Given the equation below, find d z / d t dz/dt d z / d t when d x / d t = 3 dx/dt=3 d x / d t = 3 , d y / d t = 2 dy/dt=2 d y / d t = 2 , x = 2 x=2 x = 2 , y = 4 , y=4, y = 4 , and z = 1 z=1 z = 1 . x 2 + y 2 − 3 z 2 = 125 x^2+y^2-3z^2=125 x 2 + y 2 − 3 z 2 = 125

− 14 3 -\frac{14}{3} − 3 14

− 16 3 -\frac{16}{3} − 3 16

A sphere is growing at a rate of 50 c m ⁡ 3 s 50\frac{\operatorname{\mathrm{cm}}^3}{s} 50 s cm 3 . At what rate is the radius of the sphere increasing when the radius is 5 c m ⁡ 5\operatorname{\mathrm{cm}} 5 cm ?

0.159 c m ⁡ s 0.159\frac{\operatorname{\mathrm{cm}}}{s} 0.159 s cm

1.57 c m s ⁡ 1.57\operatorname{\frac{\mathrm{cm}}{s}} 1.57 s cm

0.637 c m s ⁡ 0.637\operatorname{\frac{\mathrm{cm}}{s}} 0.637 s cm

0.318 c m s ⁡ 0.318\operatorname{\frac{cm}{s}} 0.318 s cm

A right tringle has a base of 10 c m ⁡ 10\operatorname{\mathrm{cm}} 10 cm and a height of 12 c m ⁡ 12\operatorname{\mathrm{cm}} 12 cm . The height of the right triangle is decreasing at a rate of 0.4 c m s ⁡ 0.4\operatorname{\frac{\mathrm{cm}}{s}} 0.4 s cm , at what rate is the area of the triangle decreasing?

− 2 c m 2 s ⁡ -2\operatorname{\frac{cm^2}{s}}^{} − 2 s c m 2

− 2 . 4 c m 2 s ⁡ -2\operatorname{.4\frac{cm^2}{s}}^{} − 2 .4 s c m 2

− 1 c m 2 s ⁡ -1\operatorname{\frac{cm^2}{s}}^{} − 1 s c m 2

− 4 c m 2 s ⁡ -4\operatorname{\frac{cm^2}{s}}^{} − 4 s c m 2

The perimeter of a rectangle is fixed at 30 c m ⁡ 30\operatorname{\mathrm{cm}} 30 cm . If the length L L L is increasing at a rate of 2 c m s ⁡ 2\operatorname{\frac{\mathrm{cm}}{s}} 2 s cm , for what value of L L L does the area start to decrease? Hint: the rectangle's area starts to decrease when the rate of change for the area is less than 0.

7.5\operatorname{\mathrm{cm}}"> L > 7.5 c m ⁡ L>7.5\operatorname{\mathrm{cm}} L > 7.5 cm

15\operatorname{\mathrm{cm}}"> L > 15 c m ⁡ L>15\operatorname{\mathrm{cm}} L > 15 cm

0\operatorname{\mathrm{cm}}"> L > 0 c m ⁡ L>0\operatorname{\mathrm{cm}} L > 0 cm

10\operatorname{cm}"> L > 10 cm ⁡ L>10\operatorname{cm} L > 10 cm

Two cars leave the same intersection and drive in perpendicular directions. Car A travels east at a speed of 60 m i h r 60\frac{mi}{hr} 60 h r mi , Car B travels north at a speed of 40 m i h r 40\frac{mi}{hr} 40 h r mi . Car A leaves the intersection at 2 p m 2pm 2 p m , while Car B leaves at 2 : 30 p m 2:30pm 2 : 30 p m . Determine the rate at which the distance between the two cars is changing at 3 p m 3pm 3 p m .

69.62 m i h r 69.62\frac{mi}{hr} 69.62 h r mi

1.58 m i h r 1.58\frac{mi}{hr} 1.58 h r mi

18.99 m i h r 18.99\frac{mi}{hr} 18.99 h r mi

50.41 m i h r 50.41\frac{mi}{hr} 50.41 h r mi

5. Applications of Derivatives

Related Rates

5. Applications of Derivatives

Related Rates: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Implicit differentiation is crucial for solving related rates problems, where two or more variables change over time. To find rates like $d y / d t$ , take time derivatives of both sides of the equation, applying the chain rule. For example, the volume of a cube is given by $V = x^3$ . Understanding these concepts helps in tackling real-world applications, such as calculating the volume change of a melting ice cube.

concept

Intro To Related Rates

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Intro To Related Rates Video Summary

Implicit differentiation is a crucial technique in calculus that allows us to find the derivative of one variable with respect to another, especially when dealing with equations involving multiple variables. This concept becomes particularly useful in solving related rates problems, which often challenge students initially. Related rates problems involve situations where two or more variables change with respect to time, and understanding how these changes relate to one another is key to finding solutions.

