# Calculus Articles

It's time to conquer calc. With your calculator in hand and these articles by your side, you're ready to take on even the scariest differential equations.

## Articles From Calculus

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Cheat Sheet / Updated 04-25-2022

Solving calculus problems is a great way to master the various rules, theorems, and calculations you encounter in a typical Calculus class. This Cheat Sheet provides some basic formulas you can refer to regularly to make solving calculus problems a breeze (well, maybe not a breeze, but definitely easier).

View Cheat SheetCheat Sheet / Updated 03-10-2022

To effectively work through calculus problems, you have to understand a number of topics: the process of evaluating limits, methods of solving various differentiation and integration problems, and the tests for convergence or divergence of infinite series.

View Cheat SheetArticle / Updated 12-21-2021

You can locate a function's concavity (where a function is concave up or down) and inflection points (where the concavity switches from positive to negative or vice versa) in a few simple steps. The following method shows you how to find the intervals of concavity and the inflection points of Find the second derivative of f. Set the second derivative equal to zero and solve. Determine whether the second derivative is undefined for any x-values. Steps 2 and 3 give you what you could call “second derivative critical numbers” of f because they are analogous to the critical numbers of f that you find using the first derivative. But this set of numbers has no special name. In any event, the important thing to know is that this list is made up of the zeros of f′′ plus any x-values where f′′ is undefined. Plot these numbers on a number line and test the regions with the second derivative. Use –2, –1, 1, and 2 as test numbers. Because –2 is in the left-most region on the number line below, and because the second derivative at –2 equals negative 240, that region gets a negative sign in the figure below, and so on for the other three regions. A second derivative sign graph A positive sign on this sign graph tells you that the function is concave up in that interval; a negative sign means concave down. The function has an inflection point (usually) at any x-value where the signs switch from positive to negative or vice versa. If you get a problem in which the signs switch at a number where the second derivative is undefined, you have to check one more thing before concluding that there’s an inflection point there. An inflection point exists at a given x-value only if there is a tangent line to the function at that number. This is the case wherever the first derivative exists or where there’s a vertical tangent. Plug these three x-values into f to obtain the function values of the three inflection points. A graph showing inflection points and intervals of concavity The square root of two equals about 1.4, so there are inflection points at about (–1.4, 39.6), (0, 0), and about (1.4, –39.6).

