Category Archives: Integration Applications

Challenge Answer

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I posted a challenge question on March 28, 2013. Only one person, “April,” replied with an explanation and it was a very good and succinct answer. You may read it here.

The conundrum resulted from trying to calculate when a parabolic bowl was half full. Using the “disk” method there was one answer; using the “shell” method resulted in a different answer. Both are correct; how can that be?

The explanation is this:

When you half fill the bowl in the usual way, by pouring water into it, the water accumulates in the bottom until the bowl is half full, as shown below. This is the result you get using the “disk” method.

Washer 2

But filling the bowl by “shells” is different. Think of spraying the y-axis with paint.  When the paint dries, spray another layer, and then another. Each layer is a “shell.” The paint accumulates as “shells” around the y-axis until the bowl is half full. The paint forms a cylinder with a rounded (parabolic) bottom that fills half the bowl, as shown below.

Shell 2

Geometry Counts

Does Simplifying Make Things Simpler?

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I taught a class today on volumes of solid of revolution; specifically, the ones with holes through them by the so-called “washer method.”  For this method you often see the formula

\displaystyle V=\pi \int_{a}^{b}{{{R}^{2}}-{{r}^{2}}}dx

Where R is “outside radius” and r is the “inside radius” and both are functions of x.

It seems to me that this “simplified” form is unduly complicated.

We first worked a problem where a region was rotated around its horizontal edge (Disk method). The region was between y=\sqrt{x}, and the line y=-3 between x = 0 and x = 4, revolved around y=-3.

\displaystyle V=\int_{0}^{4}{\pi {{\left( \sqrt{x}-\left( -3 \right) \right)}^{2}}dx}.

Then I asked them to change the region to that between the graph of y=\sqrt{x} and the line y=\tfrac{1}{2}x again revolved around y=-3. These graphs intersect at x = 0 and x = 4. (How convenient!).

Someone immediately had the idea to revolve the line only and subtract the answer from the last answer:

 \displaystyle V=\int_{0}^{4}{\pi {{\left( \sqrt{x}-\left( -3 \right) \right)}^{2}}dx}-\int_{0}^{4}{\pi {{\left( \tfrac{1}{2}x-\left( -3 \right) \right)}^{2}}dx}

Done!

Isn’t that good enough? Is there any need, ever, to set up the washer? Can’t you always subtract the inside volume from the outside volume?

Now I know that

\displaystyle \int_{0}^{4}{\pi {{\left( \sqrt{x}-\left( -3 \right) \right)}^{2}}dx}-\int_{0}^{4}{\pi {{\left( \tfrac{1}{2}x-\left( -3 \right) \right)}^{2}}dx}=

\displaystyle \pi \int_{0}^{4}{{{\left( \sqrt{x}-\left( -3 \right) \right)}^{2}}-}{{\left( \tfrac{1}{2}x-\left( -3 \right) \right)}^{2}}dx

And the latter is shorter and simpler to look at – only one \pi and only one integral sign, but which is really easier to understand and set up? Which shows you really what you’re doing?

Just sayin’.

Challenge

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A student was asked to find the volume of the bowl-shaped figure generated when the curve y = x2 from x = 0 to x = 2 is revolved around the y-axis. She used the disk method and found the volume to be \displaystyle \int_{0}^{4}{\pi ydy}=8\pi . To check her work she used the method of cylindrical shells and found the same answer: \displaystyle\int_{0}^{2}{2\pi x\left( 4-{{x}^{2}} \right)dx}=8\pi .

The second part of the question asked for the value of x for which the bowl would be half full. So she first solved the equation \displaystyle \int_{0}^{k}{\pi ydy=\frac{1}{2}\cdot 8\pi }  and found k=2\sqrt{2}. This is a y-value so the corresponding x-value is \sqrt{2\sqrt{2}}\approx 1.682. She again checked her work by shells by solving \displaystyle \int_{0}^{k}{2\pi x\left( 4-{{x}^{2}} \right)dx=\frac{1}{2}\cdot 8\pi } and found that k=\sqrt{-2\left( \sqrt{2}-2 \right)}\approx 1.082. (This is the x-value.)

Both computations are correct. Can you explain to her why her answers are different? Use the comment box below to share your explanations. I will post mine in a week or so.

Motion on a Line

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AP Type Questions 4

These questions may give the position equation, the velocity equation or the acceleration equation of something that is moving, along with an initial condition. The questions ask for information about motion of the particle: its direction, when it changes direction, its maximum position in one direction (farthest left or right), its speed, etc.  

The particle may be a “particle,” a person, car, a rocket, etc.  Particles don’t really move in this way, so the equation or graph should be considered to be a model. The question is a versatile way to test a variety of calculus concepts.

