\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

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If you haven't already, we recommend that you read the related rates basics page for information on how to get started on related rates problems.

This page covers related rates problems specifically involving 2-dim areas. [3-dim areas, like surface area of spheres, are covered on the volumes page.] When solving these types of problems, you first draw a picture and pick out the type of geometric figure involved. Once you have a figure with all the distances labeled, you can write down the equations involved.

What To Do With Constants In Related Rates Problems

What do you do with constants that are given in the problem? First of all, you never want to just go in and plug in all your constants before you take the derivative.

Safe Answer - - Wait and plug in your constants only after you have the derivative. So, you would label all distances with variables, take the derivative with respect to t and then plug in all your given constants. This is what you need to do when you first start learning to work related rates problems. After you have some experience, you can go on to the more experienced technique.

Experienced Answer - - Once you learn the basics of related rates problems, you will have a feel for which constants you can plug in right away and which ones you can't. The difference you need to look for is
- if the variable is NOT changing, then you can substitute the constant in before taking the derivative;
- but, if the variable is changing over time, then you must wait until after you take the derivative before you can substitute the constant into the equation.

At this point, it will just confuse you more if we write down a bunch of theory on how to work these problems. You need to actually see one, then work many in order to see patterns. If you haven't already, read the main related rates page. Then, come back here, watch a video or two and try your hand at a few problems.

Circles

When you have a circle and the area of a circle is involved, you need the equation \(A=\pi r^2\) where \(r\) is the radius of the circle. Sometimes circumference is involved and that equation is \(C=2\pi r\).

Practice

The radius of a circle is decreasing at a rate of 4cm/min. How fast are the area and circumference of the circle changing when the radius is 8cm?

Problem Statement

The radius of a circle is decreasing at a rate of 4cm/min. How fast are the area and circumference of the circle changing when the radius is 8cm?

Solution

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When a circular metal plate is heated in an oven, it's radius increases at the rate of 0.02cm/min. At what rate is the plate's area increasing when the radius is 60cm?

Problem Statement

When a circular metal plate is heated in an oven, it's radius increases at the rate of 0.02cm/min. At what rate is the plate's area increasing when the radius is 60cm?

Solution

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A stone is dropped into a pond sending out circular ripples at a rate of 2ft/sec. How fast is the area enclosed by the ripples changing 10 seconds later?

Problem Statement

A stone is dropped into a pond sending out circular ripples at a rate of 2ft/sec. How fast is the area enclosed by the ripples changing 10 seconds later?

Final Answer

\(80\pi ft^2/sec\)

Problem Statement

A stone is dropped into a pond sending out circular ripples at a rate of 2ft/sec. How fast is the area enclosed by the ripples changing 10 seconds later?

Solution

2244 video

video by Michel vanBiezen

Final Answer

\(80\pi ft^2/sec\)

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After dropping a rock in a pond, the ripples form circles where the radius is changing at a rate of 2 meters per second. Calculate the rate of change of the area when the radius is 3 meters.

Problem Statement

After dropping a rock in a pond, the ripples form circles where the radius is changing at a rate of 2 meters per second. Calculate the rate of change of the area when the radius is 3 meters.

Solution

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Triangles

When the area of a triangle is involved, your first task is to determine if the triangle is a right triangle. If it is, then the area formula is easy, \( A = bh/2 \). If not, you will need to determine what formula will help you the most, usually one of the law of cosines or law of sines.

Practice

Two sides of a triangle (not necessarily a right triangle) are 5m and 8m in length and the angle between them is increasing at a rate of 0.06rad/sec. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is \(\pi/3\) rad.

Problem Statement

Two sides of a triangle (not necessarily a right triangle) are 5m and 8m in length and the angle between them is increasing at a rate of 0.06rad/sec. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is \(\pi/3\) rad.

Solution

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A 20ft ladder is leaning against a wall. The foot of the ladder is being pulled away from the wall at a rate of 5 ft/sec while the top of the ladder remains on the wall. What is the rate of change in the area beneath the ladder when the foot is 8ft from the wall?

Problem Statement

A 20ft ladder is leaning against a wall. The foot of the ladder is being pulled away from the wall at a rate of 5 ft/sec while the top of the ladder remains on the wall. What is the rate of change in the area beneath the ladder when the foot is 8ft from the wall?

Solution

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A 13ft ladder is leaning against a house. The ladder slides down the wall at a rate of 3ft/min. How fast is the ladder moving away from the base of the wall when the foot of the ladder is 5ft from the wall? How fast is the area of the triangle changing? How fast is the angle between the ladder and the ground changing?

Problem Statement

A 13ft ladder is leaning against a house. The ladder slides down the wall at a rate of 3ft/min. How fast is the ladder moving away from the base of the wall when the foot of the ladder is 5ft from the wall? How fast is the area of the triangle changing? How fast is the angle between the ladder and the ground changing?

Solution

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We have a non-right triangle. Side a, which is not changing, has length 10m and side b, also not changing, has length 15m. However, the angle \(\theta\) between sides a and b is changing at a rate of 0.02rad/sec. How fast is the area of the triangle changing when \(\theta\) is \(\pi/3\)?

Problem Statement

We have a non-right triangle. Side a, which is not changing, has length 10m and side b, also not changing, has length 15m. However, the angle \(\theta\) between sides a and b is changing at a rate of 0.02rad/sec. How fast is the area of the triangle changing when \(\theta\) is \(\pi/3\)?

Hint

\(\text{Area}=(1/2)ab\sin(C)\)

Problem Statement

We have a non-right triangle. Side a, which is not changing, has length 10m and side b, also not changing, has length 15m. However, the angle \(\theta\) between sides a and b is changing at a rate of 0.02rad/sec. How fast is the area of the triangle changing when \(\theta\) is \(\pi/3\)?

Final Answer

0.7 m2/sec

Problem Statement

We have a non-right triangle. Side a, which is not changing, has length 10m and side b, also not changing, has length 15m. However, the angle \(\theta\) between sides a and b is changing at a rate of 0.02rad/sec. How fast is the area of the triangle changing when \(\theta\) is \(\pi/3\)?

Hint

\(\text{Area}=(1/2)ab\sin(C)\)

Solution

2250 video

video by Michel vanBiezen

Final Answer

0.7 m2/sec

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Rectangles

Rectangles are probably the easiest of all. The area of a rectangle is length times width, i.e. \(A=lw\).

Practice

The side length of a square increases at a rate of 3in/s. How fast are the area and perimeter of the square changing when the side length is 5in?

Problem Statement

The side length of a square increases at a rate of 3in/s. How fast are the area and perimeter of the square changing when the side length is 5in?

Solution

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A screen saver displays the outline of a 3 cm by 2 cm rectangle and then expands the rectangle in such a way that the 2 cm side is expanding at the rate of 5 cm/sec and the proportions of the rectangle never change. How fast is the area of the rectangle increasing when the longer side is 12 cm?

Problem Statement

A screen saver displays the outline of a 3 cm by 2 cm rectangle and then expands the rectangle in such a way that the 2 cm side is expanding at the rate of 5 cm/sec and the proportions of the rectangle never change. How fast is the area of the rectangle increasing when the longer side is 12 cm?

Final Answer

120 cm2/sec

Problem Statement

A screen saver displays the outline of a 3 cm by 2 cm rectangle and then expands the rectangle in such a way that the 2 cm side is expanding at the rate of 5 cm/sec and the proportions of the rectangle never change. How fast is the area of the rectangle increasing when the longer side is 12 cm?

Solution

First, we assign variables and get equations for the terms given in the problem statement. It also may help to draw a figure.
Let's label the sides x and y and assign x to the longer side. This means that the side labeled y is expanding at the given rate of \(dy/dt = 5\) cm/sec.

Secondly, we set up the equations based on what they want us to calculate. They ask us for the rate of change of the area, so we need an area equation. Since this is a rectangle, the area is length times width. In terms of our variables, we have \(A=xy\). They tell us to determine \(dA/dt\) when \(x=12\) cm.

Now we can start to solve this equation. Notice that we can't substitute \(x=12\) and \(y=8\) until after we take the derivative since these variables are changing. If we look at the area equation, \(A=xy\), we have two variables. It is certainly possible to take the derivative at this point, we would need to use the product rule and we would end up with quite a complicated expression. It is easier to reduce this equation to one variable before taking the derivative. To do this, we need to know the relationship between x and y. This information was given in the problem statement saying that the rectangle is initially 3 cm by 2 cm. So we can set up a ratio.
\(\displaystyle{ \frac{3}{x} = \frac{2}{y} }\)
Now we need to determine which variable to solve for. Either one will work but, since we need to eventually solve for \(dA/dt\) and we are given \( dy/dt \), it will save time to keep the variable y and get rid of x. So we solve the ratio for x, giving us \(x = 3y/2\). Substituting this into the area equation, we have
\(\displaystyle{ A = xy = \frac{3y}{2}y = \frac{3y^2}{2} }\)

It is time to take the derivative with respect to t (since this is a related rates problem).
\(\displaystyle{ A = \frac{3y^2}{2} ~~~ \to ~~~ \frac{dA}{dt} = {3y}\frac{dy}{dt} }\).

Now we can substitute the given values, \(\displaystyle{ \frac{dy}{dt} = 5 }\). We are given that \(x=12\) but we don't have any x's in the equation. However, we can use the ratio to find the y value, i.e. \(y = 2x/3 = 2(12)/3 = 8\). So now we are ready to calculate the answer.

\(\displaystyle{ \frac{dA}{dt} = {3y}\frac{dy}{dt} = 3(8~cm)(5cm/sec) = 120 }\) cm2/sec

Final Answer

120 cm2/sec

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You CAN Ace Calculus

Topics You Need To Understand For This Page

derivatives

chain rule

implicit differentiation

basics of related rates

precalculus: word problems

For related rates problems involving similar triangles, it may help you to review how to set up the ratios. You can find a discussion of this on the similar triangles precalculus page.

Related Topics and Links

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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