Master AP Calculus AB & BC
Part II. AP CALCULUS AB & BC REVIEW
CHAPTER 6. Applications of the Derivative
• Related rates
• Hands-On Activity 6.1: Rolie’s and mean value theorems
• Hands-On Activity 6.2: The first derivative test
• Motion in the plane (BC topic only)
• Technology: Modeling a particle’s movement with a graphing calculator
• Summing it up
By this time, you are beginning to form an opinion about derivatives. Either you like them or you don’t. Hopefully, the two of you are at least on speaking terms. If you are, you are going to be very impressed by the things derivatives can do. They can even help you in the real world, which sometimes surprises math students. Before we get into these topics, it’s time to decide whether or not you are a derivative fan.
You Might Love Derivatives If...
• You have an oversized foam hand that reads “Derivatives #1!”
• When your friend flipped a coin the other day, you said, “That’s what I call a rate of change.” And no one laughed but you.
• You love it when people go off on tangents.
• You took up skiing so you could learn more about slopes.
• You loudly commented at the grocery store the other day that, “This express line looks more like a linear approximation.”
• You always respond to chain letters because they remind you of the Chain Rule.
• You have a giant tattoo of the Quotient Rule on your back.
You already know that a function or equation shows a clear relationship between the variables involved. For example, in the linear equation y = 3x — 2, each ordered pair has an x that is 2 less than 3 times as large as y. What you may not know is that when you find the derivative of such an equation with respect to time, you find another relationship—one between the rates of change of the variables. Back to our equation: If you find the derivative of y = 3x — 2 with respect to t, you get
This means that y is changing at a rate 3 times faster than x is changing. This makes sense, because the derivative, dy/dx, is 3/1. For every one unit you travel to the right, you must travel up 3 to stay on the graph. These types of problems are called related rates (for obvious reasons).
As we progress through the following examples, we will be closely following the plan below. Get used to the chronology of these steps—the method of solving related rates problems always follows the same pattern.
TIP. Of all the topics in AP Calculus, students often forget how to do related rates by test time. Take some extra time and make sure that your understanding is complete.
5 Steps to Success with Related Rates
1. Identify which rate you are trying to find and what information is given to you.
2. Find an equation that relates the variables to one another if you’re not given one.
3. Eliminate extra variables, if at all possible, by substituting in for them (see Example 3).
4. Find the derivative of the equation with respect to t.
5. Plug in what you know, and solve for the required rate.
Example 1: My brother, Dave, and I recently went golfing. After a promising start, he landed 3 consecutive balls in the lake in front of the second green. As the first ball entered the water, it caused a multitude of ripples in the form of concentric circles emanating from the point of impact at a steady rate of 3/4 ft/sec.
(a) What was the rate of change of the area of the outermost ripple when its radius was 3 feet?
NOTE. dy/dt is interpreted as the rate of change of y with respect to time, and dx/dt is the rate of change of x with respect to time.
This question concerns area and radius; both of these elements are contained by the formula for area of a circle: A = πr2. To find the relationship between the rates, find the derivative with respect to t:
You are trying to find dA/dt, the rate of increase of A. You know that according to the given information, and the problem prompts you that r = 3 in this instance. Substitute these values into the equation to solve:
(b) What was the rate of change of the golf club he threw with a mighty initial velocity after drowning the third ball? No one knows, but everyone knew to stay out of the way.
ALERT! If something is decreasing or becoming smaller, its rate of change will be negative.
As Example 1 illustrates, make sure that you include the correct units in your final answer when units are included in the problem. Following are the most commonly requested rates (assuming that the problem includes meters and seconds): area (m2/sec), volume (m 3/sec), length or velocity (m/sec), and acceleration (m/sec2). If the problem contains units other than meters and seconds, the format is still the same.
ALERT! You can only insert a constant into your primary equation in related rates if that constant cannot change throughout the problem. For example, the ladder is constant in Example 2.
Example 2: While painting my house and atop a 25-foot ladder, I was horrified to discover that the ladder began sliding away from the base of my home at a constant rate of 2 ft/sec (don’t ask me how I knew that, I just did). At what rate was the top of the ladder carrying me, screaming like a 2-year-old child, toward the ground when the base of the ladder was already 17 feet away from the house?
ALERT! Don’t forget to take the derivative of the constant, 252, to get 0.
Solution: You first need to set up a relationship that contains your given information and what you need to find. The right triangle made by the ladder and my house contains all of this information (although you wouldn’t have to have the same variables, of course), so by the Pythagorean Theorem:
y2 + x2 = 252
Notice that the values of x and y will change as the ladder slides, but the ladder will always be 25 feet long, so I can use this constant rather than a third variable.
Now, find the derivative with respect to t:
The base of the ladder is sliding away from the house at 2 ft/sec, so the problem also states that x = 17. You’ll have to use the Pythagorean Theorem to find the value of y for this specific value of x:
Now, you have all the variables in question except for dy/dt, the rate that you are trying to find. Substitute all your values into the derivative you found earlier to find dy/dt:
Notice that the length of y is decreasing, since the ladder is sliding downward; therefore, dy/dt must be negative.
