for a quick review of integration (or, antidifferentiation), you might want to check out the following articles first.

• ap calculus exam review: integrals
• ap calculus exam review: antidifferentiation

and now, without further ado, here are some of the most common integrals found on the ap calculus exams!

common integrals

the following seven integrals (or their close cousins) seem to pop up all the time on the ap calculus ab and bc exams.

1. remember your trig integrals!

trigonometric functions are popular on the exam!

2. simple substitutions

you need to recognize when to use the substitution u = kx, for constant k. this substitution generates a factor of 1/k because du = k dx.

for example,

3. common integration by parts

integrands of the form x f(x) often lend themselves to integration by parts (ibp).

in the following integral, let u = x and dv = sin x dx, and use ibp.

4. linear denominators

integrands of the form a/(bx + c) pop up as a result of partial fractions decomposition. (see ap calculus bc review: partial fractions). while partial fractions is a bc test topic, it’s not rare to see an integral with linear denominator showing up in the ab test as well.

the key is that substituting u = bx + c (and du = b dx) turns the integrand into a constant times 1/u. let’s see how this works in general. keep in mind that a, b, and c must be constants in order to use this rule.

5. integral of ln x

the antiderivative of f(x) = ln x is interesting. you have to use a tricky integration by parts.

let u = ln x, and dv = dx.

by the way, this trick works for other inverse functions too, such as the inverse trig functions, arcsin x, arccos x, and arctan x. for example,

6. using trig identities

for some trigonometric integrals, you have to rewrite the integrand in an equivalent way. in other words, use a trig identity before integrating. one of the most popular (and useful) techniques is the half-angle identity.

7. trigonometric substitution

it’s no secret that the ap calculus exams consist of challenging problems. perhaps the most challenging integrals are those that require a trigonometric substitution.

the table below summarizes the trigonometric substitutions.

for example, find the integral:

here, the best substitution would be x = (3/2) sin θ.

now we’re not out of the woods yet. use the half-angle identity (see point 6 above). we also get to use the double-angle identity for sine in the second line.

note, the third line may seem like it comes out of nowhere. but it’s based on the substitution and a right triangle.

if x = (3/2) sin θ, then sin θ = (2x) / 3. draw a right triangle with angle θ, opposite side 2x, and hypotenuse 3.

by the pythagorean theorem, we find the adjacent side is equal to:

that allows us to identify cos θ in the expression (adjacent over hypotenuse).

finally, θ by itself is equal to arcsin(2x/3).

the post common integrals on the ap calc exam appeared first on magoosh blog | high school.

a slope field shows the direction of flow for solutions to a differential equation.

what is a slope field?

a slope field is a visual representation of a differential equation of the form dy/dx = f(x, y). at each sample point (x, y), there is a small line segment whose slope equals the value of f(x, y).

that is, each segment on the graph is a representation of the value of dy/dx. (check out ap calculus review: differential equations for more about differential equations on the ap calculus exams.)

because each segment has slope equal to the derivative value, you can think of the segments as small pieces of tangent lines. any curve that follows the flow suggested by the directions of the segments is a solution to the differential equation.

each curve represents a particular solution to a differential equation.

example — building a slope field

consider the differential equation dy/dx = x – y. let’s sketch a slope field for this equation. although it takes some time to do it, the best way to understand what a slope field does is to construct one from scratch.

first of all, we need to decide on our sample points. for our purposes, i’m going to choose points within the window [-2, 2] × [-2, 2], and we’ll sample points in increments of 1. just keep in mind that the window could be anything, and increments are generally smaller than 1 in practice.

now plug in each sample point (x, y) into the (multivariable) function x – y. we will keep track of the work in a table.

ok, now let’s draw the slope field. remember, the values in the table above represent slopes — positive slopes mean go up; negative ones mean go down; and zero slopes are horizontal.

spend some time matching each slope value from the table with its respective segment on the graph.

