
Then
\[ 2 = \frac{\pi}{9}(3)^2\frac{dh}{dt} = \pi\frac{dh}{dt}. \]Therefore,
\[ \boxed{ \frac{dh}{dt} = \frac{2}{\pi}\text{ m/min} }. \]Why Similarity Must Be Used Correctly
In the conical tank, the fixed dimensions of the entire tank are
\[ R=4, \qquad H=12. \]The water radius \(r\) and water depth \(h\) change.
Similarity connects them:
\[ \frac{r}{h}=\frac{R}{H}. \]We should not set \(r=4\) or \(h=12\) unless the tank is actually full at the instant being studied.
Example: A Growing Shadow
A \(6\)-foot person walks directly away from a \(15\)-foot lamp at
\[ \frac{dx}{dt}=4\text{ ft/s}. \]Let \(x\) be the person's distance from the lamp and \(s\) be the shadow length.
Similar triangles give
\[ \frac{15}{x+s} = \frac{6}{s}. \]Solving the Shadow Relationship
From
\[ \frac{15}{x+s} = \frac{6}{s}, \]cross multiply:
\[ 15s=6(x+s). \]Therefore,
\[ 15s=6x+6s, \] \[ 9s=6x, \] \[ s=\frac23x. \]Differentiate:
\[ \frac{ds}{dt} = \frac23\frac{dx}{dt}. \]Since \(\frac{dx}{dt}=4\),
\[ \boxed{\frac{ds}{dt}=\frac83\text{ ft/s}}. \]The Shadow Tip Has a Different Speed
The distance from the lamp to the shadow tip is
\[ x+s. \]Its rate is
\[ \frac{d}{dt}(x+s) = \frac{dx}{dt} + \frac{ds}{dt}. \]Therefore,
\[ 4+\frac83 = \frac{20}{3}. \]The shadow tip moves away from the lamp at
\[ \boxed{\frac{20}{3}\text{ ft/s}}. \]The speed of the shadow's length and the speed of its tip are not the same quantity.
Distances Between Moving Objects
Suppose two objects move along perpendicular paths. If their horizontal and vertical separations are \(x\) and \(y\), and their direct separation is \(s\), then
\[ s^2=x^2+y^2. \]Differentiating gives
\[ 2s\frac{ds}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}. \]Thus,
\[ \boxed{ s\frac{ds}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt} }. \]Example: Separation of Two Moving Objects
Suppose one object is \(3\) km east of an intersection and moving east at \(4\) km/h. Another is \(4\) km north of the intersection and moving north at \(3\) km/h.
At that instant,
\[ x=3,\qquad y=4,\qquad s=5, \] \[ \frac{dx}{dt}=4, \qquad \frac{dy}{dt}=3. \]Use
\[ s\frac{ds}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}. \]Then
\[ 5\frac{ds}{dt} = 3(4)+4(3) = 24. \]Therefore,
\[ \boxed{ \frac{ds}{dt} = \frac{24}{5}\text{ km/h} }. \]Angular Related Rates
An angle can also change with time.
Suppose a fixed height \(H\), horizontal distance \(x\), and angle of elevation \(\theta\) satisfy
\[ \tan\theta=\frac{H}{x}. \]If \(x\) changes, then \(\theta\) changes as well.
Differentiating gives
\[ \sec^2\theta\frac{d\theta}{dt} = -\frac{H}{x^2}\frac{dx}{dt}. \]This equation relates linear motion to angular motion.
Radians Matter for Angular Rates
Standard derivative formulas for trigonometric functions assume angles are measured in radians.
Therefore, an angular rate such as
\[ \frac{d\theta}{dt} \]is naturally expressed in radians per unit time unless another unit is explicitly converted.
Example: Angle of Elevation
A vehicle moves away from the base of a \(100\)-meter tower at
\[ \frac{dx}{dt}=20\text{ m/s}. \]Let \(\theta\) be the angle of elevation from the vehicle to the top of the tower. Then
\[ \tan\theta=\frac{100}{x}. \]Differentiate:
\[ \sec^2\theta\frac{d\theta}{dt} = -\frac{100}{x^2}\frac{dx}{dt}. \]Finishing the Angle-of-Elevation Example
Suppose \(x=100\) m at the instant of interest.