To tackle a related rates problem, the first step is to differentiate both sides of the equation with respect to time. For example, consider the equation $ y = x^3 $. If we want to find the rate of change $ \frac{dy}{dt} $, we also need the rate of change of $ x $, denoted as $ \frac{dx}{dt} $. Given a specific value of $ x $, such as $ x = 4 $, we can proceed with the differentiation.

Applying the time derivative to both sides, we have:

\[\frac{dy}{dt} = \frac{d}{dt}(x^3) = 3x^2 \cdot \frac{dx}{dt}\]

Here, we used the chain rule, which is essential when differentiating with respect to time. The derivative of $ x^3 $ is $ 3x^2 $, and since $ x $ is also a function of time, we multiply by $ \frac{dx}{dt} $.

In this case, we can isolate $ \frac{dy}{dt} $ directly, as it is already on one side of the equation. Next, we substitute the known values into the equation. If $ x = 4 $ and $ \frac{dx}{dt} = 2 $, we can calculate:

\[\frac{dy}{dt} = 3(4^2)(2) = 3(16)(2) = 96\]

This result indicates that the rate of change of $ y $ with respect to time is 96. It’s important to note that while this example is straightforward, related rates problems can often be more complex, requiring careful analysis of the relationships between variables and their rates of change. As you continue to practice, you will become more comfortable with these types of problems and the underlying principles of implicit differentiation and related rates.

Problem

Given the equation below, find $dy/dt$ when $dx/dt=12$ and $x=\frac{15}{2}$ .
$y=\sqrt{2x+1}$

1.5

Problem

Given the equation below, find $dz/dt$ when $dx/dt=3$ , $dy/dt=2$ , $x=2$ , $y=4,$ and $z=1$ .
$x^2+y^2-3z^2=125$

$\frac{14}{3}$

$\frac{16}{3}$

$-\frac{14}{3}$

$-\frac{16}{3}$

concept

Changing Geometries

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Changing Geometries Video Summary

In related rates problems, we often need to find how one quantity changes with respect to time based on the relationship between different quantities. A common scenario involves changing geometries, such as a growing cube. For instance, if a cube's volume increases at a rate of 2 cubic centimeters per second, we may want to determine how fast the side length of the cube is growing when it measures 0.8 centimeters.

To tackle this problem, we start by identifying the relevant variables and relationships. The volume $ V $ of a cube is given by the formula:

\[ V = x^3 \]

where $ x $ is the length of one side of the cube. Since we know the rate of change of volume with respect to time, $ \frac{dV}{dt} = 2 \, \text{cm}^3/\text{s} $, we can differentiate both sides of the volume equation with respect to time:

\[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \]

Here, $ \frac{dV}{dt} $ represents the rate of change of volume, and $ \frac{dx}{dt} $ is the rate at which the side length is changing. To isolate $ \frac{dx}{dt} $, we rearrange the equation:

\[ \frac{dx}{dt} = \frac{\frac{dV}{dt}}{3x^2} \]

Next, we substitute the known values into this equation. We have $ \frac{dV}{dt} = 2 $ and $ x = 0.8 $ cm:

\[ \frac{dx}{dt} = \frac{2}{3(0.8)^2} \]

Calculating this gives:

\[ \frac{dx}{dt} = \frac{2}{3 \times 0.64} = \frac{2}{1.92} \approx 1.04 \, \text{cm/s} \]

This result indicates that the side length of the cube is growing at approximately 1.04 centimeters per second when the side length is 0.8 centimeters. The key takeaway is that in related rates problems, it is crucial to identify all relevant variables, establish the correct relationships, and apply differentiation appropriately to find the desired rates of change.

Problem

A sphere is growing at a rate of $50\frac{\operatorname{\mathrm{cm}}^3}{s}$ . At what rate is the radius of the sphere increasing when the radius is $5\operatorname{\mathrm{cm}}$ ?

$1.57\operatorname{\frac{\mathrm{cm}}{s}}$

$0.637\operatorname{\frac{\mathrm{cm}}{s}}$

$0.159\frac{\operatorname{\mathrm{cm}}}{s}$

$0.318\operatorname{\frac{cm}{s}}$

Problem

A right tringle has a base of $10\operatorname{\mathrm{cm}}$ and a height of $12\operatorname{\mathrm{cm}}$ . The height of the right triangle is decreasing at a rate of $0.4\operatorname{\frac{\mathrm{cm}}{s}}$ , at what rate is the area of the triangle decreasing?

$-2\operatorname{.4\frac{cm^2}{s}}^{}$

$-1\operatorname{\frac{cm^2}{s}}^{}$

$-2\operatorname{\frac{cm^2}{s}}^{}$

$-4\operatorname{\frac{cm^2}{s}}^{}$

Problem

The perimeter of a rectangle is fixed at $30\operatorname{\mathrm{cm}}$ . If the length $L$ is increasing at a rate of $2\operatorname{\frac{\mathrm{cm}}{s}}$ , for what value of $L$ does the area start to decrease? Hint: the rectangle's area starts to decrease when the rate of change for the area is less than 0.

$L>15\operatorname{\mathrm{cm}}$

$L>7.5\operatorname{\mathrm{cm}}$

$L>0\operatorname{\mathrm{cm}}$

$L>10\operatorname{cm}$

concept

Real World Application

Video duration:

Real World Application Video Summary

Related rates problems often involve real-world applications where a quantity changes over time. In these scenarios, it is essential to identify the relationships between the changing quantities and apply calculus concepts to find the rates of change. A common example is the melting of an ice cube, where we can analyze how the volume changes as the sides shrink.

Consider an ice cube melting uniformly, where each side decreases at a rate of -3 centimeters per minute. To find the rate of change of the ice cube's volume when each side measures 0.9 centimeters, we first recognize that the volume $ V $ of a cube is given by the formula:

\[ V = x^3 \]

where $ x $ is the length of one side of the cube. Since the ice cube is melting, we know that $ \frac{dx}{dt} = -3 $ cm/min, indicating that the side length is decreasing over time.

To find the rate of change of the volume $ \frac{dV}{dt} $, we differentiate the volume equation with respect to time using implicit differentiation:

\[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \]

Substituting the known values into this equation, we set $ x = 0.9 $ cm and $ \frac{dx}{dt} = -3 $ cm/min:

\[ \frac{dV}{dt} = 3(0.9)^2(-3) \]

Calculating this gives:

\[ \frac{dV}{dt} = 3(0.81)(-3) = -7.29 \text{ cm}^3/\text{min} \]

This negative result indicates that the volume of the ice cube is decreasing, which aligns with our expectation since the ice cube is melting. Thus, the rate of change of the volume of the ice cube is -7.29 cubic centimeters per minute.

In summary, when tackling related rates problems, it is crucial to identify the relationships between the variables, differentiate appropriately, and substitute known values to find the desired rates of change. This methodical approach allows for solving complex real-world problems effectively.

Problem

A 15-foot plank leans against a vertical pole. The top of the plank begins to slide down the pole at a steady speed of 2 inches per second. How fast is the bottom of the plank moving away from the pole when it is 8 feet away from the base of the pole (in inches per second)?

$dx/dt=1.58\frac{in}{s}$

$dx/dt=3.46\frac{in}{s}$

$dx/dt=5.04\frac{in}{s}$

$dx/dt=3.17\frac{in}{s}$

Problem

Two cars leave the same intersection and drive in perpendicular directions. Car A travels east at a speed of $60\frac{mi}{hr}$ , Car B travels north at a speed of $40\frac{mi}{hr}$ . Car A leaves the intersection at $2pm$ , while Car B leaves at $2:30pm$ . Determine the rate at which the distance between the two cars is changing at $3pm$ .

$69.62\frac{mi}{hr}$

$1.58\frac{mi}{hr}$

$18.99\frac{mi}{hr}$

$50.41\frac{mi}{hr}$

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Related Rates

5. Applications of Derivatives

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5. Applications of Derivatives

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Chapter

Here’s what students ask on this topic:

What is implicit differentiation and how is it used in related rates problems?

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. In related rates problems, implicit differentiation is crucial because it allows us to take the derivative of both sides of an equation with respect to time (t), even when the variables are interdependent. By applying the chain rule, we can find the rate of change of one variable with respect to time, given the rate of change of another variable. For example, if we have an equation like y = x³, we can differentiate both sides with respect to t to find dy/dt in terms of dx/dt.

How do you solve a related rates problem involving the volume of a cube?

