Robert FusonChris Adams

Orchestra Gli Armonici, 100908 Concerto della Madonna dei fiori, 17 W.A.Mozart, KV618, Ave Verum Corpus

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Formulas You Need to KnowVolumes:

Cone: V=(1/3)∏r²hSphere: V=(4/3)∏r³

Areas:Circle: A=∏r²

Pythagorean Theorem:a²+b²=c²

Properties of Similar Triangles

What is Related Rates?In related rates problems, we are given the

rate of change of certain quantities, and are required to find the rate of change of related quantities.

So, in layman’s terms, related rates are problems where something (usually a distance or volume/area) is being changed over time.

General Steps for Related Rates1: Identify the shape being changed and sketch

it, labeling the correct parts with given information.

2: Identify the formula that relates the changing quantities.

3: Differentiate both sides with respect to time. (d_/dt)

4: Substitute the values given to you by the initial problem and solve for the desired differentiation (be sure to use the correct units for your solution).

When dealing with circles (as well as spheres since they are solved very similarly), all you need to know is the radius (or diameter, from which you can easily find the radius) and the rate of change for the area or the rate of change for the radius, depending on which you are being told to find.

First, you take the derivative of both sides form the initial equation.

Then you plug in ‘r’ and ‘dr/dt’ or ‘dA/dt’ and solve for the remaining rate of change.

A rock is dropped into a lake. Circular ripples spread over the surface of the water, with the radius of each circle increasing at the rate of 3/2 feet per second. Find the rate of change of the area inside the circle formed by a ripple at the instant the radius is 4 feet.

First step, notice that this problem is dealing with a circle. Now we, draw it, labeling the parts.

It gives us dr/dt=3/2. It tells us we are looking

for dA/dt when r=4 Obviously, we’ll use the

equation for area of a circle, A=∏r²

4ft

4ft

Right Triangles

When dealing with triangles, the Pythagorean theorem will be used frequently.

First, use it to find the missing side, since you will almost never be given all three sides of the triangle.

After finding the missing side, take the derivative of the Pythagorean theorem, then divide everything by 2 to simplify the equations.

Plug in the sides and the rates that you are given or can infer to be zero.

Solve for the desired rate

Problem #2

A 50–foot ladder is placed against a large building. The base of the ladder is resting on an oil spill, and it slips at the rated of 3 feet per minute. Find the rated of change of the height of the top of the ladder above the ground at the instant when the base of the ladder is 30 feet from the base of the building.

Problem Examination

This problem is dealing with the changing sides of a triangle, which we draw using given information.

dy/dt=5Since the length of the

ladder cannot change, dz/dt=0

First, we will have to use the Pythagorean theorem to find x. Then, we differentiate both sides and solve to find dx/dt.

50ft

30ft

x

Finding “x”

Because this is a 3²+4²=5² triangle, we can assume that x=40 ft.

However, to refresh, let’s run-through the old fashioned way.

30²+x²=50²900+x²=2500

X²=1600X=40

30ft

x

50ft

Solving for dx/dt

x²+y²=z²x(dy/dt)+y(dx/

dt)=z(dz/dt)30(3)+40(dx/dt)=50(0)

90+40(dx/dt)=0dx/dt=(-90)/40dx/dt=-(9/4)

dx/dt is negative because the distance between the top of the ladder and the ground is shrinking

30ft

40ft

50ft

Cones are easily the most confusing of the related rates, relying on the formula for volume of a cone, V=(1/3)∏r²h.

When given a problem for a cone, you will often receive the radius and height of the container and either the current radius or the current height of the contents of the container.

If this is not the case, you will receive the radius and height of a changing object, such as an icicle, which are you are usually asked to evaluate whether of not it is increasing or decreasing in volume, after being given an increasing rate of change for height, and a decreasing rate of change for radius, or visa-versa.

When you are given similar triangles, you have to compare the two radiuses to the two heights in order to find the relationship between the smaller two.

While you may have the number for either the smaller height or radius, do not plug it in at this point, for the important thing is finding the relationship between r and h.

To solve, set up as a proportion, 5/r=14/h, then cross multiply.

14r=5h From here, solve for either r or h,

usually r since most of the time in these problems, you are asked to find dh/dt.

r=(5/14)h or h=(14/5)r

http://tutorial.math.lamar.edu/Classes/CalcI/RelatedRates_files/image003.gif

After knowing the radius and height of the cone you’re solving for, whether it be a cone inside of a larger cone or the large cone, you are usually asked to solve for dA/dt, or dh/dt.

At this point, you either need to use your relationship between r and h from earlier to eliminate either h or r (usually r), or solve take the derivative of both sides, using the chain rule for the right side of the equation. Then, plug in your values and you will be able to easily determine your answer using simple algebra.

A cone-shaped icicle is dripping from the roof. The radius of the icicle is decreasing at a .2 centimeter per hour, while the length is increasing at a rate of .8 centimeter per hour. If the icicle is currently 4 centimeters in radius and 20 centimeters long, what is the volume rate of change of the icicle?

Here, we are dealing with a cone, so we draw and label the picture.

We know that dh/dt=.8cm/hr and dr/dt=-.2cm/hr (dr/dt is negative because it is decreasing)

We will use the volume of a cone formula V=(1/3)r²h, where we are looking for dV/dt

4cm

20cm

20cm

4cm

One car leaves a given point and travels north at 30 mph. Another car leaves the same point at the same time and travels west at 40 mph. at what rate is the distance between the two cars changing at the instant when the cars have traveled 2 hours?

Try Me! 1 (work page)

A pulley is on the edge of a dock, 15 feet above the water level. A rope is being used to pull in a boat. The rope is attached to the boat at water level. The rope is being pulled in at the rate of 1 foot per second. Find the rate at which the boat is approaching the dock at the instant the boat is 20 ft from the dock.

Try Me! 2 (work page)

A spherical snowball is placed in the sun. The sun melts the snowball so that its radius decreases ¼ inch per hour. Find the rate of change of the volume with respect to time at the instant the radius is 4 inches.

Try Me! 3 (work page)

1971 AB 2

Let y = 2ecos(x)

Calculate and

If x and y both vary with time in such a way that y increases at a steady rate of 5 units per second, at what rate is x changing when

Part 1

Part 1 Explanation

For the first derivative you derive it like you would for any exponential function.

For the second derivative you use the Product Rule with a combination of an exponential function.

Part 2

Part 2 Explanation

Use the first derivative that was found in part 1 but with a respect to time.

Then plug what was given in the second part of the FRQ into the derivative.

Finally solve for the rate of change of x.

BibliographyBibliography

http://people.hofstra.edu/stefan_waner/realworld/tutorials/frames4_4.html

http://tutorial.math.lamar.edu/Classes/CalcI/RelatedRates.aspx

Lial, Greenwell, Ritchey, First. Calculus With Application. Seventh. 2002. Print.

©Chris Adams, Robert Fuson