The position, velocity or acceleration may be given as an equation, a graph or a table; be sure to use examples of all three forms during the review. 

Many of the concepts related to motion problems are the same as those related to function and graph analysis. Stress the similarities and show students how the same concepts go by different names. For example, finding when a particle is “farthest right” is the same as finding the when a function reaches its “absolute maximum value.” See my post for November 16, 2012 for a list of these corresponding terms.

The position, s(t), is a function of time. The relationships are

  • The velocity is the derivative of the position, {s}'\left( t \right)=v\left( t \right). Velocity is has direction (indicated by its sign) and magnitude. Technically, velocity is a vector; the term “vector” will not appear on the AB exam.
  • Speed is the absolute value of velocity; it is a number, not a vector. See my post for November 19, 2012.
  • Acceleration is the derivative of velocity and the second derivative of position, \displaystyle a\left( t \right)={v}'\left( t \right)={{s}'}'\left( t \right). It, too, has direction and magnitude and is a vector.
  • Velocity is the antiderivative of the acceleration
  • Position is the antiderivative of velocity.

What students should be able to do:

  • Understand and use the relationships above.
  • Distinguish between position at some time and the total distance traveled during the time
    • The total distance traveled is the definite integral of the speed \displaystyle \int_{a}^{b}{\left| v\left( t \right) \right|}\,dt:
    • The net distance (displacement) is the definite integral of the velocity (rate of change): \displaystyle \int_{a}^{b}{v\left( t \right)}\,dt
    • The final position is the initial position plus the definite integral of the rate of change from x = a to x = t: \displaystyle s\left( t \right)=s\left( a \right)+\int_{a}^{t}{v\left( x \right)}\,dx Notice that this is an accumulation function equation.
  • Initial value differential equation problems: given the velocity or acceleration with initial condition(s) find the position or velocity. These are easily handled with the accumulation equation in the bullet above.
  • Find the speed at a particular time. The speed is the absolute value of the velocity.
  • Find average speed, velocity, or acceleration
  • Determine if the speed is increasing or decreasing.
    • If at some time, the velocity and acceleration have the same sign then the speed is increasing.
    • If they have different signs the speed is decreasing.
    • If the velocity graph is moving away from (towards) the t-axis the speed is increasing (decreasing).
  • Use a difference quotient to approximate derivative
  • Riemann sum approximations
  • Units of measure
  • Interpret meaning of a derivative or a definite integral in context of the problem

Shorter questions on this concept appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

For some previous posts on this subject see November 16, 19, 2012, January 21, 2013. There is also a worksheet on speed here and on the Resources pages (click at the top of this page).

The BC topic of motion in a plane, (parametric equations and vectors) will be discussed in a later post (March 15, 2013, tentative date)

Area and Volume Questions

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AP Type Questions 4

Given equations that define a region in the plane students are asked to find its area and the volume of the solid formed when the region is revolved around a line or used as a base of a solid with regular cross-sections. This standard application of the integral has appeared every year since 1969 on the AB exam and all but one year on the BC exam.

If this appears on the calculator active section, it is expected that the definite integrals will be evaluated on a calculator. Students should write the definite integral with limits on their paper and put its value after it.  It is not required to give the antiderivative and if students give an incorrect antiderivative they will lose credit even if the final answer is (somehow) correct.

There is a calculator program available that will give the set-up and not just the answer so recently this question has been on the no calculator allowed section. (The good news is that the integrals will be easy or they will be set-up but do not integrate questions.)

What students should be able to do:

  • Find the intersection(s) of the graphs and use them as limits of integration (calculator equation solving). Write the equation followed by the solution; showing work is not required. Usually no credit is earned until the solution is used in context (as a limit of integration).
  • Find the area of the region between the graph and the x-axis or between two graphs.
  • Find the volume when the region is revolved around a line, not necessarily an axis, by the disk/washer method. (Shell method is never necessary, but is eligible for full credit if properly used).
  • Find the volume of a solid with regular cross-sections whose base is the region between the curves. But see 2009 AB 4(b)
  • Find the equation of a vertical line that divides the region in half (area or volume). This involves setting up and solving an integral equation where the limit is the variable for which the equation is solved.
  • For BC only – find the area of a region bounded by polar curves:

\displaystyle A=\tfrac{1}{2}{{\int_{{{t}_{1}}}^{{{t}_{2}}}{\left( r\left( t \right) \right)}}^{2}}dt

Shorter questions on this concept appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

For some previous posts on this subject see January 9, 11, 2013

The Rate/Accumulation Question

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AP Type Questions 1

These questions are often in context with a lot of words describing a situation in which some things are changing. There are usually two rates given acting in opposite ways. Students are asked about the change that the rates produce over some time interval either separately or together.