it’s important to realize that this is just a sketch. a more accurate picture would result from sampling many more points. for example, here is a slope field for dy/dx = x – y generated by a computer algebra system. the viewing window is the same, but now there are 400 sample points (rather than the paltry 25 samples in the first graph).

slope field for dy/dx = x – y

analyzing slope fields

now let’s get down to the heart of the matter. what do i need to know about slope fields on the ap calculus ab or bc exams?

you’ll need to master these basic skills:

• given a slope field, select the differential equation that best matches it.
• given a slope field, estimate values of a solution with given initial condition.
• sketch a slope field on indicated sample points, from a given differential equation.

we’ve already seen above how to sketch a slope field, so let’s get some practice with the first two skills in the list instead.

sample problem 1

the slope field shown above corresponds to which of the following differential equations.

  a. dy/dx = y²

  b. dy/dx = sin y

  c. dy/dx = -sin y

  d. dy/dx = sin x

solution

look for the clues. the segments have the same slopes in any given row (left to right across the graph). therefore, since the slopes do not change with respect to x, we can assume that dy/dx is a function of y alone. that eliminates choice d.

the horizontal segments occur when y = 0, π, and -π. however, the only point at which y² equals 0 is y = 0 (not π or -π). that narrows it down to a choice between b and c.

finally, notice that slopes are positive when 0 < y < π and negative when -π < y < 0. this pattern corresponds to the values of sin y. (the signs are opposite for -sin y, ruling out choice c).

the correct choice is b.

sample problem 2

suppose y = f(x) is a particular solution to the differential equation dy/dx = x – y such that f(0) = 0. use the slope field shown earlier to estimate the value of f(2).

  a. -2.5

  b. -1.3

  c. 0.2

  d. 1.1

solution

because f(0) = 0, the solution curve must begin at (0, 0). then sketch the curve carefully following the directions of the segments. it helps to imagine that the segments are showing the currents in a river. your solution should be like a raft carried along by the currents.

then find the approximate value of f(2) on your solution curve.

the best choice from among the answers is: d. 1.1.

slope fields and euler’s method

often if you see a slope field problem in the free response section of the exam, one part of the problem might be to use euler’s method to estimate a value of a solution curve.

while the slope field itself can be used to estimate solutions, euler’s method is much more precise and does not rely on the visual representation. check out this review article for practice using the method: ap calculus bc review: euler’s method.

summary

• slope fields are visual representations of differential equations of the form dy/dx = f(x, y).
• at each sample point of a slope field, there is a segment having slope equal to the value of dy/dx.
• any curve that follows the flow suggested by the directions of the segments is a solution to the differential equation.

the post interpreting slope fields: ap calculus exam review appeared first on magoosh blog | high school.

properties of logarithms

first of all, let’s review what a logarithm is. more specifically, we need to understand how the logarithm function can be used to break down complicated expressions.

you might want to check out the following article before getting started: ap calculus review: properties of exponents and logartithms. however, the most important properties for us will be the product, quotient, and power properties for logarithms. here, we focus on a particular logarithm: the natural logarithm, ln x, though the properties remain true in any base.

in other words, logarithms change…

1. products into sums,
2. quotients into differences, and
3. exponents into multiplication.

using the properties

i like to think of the logarithm as a powerful acid that can dissolve a complicated algebraic expression.

let me illustrate the point with an example.

notice how the original expression involves a huge fraction with roots and powers all over the place. after applying the properties of logarithms, the resulting expression mostly has only plus and minus. (of course, there is a trade-off — there are now three natural logs in the simplified expression.)

logarithmic differentiation

now let’s get down to business! how can we exploit these logarithmic simplification rules to help find derivatives?

the most straightforward case is when the function already has a logarithm involved.

example 1

find the derivative of .

solution

first simplify using the properties of logarithms (see work above). then you can take the derivative of each term. but be careful — the final term requires a product rule!

the general method

in the above example, there was already a logarithm in the function. but what if we want to use logarithmic differentiation when our function has no logarithm?

suppose f(x) is a function with a lot of products, quotients, and/or powers. then you might use the method of logarithmic differentiation to find f ‘(x).

