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}} . If the length L L is increasing at a rate of 2 c m s ⁡ 2\operatorname{\frac{\mathrm{cm}}{s}} , for what value of 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}}

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

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

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

4. Applications of Derivatives

Related Rates

4. 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, in a cube's volume problem, use the equation $V = x^3$ and differentiate to find $d V / d t$ . This method applies to real-world scenarios, such as melting ice cubes, where understanding the relationship between changing dimensions and volume is essential.

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, particularly when dealing with equations involving multiple variables. This concept becomes especially important when tackling 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 solving them.

To approach 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 $ (for instance, $ x = 4 $) and the rate $ \frac{dx}{dt} = 2 $, we can proceed with 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} $.

Next, we can substitute the known values into the equation. With $ x = 4 $ and $ \frac{dx}{dt} = 2 $, we 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 relatively straightforward, related rates problems can become more complex, often requiring additional steps to isolate the target rate of change or to derive the necessary relationships between variables.

As you continue to explore related rates, remember that the fundamental process remains consistent: differentiate with respect to time, isolate the desired rate, and substitute known values to solve. With practice, these problems will become more manageable, and you'll develop a deeper understanding of how variables interact over time.

Problem

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

1.5

Problem

Given the equation below, find $d z / d t$ when $d x / d t = 3$ , $d y / d t = 2$ , $x equals 2$ , $y equals 4 comma$ and $z equals 1$ .
$x^{2} + y^{2} - 3 z^{2} = 125$

$\frac{14}{3}$

$\frac{16}{3}$

$- \frac{14}{3}$

$- \frac{16}{3}$

concept

Changing Geometries

Video duration:

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 the side length is 0.8 centimeters.

To tackle this problem, we start by identifying the relevant variables. 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. The rate of change of volume with respect to time is denoted as $ \frac{dV}{dt} $, and in this case, it is given as 2 cubic centimeters per second. We are looking for the rate of change of the side length with respect to time, represented as $ \frac{dx}{dt} $.

Next, we differentiate both sides of the volume equation with respect to time:

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

Here, we apply the chain rule, where $ \frac{dV}{dt} $ is the derivative of volume with respect to time, and $ 3x^2 \frac{dx}{dt} $ comes from differentiating $ x^3 $. Now, we can isolate $ \frac{dx}{dt} $:

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

Substituting the known values into this equation, we have:

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

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 while related rates problems can sometimes seem complex, breaking them down into identifiable variables and relationships can simplify the process significantly. Drawing a diagram and clearly labeling the known quantities can also aid in visualizing the problem and finding the correct equations to use.

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 $30 c m$ . If the length $L$ is increasing at a rate of $2 \frac{c m}{s}$ , 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.

$L is greater than 15$

$L is greater than 7.5$

$L is greater than 0$

$L is greater than 10 cm$

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 different variables and how they evolve. A common example is the melting of an ice cube, where we can analyze the rate of change of its volume as its sides shrink.

Consider an ice cube melting uniformly, with each side decreasing at a rate of -3 centimeters per minute. At a specific moment, each side measures 0.9 centimeters. The negative sign indicates that the ice cube is indeed shrinking, which aligns with our understanding of the melting process.

To solve this problem, we first need to visualize the situation. Drawing a diagram helps clarify the relationships between the variables involved. Here, we focus on one side of the cube, denoted as x, which is changing over time. The rate of change of this side is represented as d x/dt = -3 cm/min, and at a particular instant, x = 0.9 cm.

Next, we need to establish the equation that relates the volume of the cube to its side length. The volume V of a cube is given by the formula:

V = x^3

To find the rate of change of the volume with respect to time, we differentiate both sides of the equation with respect to time t using implicit differentiation:

dV/dt = 3x^2 (dx/dt)

In this equation, dV/dt represents the rate of change of the volume, while dx/dt is the rate of change of the side length. We can now substitute the known values into the equation. At the moment when x = 0.9 cm, we have:

dV/dt = 3(0.9)^2 (-3)

Calculating this gives:

dV/dt = 3(0.81)(-3) = -7.29 cm³/min

Thus, the rate of change of the volume of the ice cube is -7.29 cm³/min. The negative result indicates that the volume is indeed decreasing, which is consistent with the physical scenario of a melting ice cube.