A few years ago, I had to undergo massive nasal surgery, the focus of which was to scrape out all of my sinus cavities to remove disgusting “mucous cysts” that had gathered there like college students waiting for a party. I wrote the next problem soon after that experience. It was, at the time, the worst thing that could happen. (This problem appeared on my Web site as it was just getting started.)
NOTE. Example 3 is quite difficult when compared to the others. A problem of this difficulty in related rates is relatively rare on the AP test.
Example 3: The nightmare has come to pass. All of Kelley’s extensive surgeries and nasal passage scrapings have (unfortunately) gone awry, and he sits in the ear, nose, and throat doctor’s office waiting area spewing bloody nose drippings into a conical paper cup at the rate of 2.5 in3/min. The cup is being held with the vertex down and has a height of 4 inches and a base of 3 inches. How fast is the mucous level rising in the cup when the “liquid” is 2 inches deep?
Solution: You should first establish what you know: a cone’s volume is V = πr2h, dV/dt = 2.5, height of the cone is 4, and the diameter of the base is 3, which makes the radius of the base, r, equal to 3/2. (You also know that the mucous itself will be in the shape of a cone since it is in a conical container.) What’s even more important is what you don’t know. You don’t know the radius of the mucous, and you don’t know its rate of change. Therefore, you should try to eliminate the variable r from the volume equation. Why include a variable you know nothing about? To do so is the most complicated part of this problem. You need to use similar triangles. Below is a cross-section of the cup.
Two similar right triangles can be formed. Look at the set of overlapping triangles on the right. The smaller triangle (representing mucous) has unknown height and radius, whereas the larger triangle (the cup) has height 4 and radius 1.5. This allows you to set up a proportion, since corresponding sides of similar triangles are in proportion.
Solve this proportion for r and you get r = 3h/8. If you substitute this for r into the volume equation, the problem of knowing nothing about r is completely solved.
Now, find the derivative with respect to t to get rolling:
and substitute in all the information you know to solve for dh/dt, the value requested by the problem:
My cup, it overflowed.
Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.
YOU MAY USE A GRAPHING CALCULATOR FOR PROBLEMS 2 THROUGH 4.
1. A particle moves along the path y = x3 — 3x2 + 2. If the particle’s horizontal rate of change when x = 4 seconds is —3 ft/sec, what is its vertical rate of change at that instant?
2. If a spherical balloon is being deflated at a rate of 5 in3/sec, at what rate is the radius of the balloon decreasing when r = 5 in?
3. Last week, I accidentally dropped a cube into a vat of nuclear waste, setting off a chain reaction of events that eventually caused the cube to possess super powers, among them the ability to eat rocks. As the cube amassed these super powers, it swelled at a rate of 7 in/sec. At what rate was the surface area of Super Cube changing when one of its sides was 2 feet long?
4. During a reality show presentation, one of the celebrities displays her athletic prowess by skydiving out of a hovering helicopter 100 feet away from a cliff. If her position, in feet, is given by s(t) = —16t2 + 15,840, find the rate of change of the angle of depression (no pun intended) in degrees/sec at t = 30.9 seconds for a viewer standing at the edge of the cliff, assuming that his head is 600 feet above the floor of the valley below.
NOTE. A fall from a helicopter 3 miles in the sky is not only dangerous but nearly possible. It should only be attempted by a trained professional or someone really famous.
ANSWERS AND EXPLANATIONS
1. The equation is already given, so find the derivative with respect to t and substitute:
BC students note: which is also dy/dx when x = 4.
2. You need to know the volume of a sphere. Find its derivative with respect to t and plug in the given information:
Notice that you have to make dV/dt negative because the volume is decreasing.
3. The surface area of a cube is the sum of the areas of its sides. The sides are all squares, so the surface area is S = 6x2, where x is the length of a side. Now find the derivative and substitute:
Note that you use 24 inches instead of 2 feet for x, since the rest of the problem is to be given in terms of inches.
4. Begin by drawing a picture.
The horizontal distance between the spectator and Angela will remain fixed, but the vertical distance will change dramatically. You need to use a variable to label any length that can change; here we used x. You’ll need an equation that contains 0, since your goal is to find dθ/dt. The perfect choice is tangent, and the equation should be tan θ = x/100. Now, find its derivative with respect to t:
You still need to find θ and dx/dt to finish this problem. To find x, you first must find s(t) when t = 30.9, so plug 30.9 into the position equation:
s(t) = -16(30.9)2 + 15,840 = 563.04
You can tell by looking at the diagram that x = 600 — s(t), so at the instant that t = 30.9, x = 600 — 563.04 = 36.96. Finally, we can find θ (in degrees as asked):
θ = 20.28431249°
Now you need to find dx/dt. We just said x = 600 — s(t), so find the derivative with respect to t when t = 30.9:
All of the required information is finally available, so substitute and finish this problem: