4.5 Solving Related Rates Problems

• Steps to solve related rates problems

  1️⃣ Identify the variables and their rates of change

  2️⃣ Draw a diagram if necessary

  3️⃣ Establish the equation relating the variables

  4️⃣ Differentiate the equation with respect to time

  5️⃣ Substitute the given values and solve for the unknown rate
• What is the notation for the rate of volume change?

  $\frac{dV}{dt}$
• Match the variable with its rate of change and notation:

  Distance ↔️ Speed, \frac{dx}{dt}</latex>

  Area ↔️ Rate of area change, $\frac{dA}{dt}$

  Volume ↔️ Rate of volume change, $\frac{dV}{dt}$
• Drawing a diagram highlights the relationships between variables and their rates of change.


  True
• Steps to solve a related rates problem using a diagram:

  1️⃣ Draw a diagram to visualize relationships

  2️⃣ Identify the variables and their rates of change

  3️⃣ Establish the equation relating the variables

  4️⃣ Differentiate the equation with respect to time
• Drawing a diagram can help visualize the relationships between variables, making it easier to set up the equation.


  True
• Match the variable with its rate of change notation:

  Distance ($x$) ↔️ $\frac{dx}{dt}$

  Area ($A$) ↔️ $\frac{dA}{dt}$

  Volume ($V$) ↔️ $\frac{dV}{dt}$
• What is the equation relating the area $A$ and radius $r$ of a circle?


  $A =$$\pi r^{2}$
• Match the shape with its area equation:

  Circle ↔️ $A =$$\pi r^{2}$

  Square ↔️ $A =$$s^{2}$

  Rectangle ↔️ $A =$$lw$

  Triangle ↔️ $A =$$\frac{1}{2}bh$
• The equation \frac{dA}{dt} = 2\pi r \frac{dr}{dt}</latex> relates the rate of change of area to the rate of change of radius for a circle.


  True
• If the radius of a circle is increasing at a rate of $\frac{dr}{dt} =$$3 \text{ cm / s}$, what is the rate of change of the area when $r =$$5 \text{ cm}$?


  $30\pi \text{ cm}^{2} / \text{s}$
• The rate of change of distance is called speed
• The variables in an expanding circle are radius and area.


  True
• What is the equation for the area of a circle?

  $A =$$\pi r^{2}$
• Drawing a diagram in related rates problems helps visualize the relationships between variables and their rates of change
• If the problem involves an expanding circle with radius r</latex> and area $A$, the relevant equation is $A =$$\pi r^{2}$, which relates the two variables
• When differentiating $A =$$\pi r^{2}$ with respect to time, the result is $\frac{dA}{dt} =$$2\pi r \frac{dr}{dt}$, which relates the rate of change of the area to the rate of change of the radius
• Drawing a diagram in related rates problems is crucial for visualizing the relationships between variables and their rates of change.


  True
• A diagram simplifies complex problems by visually representing the scenario, making it easier to understand.


  True
• What is the purpose of differentiating an equation with respect to time in related rates problems?

  Express rates of change
• Identify the initial values of the variables and any known rates of change
• What is the derivative of a variable with respect to time denoted as?

  $\frac{dx}{dt}$
• Drawing a diagram in related rates problems helps simplify complex problems.
• What is one benefit of drawing a diagram in related rates problems?

  Simplifies complex problems
• What is the first step in solving a related rates problem?

  Establish the equation
• Why is it necessary to differentiate the equation with respect to time in related rates problems?

  Express rates of change
• What are the rates of change in an expanding circle problem with radius $r$ and area $A$?


  $\frac{dr}{dt}$ and $\frac{dA}{dt}$
• Drawing a diagram is crucial in solving related rates problems because it helps visualize the relationships between the variables
• Establishing an equation to relate variables is crucial in solving related rates problems
• What are the steps to substitute given values and solve for the unknown rate in related rates problems?

  Identify givens, substitute, solve
• The rate of change of the area of a circle is $30\pi \text{ cm}^{2} / \text{s}$ when the radius is $5 \text{ cm}$ and increasing at a rate of $3 \text{ cm / s}$.


  True