In summary, solving related rates problems requires careful identification of the variables involved, establishing the appropriate relationships through equations, and applying differentiation to find the desired rates of change. This process, while sometimes complex, follows a systematic approach that can be applied to various real-world situations.

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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More sets

Related Rates

4. Applications of Derivatives

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4. Applications of Derivatives

5 topics 10 problems

Chapter

Here’s what students ask on this topic:

What is the process for solving related rates problems in calculus?

To solve related rates problems, follow these steps: First, identify all the variables involved and how they change over time. Draw a diagram if necessary. Next, write an equation that relates these variables. Then, take the time derivative of both sides of the equation using implicit differentiation. This will involve applying the chain rule. After that, isolate the target rate of change. Finally, substitute the known values and solve for the unknown rate. This process helps in understanding how different variables are interrelated and how they change with respect to time.

How do you use implicit differentiation in related rates problems?

Implicit differentiation is used in related rates problems to find the derivative of one variable with respect to another when the variables are related by an equation. To use implicit differentiation, take the derivative of both sides of the equation with respect to time (t). Apply the chain rule to differentiate each term. For example, if you have an equation y = x^3, taking the derivative with respect to t gives you dy/dt = 3x^2 * dx/dt. This allows you to relate the rates of change of the variables involved.

Can you explain a related rates problem involving a growing cube?

Consider a cube whose volume is increasing at a rate of 2 cubic centimeters per second. We want to find the rate at which the side length of the cube is increasing when the side length is 0.8 cm. First, note that the volume V of a cube with side length x is given by V = x^3. Taking the derivative with respect to time t, we get dV/dt = 3x^2 * dx/dt. Given dV/dt = 2 cm³/s and x = 0.8 cm, we can solve for dx/dt: 2 = 3(0.8)² * dx/dt, which simplifies to dx/dt = 2 / (3 * 0.64) ≈ 1.04 cm/s.

How do you solve a related rates problem involving a melting ice cube?

To solve a related rates problem involving a melting ice cube, start by identifying the variables and their rates of change. Suppose each side of the ice cube is shrinking at a rate of 3 cm/min, and we want to find the rate at which the volume is decreasing when each side is 0.9 cm. The volume V of a cube with side length x is V = x^3. Taking the derivative with respect to time t, we get dV/dt = 3x^2 * dx/dt. Given dx/dt = -3 cm/min and x = 0.9 cm, we can solve for dV/dt: dV/dt = 3(0.9)² * (-3) = -4.86 cm³/min.

What are some common mistakes to avoid when solving related rates problems?

Common mistakes in related rates problems include: 1) Not correctly identifying all the variables and their relationships. 2) Forgetting to apply the chain rule when differentiating implicitly. 3) Not correctly isolating the target rate of change. 4) Plugging in values too early before differentiating. 5) Ignoring the units of measurement, which can lead to incorrect interpretations of the rates. Always ensure you follow the steps methodically and double-check your work for these common errors.

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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/dt when dx/dt=12dx/dt=12 and x=152x=\frac{15}{2}.y=2x+1y=\sqrt{2x+1}

Given the equation below, find dz/dtdz/dt when dx/dt=3dx/dt=3 , dy/dt=2dy/dt=2 , x=2x=2, y=4,y=4, and z=1z=1.x2+y2−3z2=125x^2+y^2-3z^2=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 the process for solving related rates problems in calculus?

How do you use implicit differentiation in related rates problems?

Can you explain a related rates problem involving a growing cube?

How do you solve a related rates problem involving a melting ice cube?

What are some common mistakes to avoid when solving related rates problems?

Your Calculus tutors

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

Given the equation below, find $d z / d t$ when $d x / d t = 3$ , $d y / d t = 2$ , $x equals 2$ , $y equals 4 comma$ and $z equals 1$ .
$x^{2} + y^{2} - 3 z^{2} = 125$

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?