1. 1. first write y = f(x)
   2. next, apply the (natural) logarithm function to both sides of the equation.
   3. then use the product, quotient, and/or power properties to break down the expression on the right. in other words, simplify ln(f(x)) algebraically.
   4. now apply the derivative operator to both sides.
   5. use your rules of differentiation to find the derivatives. note that the left-hand side requires implicit differentiation. (check out: ap calculus review: implicit variation for details.)
      
   6. solve for dy/dx by multiplying both sides by y. in your answer, don’t forget to replace y by f(x).

example 2

use logarithmic differentiation to find the derivative of:  

solution

let’s follow the steps outlined above. the first two steps are routine.

on the other hand, step 3 requires us to break down the logarithmic expression using the properties. the work in this step depends on the function. in our case, there is a product of two factors, so we’ll start with the product property. the power property helps to break down the radical. finally, don’t forget the cancellation rule: ln(e^x) = x.

next, in steps 4 and 5 apply the derivative and work out the right-hand side.

finally, in step 6, we solve for the unknown derivative by multiplying both sides by y. don’t forget to substitute back the original function f(x) in place of y.

functions raised to a function power

a very famous question in calculus class is: what is the derivative of x^x ?

• the answer is not x x^{x – 1}, because the power rule for derivatives cannot be used when the power includes a variable.
• the answer is not x^x ln x, because the exponential rule for derivatives cannot be used when the base includes a variable.

so what is the derivative of x^x ?

well, it turns out that only logarithmic differentiation can decide this one for us!

in fact any time there is a function raised to a function power (that is, neither the exponent nor the base is constant), then you will have to use logarithms to break it down before you can take a derivative.

let’s see how it works in the simplest case: x^x.

first, write y = x^x.

then, apply the logarithm to both sides:   ln y = ln x^x.

break down the right-hand side of the equation using the algebraic properties of logarithms. in this case, only the power property plays a role.

ln y = x ln x

now you can take derivatives of the functions on both sides. but be careful… the function on the right requires a product rule.

(1/y)(dy/dx) = (1) ln x + x(1/x) = ln x + 1

finally, multiply both sides by the original function (y = x^x) to isolate dy/dx.

dy/dx = x^x(ln x + 1)

and there you have it! the derivative of x^x turns out to be trickier than you might have thought at first, but it’s not impossible.

summary

• logarithmic differentiation is a method for finding derivatives of complicated functions involving products, quotients, and/or powers.
• you can use the algebraic properties of logarithms to break down functions into simpler pieces before taking the derivative.

the post what is logarithmic differentiation? ap calc review appeared first on magoosh blog | high school.

in this review article, we’ll explore the methods and applications of linear approximation. we’ll also take a look at plenty of examples along the way to better prepare you for the ap calculus exams.

linear approximation and tangent lines

by definition the linear approximation for a function f(x) at a point x = a is simply the equation of the tangent line to the curve at that point. and that means that derivatives are key! (check out how to find the slope of a line tangent to a curve or is the derivative of a function the tangent line? for some background material.)

formula for the linear approximation

given a point x = a and a function f that is differentiable at a, the linear approximation l(x) for f at x = a is:

l(x) = f(a) + f '(a)(x – a)

the main idea behind linearization is that the function l(x) does a pretty good job approximating values of f(x), at least when x is near a.

in other words, l(x) ≈ f(x) whenever x ≈ a.

example 1 — linearizing a parabola

find the linear approximation of the parabola f(x) = x² at the point x = 1.

  a. x² + 1

  b. 2x + 1

  c. 2x – 1

  d. 2x – 2

solution

c.

note that f '(x) = 2x in this case. using the formula above with a = 1, we have:

l(x) = f(1) + f '(1)(x – 1)

l(x) = 1² + 2(1)(x – 1) = 2x – 1

follow-up: interpreting the results

clearly, the graph of the parabola f(x) = x² is not a straight line. however, near any particular point, say x = 1, the tangent line does a pretty good job following the direction of the curve.