View ArticleArticle / Updated 12-21-2021

Every time you get in your car, you witness differentiation firsthand. Your speed is the first derivative of your position. And when you step on the accelerator or the brake — accelerating or decelerating — you experience a second derivative. If a function gives the position of something as a function of time, the first derivative gives its velocity, and the second derivative gives its acceleration. So, you differentiate position to get velocity, and you differentiate velocity to get acceleration. Here’s an example. A yo-yo moves straight up and down. Its height above the ground, as a function of time, is given by the function, where t is in seconds and H(t) is in inches. At t = 0, it’s 30 inches above the ground, and after 4 seconds, it’s at height of 18 inches. The yo-yo’s height, from 0 to 4 seconds Velocity, V(t) is the derivative of position (height, in this problem), and acceleration, A(t), is the derivative of velocity. Thus The graphs show the yo-yo’s height, velocity, and acceleration functions from 0 to 4 seconds. Velocity versus speed Your friends won’t complain — or even notice — if you use the words “velocity” and “speed” interchangeably, but your friendly mathematician will complain. Here’s the difference. For the velocity function in Figure 2, upward motion is defined as a positive velocity and downward velocity is defined as a negative velocity — this is the standard way velocity s treated in most calculus and physics problems. (If the motion is horizontal, going right is a positive velocity and going left is a negative velocity.) Speed, on the other hand, is always positive (or zero). If a car goes by at 50 mph, for instance, you say its speed is 50, and you mean positive 50, regardless of whether it's going to the right or the left. For velocity, the direction matters; for speed it doesn’t. In everyday life, speed is a simpler idea than velocity because it agrees with common sense. But in calculus, speed is actually the trickier idea because it doesn’t fit nicely in the three-function scheme shown in Figure 2. You need to keep the velocity-speed distinction in mind when analyzing velocity and acceleration. For example, if an object is going down (or to the left) faster and faster, its speed is increasing, but its velocity is decreasing because its velocity is becoming a bigger negative (and bigger negatives are smaller numbers). This seems weird, but that’s the way it works. And here’s another strange thing: Acceleration is defined as the rate of change of velocity, not speed. So, if an object is slowing down while going in the downward direction, and thus has an increasing velocity — because the velocity is becoming a smaller negative — the object has a positive acceleration. In everyday English, you’d say the object is decelerating (slowing down), but in calculus class, you say that the object has a negative velocity and a positive acceleration. By the way, “deceleration” isn’t exactly a technical term, so you should probably avoid it in calculus class. It’s best to use the following vocabulary: “positive acceleration,” “negative acceleration,” “speeding up,” and “slowing down.” Maximum and minimum height Maximum and minimum height of H(t) occur at the local extrema you see in Figure 1. To locate them, set the derivative of H(t) — that’s V(t) — equal to zero and solve. These two numbers are the zeros of V(t) and the t-coordinates — that’s time-coordinates — of the max and min of H(t), which you can see in Figure 2. In other words, these are the times when the yo-yo reaches its maximum and minimum heights. Plug these numbers into H(t) to obtain the heights: H (0.47) ≈ 31.1 H (3.53) ≈ 16.9 So, the yo-yo gets as high as about 31.1 inches above the ground at t ≈ 0.47 seconds and as low as about 16.9 inches at t ≈ 3.53 seconds. Total displacement is defined as the final position minus the initial position. So, because the yo-yo starts at a height of 30 and ends at a height of 18, Total displacement = 18 – 30 = –12. This is negative because the net movement is downward. Average velocity is given by total displacement divided by elapsed time. Thus, This negative answer tells you that the yo-yo is, on average, going down 3 inches per second. Maximum and minimum velocity Maximum and minimum velocity of the yo-yo during the interval from 0 to 4 seconds are determined with the derivative of V(t): Set the derivative of V(t) — that’s A(t) — equal to zero and solve: Now, evaluate V(t) at the critical number, 2, and at the interval’s endpoints, 0 and 4: So, the yo-yo has a maximum velocity of 5 inches per second twice — at both the beginning and the end of the interval. It reaches a minimum velocity of — 7 inches per second at t = 2 seconds. Total distance traveled Total distance traveled is determined by adding up the distances traveled on each leg of the yo-yo’s trip: the up leg, the down leg, and the second up leg. First, the yo-yo goes up from a height of 30 inches to about 31.1 inches (where the first turn-around point is). That’s a distance of about 1.1 inches. Next, it goes down from about 31.1 to about 16.9 (the height of the second turn-around point). That’s a distance of 31.1 minus 16.9, or about 14.2 inches. Finally, the yo-yo goes up again from about 16.9 inches to its final height of 18 inches. That’s another 1.1 inches. Add these three distances to obtain the total distance traveled: ~1.1 + ~14.2 + ~1.1 ≈ 16.4 inches. Average speed is given by the total distance traveled divided by the elapsed time. Thus, Maximum and minimum speed You previously determined the yo-yo’s maximum velocity (5 inches per second) and its minimum velocity (-7 inches per second). A velocity of -7 is a speed of 7, so that’s the yo-yo’s maximum speed. Its minimum speed of zero occurs at the two turnaround points. For a continuous velocity function, the minimum speed is zero whenever the maximum and minimum velocities are of opposite signs or when one of them is zero. When the maximum and minimum velocities are both positive or both negative, then the minimum speed is the lesser of the absolute values of the maximum and minimum velocities. In all cases, the maximum speed is the greater of the absolute values of the maximum and minimum velocities. Is that a mouthful or what? Maximum and minimum acceleration Maximum and minimum acceleration may seem pointless when you can just look at the graph of A(t) and see that the minimum acceleration of –12 occurs at the far left when t = 0 and that the acceleration then goes up to its maximum of 12 at the far right when t = 4. But it’s not inconceivable that you’ll get one of those incredibly demanding calculus teachers who has the nerve to require that you actually do the math and show your work — so bite the bullet and do it. To find the acceleration’s min and max from t = 0 to t = 4, set the derivative of A(t) equal to zero and solve: This equation, of course, has no solutions, so there are no critical numbers and thus the absolute extrema must occur at the interval’s endpoints, 0 and 4. You arrive at the answers you already knew. Note that when the acceleration is negative — on the interval [0, 2) — that means that the velocity is decreasing. When the acceleration is positive — on the interval (2, 4] — the velocity is increasing. Speeding up and slowing down Figuring out when the yo-yo is speeding up and slowing down is probably more interesting and descriptive of its motion than the info above. An object is speeding up (what we call “acceleration” in everyday speech) whenever the velocity and the calculus acceleration are both positive or both negative. And an object is slowing down (what we call “deceleration”) when the velocity and the calculus acceleration are of opposite signs. Look at all three graphs in Figure 2 again. From t = 0 to about t = 0.47 (when the velocity is zero), the velocity is positive and the acceleration is negative, so the yo-yo is slowing town (until it reaches its maximum height). When t = 0, the deceleration is greatest (12 inches per second per second; the graph shows an acceleration of negative 12, but here we’re calling it a deceleration so the 12 is positive). From about t = 0.47 to t = 2, both velocity and acceleration are negative, so the yo-yo is slowing down again (until it bottoms out at the lowest height). Finally, from about t = 3.53 to t = 4, both velocity and acceleration are positive, so the yo-yo is speeding up again. The yo-yo reaches its greatest acceleration of 12 inches per second at t = 4 seconds. Tying it all together Note the following connections among the three graphs in Figure 2. The negative section of the graph of A(t) – from t = 0 to t = 2 – corresponds to a decreasing section of the graph of V(t) and a concave down section of the graph H(t). The positive interval of the graph of A(t) – from t = 2 to t = 4 – corresponds to an increasing interval on the graph of V(t) and a concave up interval on the graph H(t). When t = 2 seconds, A(t) has a zero, V(t) has a local minimum, and H(t) has an inflection point.

View ArticleCheat Sheet / Updated 08-31-2021

By its nature, calculus can be intimidating. But you can take some of the fear of studying calculus away by understanding its basic principles, such as derivatives and antiderivatives, integration, and solving compound functions. Also discover a few basic rules applied to calculus like Cramer's Rule, the Constant Multiple Rule, and a few others, and you'll be on your way to acing the course.

View Cheat SheetArticle / Updated 07-16-2021

You can solve most limit problems by using your calculator. There are two basic methods. For example, say you want to evaluate the following limit: Method one Here’s what you do. Take a number extremely close to 5 and plug it into x. If you have a Texas Instruments TI-84 calculator, follow these steps: Enter your number, say 4.9999, on the home screen. Press the Sto (store) button, then the x button, and then the Enter button. This stores the number into x. Enter the function: Hit Enter. The result, 9.9999, is extremely close to a round number, 10, so that’s your answer. For good measure, store 4.999999 into x. Follow the procedure in Step 2. Scroll back up to the function by hitting 2nd, Enter, 2nd, Enter. Hit Enter one more time. You get 9.999999 — even closer to 10. If you still have any doubts, try one more number. Store 4.99999999 into x, scroll up to the function, and Hit Enter. The result, 10, clinches it. (The function value at 4.99999999 isn’t actually 10, but it’s so close that the calculator rounds it off to 10.) By the way, if you’re using a different calculator model, you can likely achieve the same result with the same technique or something very close to it. Method two The second calculator method is to produce a table of values: In your calculator’s graphing mode, enter the following: Go to “table set up” and enter the limit number, 5, as the “table start” number. Enter a small number, say 0.001, for ∆Tbl. That’s the size of the x-increments in the table. Hit the Table button to produce the table. Scroll up so you can see a couple numbers less than 5. You should see a table of values like the one in this table. X Y 4.998 9.998 4.999 9.999 5 Error 5.001 10.001 5.002 10.002 5.003 10.003 Because y gets very close to 10 as x zeros in on 5 from above and below, 10 is the limit. These calculator techniques are useful for a number of reasons: Your calculator can give you the answers to limit problems that are impossible to do algebraically. It can solve limit problems that you could do with paper and pencil except that you’re stumped. For problems that you do solve on paper, you can use your calculator to check your answers. Many calculus problems can be done algebraically, graphically, and numerically. When possible, use two or three of the approaches. Each approach gives you a different take on a problem and enhances your grasp of the relevant concepts. Use the calculator methods to supplement algebraic methods, but don’t rely too much on them. First of all, the calculator techniques won’t give you an exact answer unless the numbers your calculator gives you are getting close to a number you recognize — like 9.99998 is close to 10, or 0.333332 is close to 1/3. However, even when you don’t recognize the exact answer in such cases, you can still learn an approximate answer, in decimal form, to the limit question. This limit equals zero, but you can’t get that result with your calculator.