To solve a related rates problem involving the volume of a cube, follow these steps: First, identify the variables and their rates of change. For a cube, the volume V is given by V = x³, where x is the side length. Next, take the time derivative of both sides of the equation using implicit differentiation: dV/dt = 3x²(dx/dt). Then, isolate the target rate of change (e.g., dV/dt) and plug in the known values for x and dx/dt. Finally, solve for the unknown rate. For example, if x = 0.8 cm and dx/dt = 2 cm/s, then dV/dt = 3(0.8)²(2) = 3.84 cm³/s.

What are some common real-world applications of related rates problems?

Related rates problems have numerous real-world applications. Some common examples include: calculating the rate at which water drains from a tank, determining the speed at which a shadow lengthens as a person walks away from a light source, finding the rate at which the volume of a melting ice cube decreases, and analyzing the rate of change of the area of a balloon as it inflates. These problems involve understanding how different quantities change over time and are essential in fields such as physics, engineering, and environmental science.

How do you approach solving a related rates problem with changing geometries?

To solve a related rates problem with changing geometries, follow these steps: First, draw and label a diagram of the scenario. Identify the variables and their rates of change. Next, write an equation that relates the variables. For example, for a sphere, the volume V is given by V = (4/3)πr³. Then, take the time derivative of both sides using implicit differentiation: dV/dt = 4πr²(dr/dt). Isolate the target rate of change and plug in the known values. Finally, solve for the unknown rate. This method helps in understanding how different geometric properties change over time.

What is the chain rule and how is it applied in related rates problems?

The chain rule is a fundamental calculus principle used to differentiate composite functions. It states that if a variable z depends on y, which in turn depends on x, then the derivative of z with respect to x is the product of the derivative of z with respect to y and the derivative of y with respect to x: dz/dx = (dz/dy)(dy/dx). In related rates problems, the chain rule is applied to differentiate equations with respect to time. For example, if y = x³, then dy/dt = 3x²(dx/dt). This allows us to find the rate of change of one variable in terms of another variable's rate of change.

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Related Rates: Videos & Practice Problems

Intro To Related Rates

Intro To Related Rates Video Summary

Given the equation below, find dy/dtdy/dtdy/dt when dx/dt=12dx/dt=12dx/dt=12 and x=152x=\frac{15}{2}x=215.y=2x+1y=\sqrt{2x+1}y=2x+1

Given the equation below, find dz/dtdz/dtdz/dt when dx/dt=3dx/dt=3 dx/dt=3, dy/dt=2dy/dt=2 dy/dt=2, x=2x=2x=2, y=4,y=4,y=4, and z=1z=1z=1.x2+y2−3z2=125x^2+y^2-3z^2=125x2+y2−3z2=125

Changing Geometries

Changing Geometries Video Summary

A sphere is growing at a rate of 50cm3s50\frac{\operatorname{\mathrm{cm}}^3}{s}50scm3. At what rate is the radius of the sphere increasing when the radius is 5cm5\operatorname{\mathrm{cm}}5cm?

A right tringle has a base of 10cm10\operatorname{\mathrm{cm}}10cm and a height of 12cm12\operatorname{\mathrm{cm}}12cm. The height of the right triangle is decreasing at a rate of 0.4cms0.4\operatorname{\frac{\mathrm{cm}}{s}}0.4scm, at what rate is the area of the triangle decreasing?

Real World Application

Real World Application Video Summary

A 15-foot plank leans against a vertical pole. The top of the plank begins to slide down the pole at a steady speed of 2 inches per second. How fast is the bottom of the plank moving away from the pole when it is 8 feet away from the base of the pole (in inches per second)?

Do you want more practice?

Here’s what students ask on this topic:

What is implicit differentiation and how is it used in related rates problems?

How do you solve a related rates problem involving the volume of a cube?

What are some common real-world applications of related rates problems?

How do you approach solving a related rates problem with changing geometries?

What is the chain rule and how is it applied in related rates problems?

Your Business Calculus tutors

Given the equation below, find $dy/dt$ when $dx/dt=12$ and $x=\frac{15}{2}$ .
$y=\sqrt{2x+1}$

Given the equation below, find $dz/dt$ when $dx/dt=3$ , $dy/dt=2$ , $x=2$ , $y=4,$ and $z=1$ .
$x^2+y^2-3z^2=125$

A sphere is growing at a rate of $50\frac{\operatorname{\mathrm{cm}}^3}{s}$ . At what rate is the radius of the sphere increasing when the radius is $5\operatorname{\mathrm{cm}}$ ?

A right tringle has a base of $10\operatorname{\mathrm{cm}}$ and a height of $12\operatorname{\mathrm{cm}}$ . The height of the right triangle is decreasing at a rate of $0.4\operatorname{\frac{\mathrm{cm}}{s}}$ , at what rate is the area of the triangle decreasing?