how good is this approximation? well, at x = 1, it’s exact! l(1) = 2(1) – 1 = 1, which is the same as f(1) = 1² = 1.

but the further away you get from 1, the worse the approximation becomes.

approximating using differentials

the formula for linear approximation can also be expressed in terms of differentials. basically, a differential is a quantity that approximates a (small) change in one variable due to a (small) change in another. the differential of x is dx, and the differential of y is dy.

based upon the formula dy/dx = f '(x), we may identify:

dy = f '(x) dx

the related formula allows one to approximate near a particular fixed point:

f(x + dx) ≈ y + dy

example 2 — using differentials with limited information

suppose g(5) = 30 and g '(5) = -3. estimate the value of g(7).

  a. 24

  b. 27

  c. 28

  d. 33

solution

a.

in this example, we do not know the expression for the function g. fortunately, we don’t need to know!

first, observe that the change in x is dx = 7 – 5 = 2.

next, estimate the change in y using the differential formula.

dy = g '(x) dx = g '(5) · 2 = (-3)(2) = -6.

finally, put it all together:

g(5 + 2) ≈ y + dy = g(5) + (-6) = 30 + (-6) = 24

example 3 — using differentials to approximate a value

approximate using differentials. express your answer as a decimal rounded to the nearest hundred-thousandth.

solution

1.03333.

here, we should realize that even though the cube root of 1.1 is not easy to compute without a calculator, the cube root of 1 is trivial. so let’s use a = 1 as our basis for estimation.

consider the function . find its derivative (we’ll need it for the approximation formula).

then, using the differential, , we can estimate the required quantity.

exact change versus approximate change

sometimes we are interested in the exact change of a function’s values over some interval. suppose x changes from x₁ to x₂. then the exact change in f(x) on that interval is:

δy = f(x₂) – f(x₁)

we also use the “delta” notation for change in x. in fact, δx and dx typically mean the same thing:

δx = dx = x₂ – x₁

however, while δy measures the exact change in the function’s value, dy only estimates the change based on a derivative value.

example 4 — comparing exact and approximate values

let f(x) = cos(3x), and let l(x) be the linear approximation to f at x = π/6. which expression represents the absolute error in using l to approximate f at x = π/12?

  a. π/6 – √2/2

  b. π/4 – √2/2

  c. √2/2 – π/6

  d. √2/2 – π/4

solution

b.

absolute error is the absolute difference between the approximate and exact values, that is, e = | f(a) – l(a) |.

equivalently, e = | δy – dy |.

let’s compute dy ( = f '(x) dx ). here, the change in x is negative. dx = π/12 – π/6 = -π/12. note that by the chain rule, we obtain: f '(x) = -3 sin(3x). putting it all together,

dy = -3 sin(3 π/6 ) (-π/12) = -3 sin(π/2) (-π/12) = 3π/12 = π/4

ok, next we compute the exact change.

δy = f(π/12) – f(π/6) = cos(π/4) – cos(π/2) = √2/2

lastly, we take the absolute difference to compute the error,

e = | √2/2 – π/4 | = π/4 – √2/2.

application — finding zeros

linear approximations also serve to find zeros of functions. in fact newton’s method (see ap calculus review: newton’s method for details) is nothing more than repeated linear approximations to target on to the nearest root of the function.

the method is simple. given a function f, suppose that a zero for f is located near x = a. just linearize f at x = a, producing a linear function l(x). then the solution to l(x) = 0 should be fairly close to the true zero of the original function f.

example 5 — estimating zeros

estimate the zero of the function using a tangent line approximation at x = -1.

  a. -1.48

  b. -1.53

  c. -1.62

  d. -1.71

solution

d.

remember, the purpose of this question is to estimate the zero. first of all, the tangent line approximation is nothing more than a linearization. we’ll need to know the derivative:

then find the expression for l(x). note that g(-1) ≈ 2.13 and g '(-1) = 3.

l(x) = 2.13 + 3(x – (-1)) = 5.13 + 3x.