Then
\[ \tan\theta=\frac{100}{100}=1, \]so
\[ \sec^2\theta=1+\tan^2\theta=2. \]Therefore,
\[ 2\frac{d\theta}{dt} = -\frac{100}{100^2}(20) = -\frac15. \]Hence
\[ \boxed{ \frac{d\theta}{dt} = -\frac{1}{10}\text{ rad/s} }. \]The angle decreases as the vehicle moves farther from the tower.
Rates Can Be Related Without Explicitly Solving First
In the ladder problem, we could solve
\[ y=\sqrt{169-x^2} \]before differentiating, but that is unnecessary.
Differentiating
\[ x^2+y^2=169 \]directly is often cleaner.
Related-rates problems frequently reward keeping the natural geometric relationship intact.
Sometimes Simplifying Before Differentiation Helps
In the conical-tank example, substituting the similarity relation
\[ r=\frac{h}{3} \]into the volume formula before differentiating removed an unnecessary changing variable.
The guiding principle is not “never substitute before differentiating.” It is:
Do not replace a genuinely changing quantity by its one-instant numerical value before differentiation.
Algebraic relationships that hold throughout the motion may be used to simplify the model first.
Common Error: Substituting an Instantaneous Value Too Early
Suppose
\[ A=\pi r^2 \]and \(r=5\) at the requested instant.
Replacing the relationship by
\[ A=25\pi \]before differentiating makes the area appear constant:
\[ \frac{dA}{dt}=0. \]That is wrong because \(r=5\) is only the current value, not a fixed radius for all time.
Common Error: Forgetting the Chain Rule
If
\[ x=x(t), \]then
\[ \frac{d}{dt}(x^2) = 2x\frac{dx}{dt}, \]not merely
\[ 2x. \]The factor \(\frac{dx}{dt}\) records that \(x\) itself is changing with time.
Common Error: Treating a Changing Quantity as Constant
If
\[ V=\pi r^2h \]and both \(r\) and \(h\) vary with time, then
\[ \frac{dV}{dt} = 2\pi rh\frac{dr}{dt} + \pi r^2\frac{dh}{dt}. \]Writing only the first term incorrectly assumes \(h\) is constant.
Common Error: Using the Wrong Sign
If the top of a ladder is sliding downward, then with upward height chosen as positive,
\[ \frac{dy}{dt}<0. \]If a radius is shrinking,
\[ \frac{dr}{dt}<0. \]Sign is part of the answer because it encodes direction.
Common Error: Solving for the Wrong Rate
In a shadow problem, the question may ask for:
- the person's speed;
- the rate at which the shadow length changes;
- the speed of the shadow tip.
These are different derivatives.
Define each variable before differentiating so the requested rate is unambiguous.
Common Error: Ignoring Units
If a sphere's radius is measured in centimeters, then
\[ \frac{dr}{dt} \]may be measured in centimeters per second, while
\[ \frac{dV}{dt} \]is measured in cubic centimeters per second.
The different dimensions are expected.
Common Error: Using a Formula That Does Not Match the Situation
A cylinder, cone, sphere, and rectangular tank have different volume formulas.
A right-triangle model is useful only when the geometry actually contains a right triangle.
Related rates cannot repair an incorrect model. The governing equation must correctly represent the physical or geometric relationship first.
Common Error: Mixing Information from Different Instants
Related rates are instantaneous.
If a problem gives values that apply when \(x=5\), those values belong to one snapshot of the motion. A value from a different time should not be combined with that snapshot unless the problem establishes that it remains valid.
A Table Can Organize the Information
Before calculating, it can help to separate values from rates.
| Quantity | Value at the instant | Rate |
|---|---|---|
| \(x\) | \(x=a\) | \(\frac{dx}{dt}\) |
| \(y\) | \(y=b\) | \(\frac{dy}{dt}\) |
This simple distinction prevents many setup errors.
A Compact Symbolic Pattern
Suppose changing quantities satisfy
\[ F(x,y)=0. \]Differentiating with respect to time gives
\[ F_x\frac{dx}{dt} + F_y\frac{dy}{dt} = 0, \]whenever the required partial derivatives exist.
This compact form shows the general structure behind many two-variable related-rates problems.
Solving the Compact Pattern for an Unknown Rate
If \(F_y\neq0\), then
\[ F_x\frac{dx}{dt} + F_y\frac{dy}{dt} = 0 \]can be solved for \(\frac{dy}{dt}\):
\[ \boxed{ \frac{dy}{dt} = -\frac{F_x}{F_y}\frac{dx}{dt} }. \]This is the time-rate analogue of the implicit derivative relationship
\[ \frac{dy}{dx} = -\frac{F_x}{F_y}. \]Related Rates and the Chain Rule
Suppose
\[ y=f(x) \]and \(x=x(t)\).
Then
\[ y=f(x(t)). \]The chain rule gives
\[ \boxed{ \frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} }. \]This formula is one of the clearest expressions of how one rate can be transmitted through a dependency relationship.
Rate Transmission Through a Function
In
\[ \frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}, \]the factor
\[ \frac{dy}{dx} \]measures how sensitively \(y\) responds to \(x\), while
\[ \frac{dx}{dt} \]measures how fast the input itself is changing.
Their product gives the resulting time rate of \(y\).
Estimation versus Exact Instantaneous Rate
Related-rates calculations use derivatives to determine exact instantaneous rates within the mathematical model.
A statement such as
\[ \frac{dV}{dt}=12\text{ L/min} \]describes the rate at one instant unless the problem states that the rate remains constant.
The quantity itself may follow a nonlinear path even when one of the component rates is constant.
Practice 1: Circle Area
The radius of a circle increases at \(2\) cm/s. How fast is its area increasing when the radius is \(7\) cm?
Practice 2: Shrinking Sphere
The radius of a sphere decreases at \(0.5\) cm/min. Find \(\frac{dV}{dt}\) when \(r=4\) cm.
Practice 3: Sliding Ladder
A \(10\)-foot ladder leans against a wall. The bottom moves away from the wall at \(3\) ft/s. How fast is the top moving when the bottom is \(6\) ft from the wall?
Practice 4: Changing Rectangle
A rectangle has \(L=8\) m and \(W=5\) m at a certain instant. Its length increases at \(2\) m/s while its width decreases at \(1\) m/s. Find the area rate.
Practice 5: Moving Separation
Two objects move away from the same intersection along perpendicular roads. One is \(6\) km east and moves east at \(5\) km/h. The other is \(8\) km north and moves north at \(2\) km/h. Find the rate at which their separation changes.
Practice 6: Conical Tank
An inverted conical tank has height \(9\) m and top radius \(3\) m. Water enters at \(4\) m\(^3\)/min. How fast is the water depth increasing when the water is \(3\) m deep?
Practice 7: Shadow Length
A \(5\)-foot person walks away from a \(15\)-foot lamp at \(6\) ft/s. How fast is the person's shadow length increasing?
Practice 8: Angle of Elevation
A vehicle moves away from a \(60\)-meter tower at \(15\) m/s. Find the rate of change of the angle of elevation to the top when the vehicle is \(80\) m from the tower.
Practice 9: Interpret the Sign
A calculation gives
\[ \frac{dh}{dt}=-1.4\text{ cm/s}. \]What does the negative sign mean if \(h\) measures height upward from the ground?
Practice 10: Identify the Error
A student studies a circle whose radius is \(5\) cm at one instant and is increasing. The student substitutes \(r=5\) into
\[ A=\pi r^2 \]first, obtains \(A=25\pi\), differentiates, and concludes
\[ \frac{dA}{dt}=0. \]Explain the error and state the correct differentiated relationship.
Practice Answers
Practice 1
Use
\[ A=\pi r^2. \]Then
\[ \frac{dA}{dt} = 2\pi r\frac{dr}{dt}. \]With \(r=7\) and \(\frac{dr}{dt}=2\),
\[ \boxed{ \frac{dA}{dt} = 28\pi\text{ cm}^2/\text{s} }. \]Practice 2
Since the radius decreases,
\[ \frac{dr}{dt}=-0.5. \]For
\[ V=\frac43\pi r^3, \]we have
\[ \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}. \]At \(r=4\),
\[ \boxed{ \frac{dV}{dt} = -32\pi\text{ cm}^3/\text{min} }. \]Practice 3
Let \(x\) be the bottom distance and \(y\) the height:
\[ x^2+y^2=100. \]When \(x=6\),
\[ y=8. \]Differentiate:
\[ x\frac{dx}{dt} + y\frac{dy}{dt} = 0. \]Substitute:
\[ 6(3)+8\frac{dy}{dt}=0. \]Therefore,
\[ \boxed{ \frac{dy}{dt} = -\frac94\text{ ft/s} }. \]Practice 4
Use
\[ A=LW. \]Then
\[ \frac{dA}{dt} = L\frac{dW}{dt} + W\frac{dL}{dt}. \]Substitute:
\[ \frac{dA}{dt} = 8(-1)+5(2)=2. \]Thus,
\[ \boxed{ \frac{dA}{dt} = 2\text{ m}^2/\text{s} }. \]Practice 5
The separation is
\[ s=\sqrt{6^2+8^2}=10. \]Differentiate
\[ s^2=x^2+y^2: \] \[ s\frac{ds}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}. \]Therefore,
\[ 10\frac{ds}{dt} = 6(5)+8(2) = 46. \]Hence,
\[ \boxed{ \frac{ds}{dt} = \frac{23}{5}\text{ km/h} }. \]Practice 6
Similarity gives
\[ \frac{r}{h} = \frac{3}{9} = \frac13, \]so
\[ r=\frac{h}{3}. \]Therefore,
\[ V = \frac13\pi r^2h = \frac{\pi}{27}h^3. \]Differentiate:
\[ \frac{dV}{dt} = \frac{\pi}{9}h^2\frac{dh}{dt}. \]With \(\frac{dV}{dt}=4\) and \(h=3\),
\[ 4 = \pi\frac{dh}{dt}. \]Thus,
\[ \boxed{ \frac{dh}{dt} = \frac4\pi\text{ m/min} }. \]Practice 7
Let \(x\) be the person's distance from the lamp and \(s\) the shadow length. Similar triangles give
\[ \frac{15}{x+s} = \frac5s. \]Therefore,
\[ 15s=5x+5s, \] \[ 10s=5x, \] \[ s=\frac12x. \]Differentiate:
\[ \frac{ds}{dt} = \frac12\frac{dx}{dt} = \frac12(6). \]Hence,
\[ \boxed{ \frac{ds}{dt}=3\text{ ft/s} }. \]Practice 8
Use
\[ \tan\theta=\frac{60}{x}. \]Differentiate:
\[ \sec^2\theta\frac{d\theta}{dt} = -\frac{60}{x^2}\frac{dx}{dt}. \]At \(x=80\),
\[ \tan\theta=\frac{60}{80}=\frac34, \]so
\[ \sec^2\theta = 1+\frac{9}{16} = \frac{25}{16}. \]Therefore,
\[ \frac{25}{16}\frac{d\theta}{dt} = -\frac{60}{80^2}(15) = -\frac{9}{64}. \]Thus,
\[ \boxed{ \frac{d\theta}{dt} = -\frac9{100}\text{ rad/s} }. \]Practice 9
Since upward height is positive,
\[ \boxed{ \frac{dh}{dt}=-1.4\text{ cm/s} } \]means the height is decreasing at \(1.4\) cm/s. The object or level is moving downward relative to the chosen coordinate.
Practice 10
The value \(r=5\) applies only at one instant. The radius is not constant throughout the motion, so substituting \(r=5\) before differentiation incorrectly removes its time dependence.
Differentiate the general relationship first:
\[ A=\pi r^2 \] \[ \boxed{ \frac{dA}{dt} = 2\pi r\frac{dr}{dt} }. \]The instantaneous value \(r=5\) can then be substituted together with the radius rate.
What to Remember
Related rates connect derivatives of quantities that change together.
The core pattern is:
- identify the changing quantities;
- write an equation relating them;
- differentiate with respect to time;
- substitute values from the requested instant;
- solve for the unknown rate;
- interpret the sign and units.
Keep these distinctions clear:
- a value is not the same as a rate;
- a quantity that changes must be treated as a function of time;
- the chain rule introduces factors such as \(\frac{dx}{dt}\);
- fixed constants may be used immediately;
- one-instant numerical values should usually be substituted after differentiation;
- similarity and geometry may be needed before calculus begins;
- positive and negative signs communicate direction;
- units are part of the final interpretation.
The reusable idea is:
\[ \boxed{\text{relationship among quantities} \;\Longrightarrow\; \text{relationship among their rates}}. \]Where We Go Next
Related rates use derivatives to describe how linked quantities change at the same instant. Derivatives can also be used to replace a curved function locally by a simpler linear model.
The next lesson, Linearization and Differentials, develops tangent-line approximations, differential notation, propagation of small changes, and local estimates built from derivative information.