• Differential Calculus
  • About This Book
  • I Functions and Models
  • II Limits and Derivatives
  • 2.2 The Limit of a Function
    • The Idea of a Limit
    • Left and Right Limits
    • Limits That Do Not Exist
    • Infinite Limits and Vertical Asymptotes
  • 2.3 Calculating Limits Using the Limit Laws
    • The Limit Laws
    • Direct Substitution
    • Limits of Trigonometric Functions
    • The Squeeze Theorem
  • 2.4 The Precise Definition of a Limit
    • Proof of Limit Laws
    • Proofs of Trigonometric Limits
    • Limits at Infinity by Definition
    • Proof of Direct Substitution Property
    • Proof of The Squeeze Theorem
  • 2.5 Continuity
    • Limits and Continuity
    • Types of Discontinuities
    • Properties of Continuous Functions
    • Proof of Continuity Laws
    • The Intermediate Value Theorem
    • Proof of The IVT
    • Continuity of Trigonometric Functions
  • 2.6 Limits at Infinity; Horizontal Asymptotes
    • Limits At Infinity
    • Limits at Infinity by Definition
  • 2.7 Derivatives and Rates of Change
    • Application: Tangent Lines
  • 2.8 The Derivative as a Function
    • Differentiability
    • Differentiability & Continuity
    • Notation
  • III Differentiation Rules
  • 3.1 Derivatives of Polynomials and Exponential Functions
    • Power Functions
    • Proof of Differentiation Rules
    • Derivatives of Exponential Functions
  • 3.2 The Product and the Quotient Rule
    • The Product Rule
    • Proof of the Product Rule
    • The Quotient Rule
    • Proof of the Quotient Rule
  • 3.3 Derivatives of Trigonometric Functions
    • Derivative of \(\sin(x)\) and \(\cos(x)\)
    • Other Trigonometric Derivatives
    • Proofs of Trigonometric Derivatives
    • Higher Derivatives
  • 3.4 The Chain Rule
    • Proof of The Chain Rule
  • 3.5 Implicit Differentiation
    • Derivatives of Inverse Trigonometric Functions
    • Proof of Inverse Trig Derivative Laws
  • 3.6 Derivatives of Logarithmic and Inverse TrigonometricFunctions
    • Logarithmic Derivatives
    • Proof of Logarithmic Derivative Laws
    • Logarithmic Differentiation
    • Proof of the Power Rule
  • 3.9 Related Rates
  • 3.10 Linear Approximations and Differentials
    • Linear Approximation
    • Differentials
  • IV Applications of Differentiation
  • 4.1 Maximum and Minimum Values
    • Critical Points
    • Maximum and Minimum Values
    • Proof of the Extreme Value Theorem
    • Finding Absolute Extrema

Calculus I

3.9 Related Rates

Related rates problems are an application of implicit differentiation. There are four basic steps to solve related rates problems:

  1. Draw a picture of the physical situation. Label any quantities you are given in the question and create a variable for any unknown quantities or quantities that change.
  2. Write an equation that relates the quantities of interest. Be sure to label as a variable any value that changes as the situation progresses. Do not substitute a number for it yet.
  3. To develop your equation, you will probably use either a simple geometric fact (ie, the equation of a circle or the Pythagorean theorem), a trigonometric function, the Pythagorean theorem, similar triangles.
  4. Take the derivative with respect to time of both sides of your equation. Use implicit differentiation. Solve for the unknown quantity.

The best way to learn how to do related rates problems is to actually do related rates problems.

Example: Air is being pumped into a spherical balloon at a rate of \(4 cm^3/min\). Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is \(18 cm\).

Solution: The value that we need to find is the rate at which the radius of the balloon is increasing. If the radius is \(r\), the rate at which the radius is increasing is \(\dfrac{dr}{dt}\).

The other things we know from the question is that the air being pumped into the balloon is \(4 cm^3/min\). If \(V\) is the volume of the balloon, then this quantity is \(\dfrac{dV}{dt} = 4\).

One last thing to note is that as time goes on is that the radius is going to changed. That is, the radius \(r\) is not fixed. We need to find \(\dfrac{dr}{dt}\) when the diameter is \(18 cm\) or when \(r = 9 cm\).

The equation that relates volume of a sphere to the diameter is \[V = \dfrac{4}{3} \pi r^3.\] Differentiating this equation with respect to time, we find \[\dfrac{dV}{dt} = \dfrac{4}{3} \pi \left(3r^2 \dfrac{dr}{dt} \right) = 4\pi r^2 \dfrac{dr}{dt}.\] We want to find \(\dfrac{dr}{dt}\) at the time when \(r = 9\). Plugging in the value of \(r\) and the value of \(\dfrac{dV}{dt}\), we get \[\begin{align*} \dfrac{dV}{dt} &= 4\pi r^2 \dfrac{dr}{dt}\\ 4 &= 4\pi (9)^2 \dfrac{dr}{dt}\\ \dfrac{1}{81\pi} &= \dfrac{dr}{dt}. \end{align*}\] Therefore, the rate at which the radius is increasing is \[\dfrac{dr}{dt} = \dfrac{1}{81\pi}.\]

Example: You are pouring pancake batter onto a hot griddle. If the pancake has is \(10 cm\) across and our pour increases the area at a rate of 10 \(cm/s\), how quickly is the pancake widening?

Solution:The value that we need to find is the rate at which the diameter of the pancake is increasing. If the diameter is \(d\), the rate at which the radius is increasing is \(\dfrac{dd}{dt}\).