finally to find the zero, set l(x) = 0 and solve for x

0 = 5.13 + 3x  →  x = -5.13/3 = -1.71

the post linear approximation: ap calculus exam review appeared first on magoosh blog | high school.

average value of functions

suppose f is a continuous function defined over an interval [a, b]. in particular f(x) exists at every one of the infinitely-many points x between and including a and b. so, if you’re looking for the average value of f on that interval, it won’t do any good to try adding up those infinitely-many data points.

instead, the way to tame the infinity is to use calculus. specifically, we define the average value of a function f as the following definite integral.

the theory behind the formula

but where does the integral formula for average come from?

the key is sampling. if you included enough of the function values, say a thousand, a million, or even more, then that should approximate the average of all infinitely-many points!

let’s illustrate with the following example.

estimate the average value of the function f(x) = √(x) + 1 over the interval [1, 3].

since we’re just estimating, let’s pick four sample points. (the more sample points you pick, the better your estimated average will be.)

divide the interval [1, 3] into four equal subintervals, and let’s agree to choose the midpoint of each subinterval. then plug those midpoints into f to find the sample values.

finally, use the familiar old averaging formula. add up the data and divide by the number of data points:

(2.12 + 2.32 + 2.50 + 2.66)/4 = 2.4

so the (approximate) average of the function is 2.4.

taking it to the limit

however, what we’ve just done will not give us an exact answer because we’ve essentially ignored most of the function! what about f(1.3) or f(2.95234)? no matter how many sample points we include, there will always be some missing… unless we can use the magic of calculus to catch them all.

the sampling process should remind you of a riemann sum. for a quick reminder, feel free to check out ap calculus review: riemann sums.

let n be any whole number and x_k^* stand for the various sample x-values. then the estimated average is the sum:

next, allow n → ∞ using a limit. we also need to get Δx into the act somehow. the trick is to multiply and divide by (b – a). remember, Δx = (b – a)/n.

examples

ok, now that you’ve seen the theory, let’s use the formula in practice!

problem 1

find the exact average value of f(x) = √(x) + 1 over the interval [1, 3].

solution

above, we only estimated the average to be 2.4. now we’ll use the integral formula to determine the average value precisely.

(it’s interesting to compare our estimate with the exact value above. √(3) + 2/3 ≈ 2.39871747, which means that our estimate of 2.4 was actually pretty good!)

problem 2

the amount of energy associated with a certain chemical reaction is given by e = x ln x, where 1 ≤ x ≤ e, and x represents the amount of one of the reactants. find the average energy of the reaction over the range of possible levels of reactant.

solution

this problem seems more like chemistry than math!

however, the keyword average tells us that mathematics plays a major role in this problem. in fact, they are simply asking for the average value of f(x) = x ln x, over the interval [1, e].

first set up the integral formula with a = 1 and b = e. then work out the integration, which involves integration by parts in this case.

thus the average energy of the reaction is (e² + 1)/[ 4(e – 1) ], or roughly 1.22.

mean value theorem for integrals

averages are also called means. so you may use the same formula to find the mean value of a function.

there is also an important result in calculus that relates the mean value to a particular function value on the given interval.

the mean value theorem for integrals (mvti). if f is continuous on a closed interval [a, b], then there is at least one point x = c in that interval such that the mean value of the function is equal to f(c). that is,

(caution: there is also a mean value theorem for derivatives. it’s important not to confuse the two.)

problem 3

let f(x) = 6x² – 8x + 1. determine the value of c at which the mean value of f on [-1, 1] is the same as f(c).

solution

according to the mean value theorem for integrals, there must be at least one such value c. let’s set up the formula and find it!

at this point, we will need to solve a quadratic equation. don’t forget your quadratic formula!