View ArticleArticle / Updated 07-13-2021

A Taylor polynomial approximates the value of a function, and in many cases, it’s helpful to measure the accuracy of an approximation. This information is provided by the Taylor remainder term: f(x) = Tn(x) + Rn(x) Notice that the addition of the remainder term Rn(x) turns the approximation into an equation. Here’s the formula for the remainder term: It’s important to be clear that this equation is true for one specific value of c on the interval between a and x. It does not work for just any value of c on that interval. Ideally, the remainder term gives you the precise difference between the value of a function and the approximation Tn(x). However, because the value of c is uncertain, in practice the remainder term really provides a worst-case scenario for your approximation. The following example should help to make this idea clear, using the sixth-degree Taylor polynomial for cos x: Suppose that you use this polynomial to approximate cos 1: How accurate is this approximation likely to be? To find out, use the remainder term: cos 1 = T6(x) + R6(x) Adding the associated remainder term changes this approximation into an equation. Here’s the formula for the remainder term: So substituting 1 for x gives you: At this point, you’re apparently stuck, because you don’t know the value of sin c. However, you can plug in c = 0 and c = 1 to give you a range of possible values: Keep in mind that this inequality occurs because of the interval involved, and because that sine increases on that interval. You can get a different bound with a different interval. This simplifies to provide a very close approximation: Thus, the remainder term predicts that the approximate value calculated earlier will be within 0.00017 of the actual value. And, in fact, As you can see, the approximation is within the error bounds predicted by the remainder term.

View ArticleArticle / Updated 07-12-2021

Every nonzero vector has a corresponding unit vector, which has the same direction as that vector but a magnitude of 1. To find the unit vector u of the vector you divide that vector by its magnitude as follows: Note that this formula uses scalar multiplication, because the numerator is a vector and the denominator is a scalar. A scalar is just a fancy word for a real number. The name arises because a scalar scales a vector — that is, it changes the scale of a vector. For example, the real number 2 scales the vector v by a factor of 2 so that 2v is twice as long as v. As you may guess from its name, the unit vector is a vector. For example, to find the unit vector u of the vector you first calculate its magnitude |q|: Now use the previous formula to calculate the unit vector: You can check that the magnitude of resulting vector u really is 1 as follows:

View ArticleArticle / Updated 07-12-2021

An ellipse is a set of points on a plane, creating an oval, curved shape, such that the sum of the distances from any point on the curve to two fixed points (the foci) is a constant (always the same). An ellipse is basically a circle that has been squished either horizontally or vertically. Graphically speaking, you must know two different types of ellipses: horizontal and vertical. A horizontal ellipse is short and fat; a vertical one is tall and skinny. Each type of ellipse has these main parts: The point in the middle of the ellipse is called the center and is named (h, v) just like the vertex of a parabola and the center of a circle. The major axis is the line that runs through the center of the ellipse the long way. The variable a is the letter used to name the distance from the center to the ellipse on the major axis. The endpoints of the major axis are on the ellipse and are called vertices. The minor axis is perpendicular to the major axis and runs through the center the short way. The variable b is the letter used to name the distance to the ellipse from the center on the minor axis. Because the major axis is always longer than the minor one, a > b. The endpoints on the minor axis are called co-vertices. The foci are the two points that dictate how fat or how skinny the ellipse is. They are always located on the major axis, and can be found by the following equation: where a and b are mentioned as in the preceding bullets, and F is the distance from the center to each focus. This figure shows a horizontal ellipse and a vertical ellipse with their parts labeled. Notice that the length of the major axis is 2a, and the length of the minor axis is 2b. This figure also shows the correct placement of the foci — always on the major axis. Two types of equations apply to ellipses, depending on whether they’re horizontal or vertical: The horizontal equation is with the center at (h, v), major axis of 2a, and minor axis of 2b. The vertical equation is with the same parts — although a and b have switched places. When the bigger number a is under x, the ellipse is horizontal; when the bigger number is under y, it’s vertical. You have to be prepared to not only graph ellipses, but also to name all their parts. If a problem asks you to calculate the parts of an ellipse, you have to be ready to deal with some ugly square roots and/or decimals. The following presents the parts for both horizontal and vertical ellipses. Horizontal Ellipse Center: (h, v) Vertices: (h ± a, v) Co-vertices: (h, v ± b) Length of Major Axis: 2a Length of Minor Axis: 2b Vertical Ellipse Center: (h, v) Vertices: (h, v ± a) Co-vertices: (h ± b, v) Length of Major Axis: 2a Length of Minor Axis: 2b To find the vertices in a horizontal ellipse, use (h ± a, v); to find the co-vertices, use (h, v ± b). A vertical ellipse has vertices at (h, v ± a) and co-vertices at (h ± b, v). For example, look at which is already in the proper form to graph. You know that h = 5 and v = –1 (switching the signs inside the parentheses). This example is a vertical ellipse because the bigger number is under y, so be sure to use the correct formula. This equation has vertices at (5, –1 ± 4), or (5, 3) and (5, –5). It has co-vertices at (5 ± 3, –1), or (8, –1) and (2, –1). The major axis in a horizontal ellipse is given by the equation y = v; the minor axis is given by x = h. The major axis in a vertical ellipse is represented by x = h; the minor axis is represented by y = v. The length of the major axis is 2a, and the length of the minor axis is 2b. You can calculate the distance from the center to the foci in an ellipse (either variety) by using the equation where F is the distance from the center to each focus. The foci always appear on the major axis at the given distance (F) from the center. What if the elliptical equation you’re given isn’t in standard form? Take a look at the example Follow these steps to put the equation in standard form: Add the constant to the other side. This gives you Complete the square. You need to factor out two different constants now — the different coefficients for Balance the equation by adding the new terms to the other side. In other words, Note: Adding 1 and 4 inside the parentheses really means adding to each side, because you must multiply by the coefficient before adding it to the right side. Factor the left side of the equation and simplify right. You now have Divide the equation by the constant on the right to get 1 and then reduce the fractions. You now have the form Determine if the ellipse is horizontal or vertical. Because the bigger number is under x, this ellipse is horizontal. Find the center and the length of the major and minor axes. The center is located at (h, v), or (–1, 2). Graph the ellipse to determine the vertices and co-vertices. Go to the center first and mark the point. Plotting these points will locate the vertices of the ellipse. Plot the foci of the ellipse. You determine the focal distance from the center to the foci in this ellipse with the equation The above figure shows all the parts of this ellipse in its fat glory.

View ArticleArticle / Updated 07-12-2021

Some differentiation rules are a snap to remember and use. These include the constant rule, power rule, constant multiple rule, sum rule, and difference rule. The constant rule: This is simple. f (x) = 5 is a horizontal line with a slope of zero, and thus its derivative is also zero. The power rule: To repeat, bring the power in front, then reduce the power by 1. That’s all there is to it. The power rule works for any power: a positive, a negative, or a fraction. Make sure you remember how to do the last function. It’s the simplest function, yet the easiest problem to miss. By the way, do you see how finding this last derivative follows the power rule? (Hint: x to the zero power equals one). You can differentiate radical functions by rewriting them as power functions and then using the power rule. The constant multiple rule: What if the function you’re differentiating begins with a coefficient? Makes no difference. A coefficient has no effect on the process of differentiation. You just ignore it and differentiate according to the appropriate rule. The coefficient stays where it is until the final step when you simplify your answer by multiplying by the coefficient. Don’t forget that ð (~3.14) and e (~2.72) are numbers, not variables, so they behave like ordinary numbers. Constants in problems, like c and k, also behave like ordinary numbers. So, for example, The sum rule: When you want the derivative of a sum of terms, take the derivative of each term separately. The difference rule: If you have a difference (that’s subtraction) instead of a sum, it makes no difference. You still differentiate each term separately. The addition and subtraction signs are unaffected by the differentiation.

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