The other things we know from the question is that the area of the pancake is increasing at a rate of \(10 cm/s\). If \(A\) is the area of the balloon, then this quantity is \(\dfrac{dA}{dt} = 10\).

One last thing to note is that as time goes on, the area is going to change. That is, the surface area of the pancake, \(A\), is not fixed. We need to find \(\dfrac{dd}{dt}\) when the diameter is \(10 cm\)..

The equation that relates area of a circle to the radius is \[A = \pi r^2.\] Since the radius is half of the diameter (\(r = d/2\)), we should actually work with the equation \[A = \pi (d/2)^2 = \dfrac{\pi}{4} d^2.\] Differentiating this equation with respect to time, we find \[\dfrac{dA}{dt} = \dfrac{\pi}{4} \left(2d \dfrac{dd}{dt} \right) = \dfrac{\pi}{2} d \dfrac{dd}{dt}.\] We want to find \(\dfrac{dd}{dt}\) at the time when \(d = 10 cm\). Plugging in the value of \(d\) and the value of \(\dfrac{dA}{dt}\), we get \[\begin{align*} \dfrac{dA}{dt} &= \dfrac{\pi}{2} d \dfrac{dd}{dt}\\ 10 &= \dfrac{\pi}{2} (10) \dfrac{dd}{dt}\\ \dfrac{2}{\pi} &= \dfrac{dd}{dt}. \end{align*}\] Therefore, the rate at which the diameter is increasing is \[\dfrac{dd}{dt} = \dfrac{2}{\pi}.\]

Example: The top of a 23 foot ladder, leaning against a vertical wall, is slipping down the wall at a rate of 4 feet per second. How fast is the bottom of the ladder sliding along the ground when the bottom of the ladder is 9 feet away from the base of the wall?

Solution: Let \(y\) denote the height of the top of the ladder from the bottom of the wall and \(x\) the distance from the bottom of the ladder to the wall. The value that we need to find is the rate at which the bottom of the ladder is sliding away from the wall, \(\dfrac{dx}{dt}\).

The other things we know from the question is that the length of the ladder is \(23\) foot long (\(h = 23\)) and that the rate at which the ladder is sliding down the wall is \(\dfrac{dy}{dt} = 4\).

One last thing to note is that as time goes on, the position of the top and bottom of the latter will change. So, both \(x\) and \(y\) are variable. However, the length of the latter remains constant. So, \(h = 23\) is constant. We need to find \(\dfrac{dx}{dt}\) when the \(x = 10\). Using Pythagoras, we know this means that \(y = \sqrt{23^2 - 10^2} = 20.71\).

The equation that relates the horizontal and vertical parts of a triangle is the Pythagorean theorem \[x^2 + y^2 = h^2.\] Since \(h\) is constant, \[x^2 + y^2 = 23^2.\] Differentiating this equation with respect to time, we find \[2x\dfrac{dx}{dt} + 2y \dfrac{dy}{dt} = 0.\] We want to find \(\dfrac{dx}{dt}\) at the time when \(x = 10\) and \(y = 20.17\). Plugging in these values, we get \[\begin{align*} 2x\dfrac{dx}{dt} + 2y \dfrac{dy}{dt} &= 0\\ 2(10)\dfrac{dx}{dt} + 2(20.71) (4) &= 0\\ \dfrac{dx}{dt} &= 8.284. \end{align*}\] Therefore, the rate at which the ladder slides away from the wall is \[\dfrac{dd}{dt} = 8.284 \text{feet per second}.\]

Practice Problems

  1. You are blowing air into a spherical balloon at a rate of \(5 in^3/s\). What is the rate of change of the surface area of the balloon at \(t=1s\) given that the balloon has a radius of \(6 in\)?

  2. A boat is pulled into a dock by a rope attached to the bow (front) of the boat and passing through a pulley that is \(1 m\) above the bow of the boat. If the rope is pulled at a rate of \(1 m/s\), how fast is the boat approaching the dock when it is \(9 m\) from the dock?

  3. Gravel is being dumped from a conveyor belt at a rate of \(28 ft^3/m\). The gravel is so coarse so that the pile forms in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is \(12 ft\) high? (Hint: the volume of a cone is given by \(V=\pi r^2 h/3\) where \(r\) is the base radius and \(h\) is the height).

  4. A \(15 ft\) ladder is resting against the wall. The bottom is initially \(12 ft\) away from the wall and is being pushed towards the wall at a rate of \(0.25 ft/sec\). How fast is the top of the ladder moving up the wall \(10 s\) after we start pushing?

  5. Two people are \(45 ft\) apart. One of them starts walking north at a rate so that the angle between the starting position of the two people and the current position of the two people is changing at a constant rate of \(0.01 rad/min\). At what rate is distance between the two people changing when \(\theta = 0.5 rads\)?

  6. A tank of water in the shape of a cone is leaking water at a constant rate of \(2ft^3/ hr\). The base radius of the tank is \(5 ft\) and the height of the tank is \(14 ft\). (a) At what rate is the depth of the water in the tank changing when the depth of the water is \(6 ft\)? (b) At what rate is the radius of the top of the water in the tank changing when the depth of the water is \(6 ft\)?

  7. Consider a weight hanging from a rope which stretches up to a pulley 10ft above the floor and then to your hand which is 3ft above the floor and 15ft horizontally to the right of the pulley. If you walk horizontally to the right at 2 ft/s, how fast does the weight rise?