The rates are often fairly complicated functions. If they are on the calculator allowed section, students should store them in the equation editor of their calculator and use their calculator to do any integration or differentiation that may be necessary.

The integral of a rate of change is the net amount of change

\displaystyle \int_{a}^{b}{{f}'\left( t \right)dt}=f\left( b \right)-f\left( a \right)

over the time interval [a, b]. If the question asked for an amount, look around for a rate to integrate.

The final accumulated amount is the initial amount plus the accumulated change:

\displaystyle f\left( x \right)=f\left( {{x}_{0}} \right)+\int_{{{x}_{0}}}^{x}{{f}'\left( t \right)}\,dt,

where {{x}_{0}} is the initial time, and  f\left( {{x}_{0}} \right) is the initial amount. Since this is one of the main interpretations of the definite integral the concept may come up in a variety of situations.

What students should be able to do:

  • Be ready to read and apply; often these problems contain a lot of writing which needs to be carefully read.
  • Recognize that rate = derivative.
  • Recognize a rate from the units given without the words “rate” or “derivative.”
  • Find the change in an amount by integrating the rate. The integral of a rate of change gives the amount of change (FTC):

\displaystyle \int_{a}^{b}{{f}'\left( t \right)dt}=f\left( b \right)-f\left( a \right).

  • Find the final amount by adding the initial amount to the amount found by integrating the rate. If x={{x}_{0}} is the initial time, and f\left( {{x}_{0}} \right)  is the initial amount, then final accumulated amount is

\displaystyle f\left( x \right)=f\left( {{x}_{0}} \right)+\int_{{{x}_{0}}}^{x}{{f}'\left( t \right)}\,dt,

  • Understand the question. It is often not necessary to as much computation as it seems at first.
  • Use FTC to differentiate a function defined by an integral.
  • Explain the meaning of a derivative or its value in terms of the context of the problem.
  • Explain the meaning of a definite integral or its value in terms of the context of the problem. The explanation should contain (1) what it represents, (2) its units, and (3) how the limits of integration apply in context.
  • Store functions in their calculator recall them to do computations on their calculator.
  • If the rates are given in a table, be ready to approximate an integral using a Riemann sum or by trapezoids.
  • Do a max/min or increasing/decreasing analysis.

Shorter questions on this concept appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

For some previous posts on this subject see January 21, 23, 2013

Interpreting Graphs

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AP Type Questions 1

The long name is “Here’s the graph of the derivative, tell me things about the function.”

Most often students are given the graph identified as the derivative of a function. There is no equation given and it is not expected that students will write the equation (although this may be possible); rather, students are expected to determine important features of the function directly from the graph of the derivative. They may be asked for the location of extreme values, intervals where the function is increasing or decreasing, concavity, etc. They may be asked for function values at points.

The graph may be given in context and student will be asked about that context. The graph may be identified as the velocity of a moving object and questions will be asked about the motion and position. (Motion problems will be discussed as a separate type in a later post.)

Less often the function’s graph may be given and students will be asked about its derivatives.

What students should be able to do:

  • Read information about the function from the graph of the derivative. This may be approached as a derivative techniques or antiderivative techniques.
  • Find where the function is increasing or decreasing.
  • Find and justify extreme values (1st  and 2nd derivative tests, Closed interval test, aka.  Candidates’ test).
  • Find and justify points of inflection.
  • Find slopes (second derivatives, acceleration) from the graph.
  • Write an equation of a tangent line.
  • Evaluate Riemann sums from geometry of the graph only.
  • FTC: Evaluate integral from the area of regions on the graph.
  • FTC: The function, g(x), maybe defined by an integral where the given graph is the graph of  the integrand, f(t), so students should know that if,  \displaystyle g\left( x \right)=g\left( a \right)+\int_{a}^{t}{f\left( t \right)dt} then  {g}'\left( x \right)=f\left( x \right)  and  {{g}'}'\left( x \right)={f}'\left( x \right).

The ideas and concepts that can be tested with this type question are numerous. The type appears on the multiple-choice exams as well as the free-response. They have accounted for almost 25% of the points available on recent test. It is very important that students are familiar with all of the ins and outs of this situation.

As with other questions, the topics tested come from the entire year’s work, not just a single unit. In my opinion many textbooks do not do a good job with these topics.

Study past exams; look them over and see the different things that can be asked.

For some previous posts on this subject see October 15, 17, 19, 24, 26, 2012 January 25, 28, 2013