6c² – 8c – 2 = 0

c = (2 ± √(7))/3

it seems as though there may be two answers. however, only one lies within the given interval [-1, 1].

c = (2 + √(7))/3 ≈ 1.549, not in the interval.

c = (2 – √(7))/3 ≈ -0.215, in the interval.

therefore, the only value that satisfies the mvti is c = (2 – √(7))/3.

summary

although average value and the mean value theorem for integrals are specialized topics and only show up in a few problems on any given ap calculus test, they are important concepts to master. for one thing, they illustrate how integral calculus can be used in applications.

moreover, working out the average value of a function is no more difficult than computing a definite integral. so now when you see these kinds of problems on the ap calculus exam, you can rise to the challenge!

the post ap calculus review: average value of functions appeared first on magoosh blog | high school.

basic differential equations: integration

you have probably worked out hundreds of differential equations without even realizing it!

it’s true! let me give you an example.

what makes this a differential equation?

well think about what the notation means. you know that integration is the opposite of differentiation. so what we’re looking for here is a function f(x) whose derivative is equal to 2x. in other words, we have to solve:

f ‘(x) = 2x

that’s a relation involving an unknown function f and its derivative.

but this differential equation is trivial to solve! just use the power rule for integrals (or guess-and-check).

f(x) = x² + c

(for a full review of integration, check out: ap calculus exam review: antidifferentiation or ap calculus exam review: integrals.)

initial value problems

now suppose you have more information about an unknown function, such as its value at a certain point. then you may be able to solve for the function explicitly, rather than getting stuck with an unknown constant of integration at the end.

an initial value problem typically gives a derivative expression along with a function value. the goal is to produce the original function.

example 1: initial value problem

if f ‘(x) = 1/x² for x > 0, and f(1) = 5, find the expression for f(x) for x > 0.

solution

because the derivative expression is given, we integrate to find the original function. don’t forget about your constant of integration!

next, plug in the known info and solve for the unknown constant of integration.

this implies that the original function must be:    f(x) = -1/x + 6.

separation of variables

of course not every differential equation (or de) problem can be handled by simple integration. often more advanced techniques must be used, especially if both x and y show up in the equation.

for example, the following de cannot be cracked by integration (…not in its current form, that is).

the key is to somehow break up the dy and dx.

the derivative notation dy/dx looks like a fraction. however, it’s not really a fraction, because the two quantities dy and dx (called differentials) are supposed to represent the idea of taking Δx to zero in the limit. in other words, if you had to identify the numerical “value” of a differential, it would make sense to say dx = dy = 0. and we all know that 0/0 is undefined.

but those differential gadgets are more subtle than that. they really don’t carry a numerical value. instead, they stand for a limiting process. and if we’re very careful, we can work with dx and dy individually.

the method of separation of variables works by manipulating dy/dx as if it were a fraction, and then using integration to get rid of the differentials.

now let’s take another look at our example above.

example 2: separation of variables

if    , and y(0) = -3, find an equation for y in terms of x.

solution

first multiply dx to both sides. then multiply or divide as necessary so that only expressions of y are on the left, and only those with x are on the right.

next, we use the magic of calculus! all you have to do is to apply the integral symbol to each side of the equation. now you have two separate integrals to work out.

now even though both sides generated a constant of integration, those can be combined into a single constant on the right hand side. the next few steps are just for isolating y.

notice that the quantity e^c is replaced by another constant k. this is because c is still arbitrary (and unknown). so just think of k as a related, but still arbitrary constant.

finally we solve for k using the known information, y(0) = -3.

(by the way, we will drop the “±” notation at this point because k itself can take care of that choice.)

therefore, replacing k by its computed value -3, we obtain the final form of the function.

setting up differential equations

fortunately you won’t encounter any de problems on the ap calculus exam that can’t be handled by either integration or separation of variables.

however, you may be asked to set up and/or analyze a de (without solving it).

example 3: setting up a de

a certain bacteria culture has p cells at time t and grows in proportion to the square root of the amount present. set up a differential equation that models a(t).

solution

because the culture grows at a certain rate with respect to time, we know we’ll be working with the derivative da/dt.

let k be the constant of proportionality. then translate the given information into mathematics:

da/dt = k × √(a)

remember, we don’t have to solve for a(t); just set up the de that could be used to solve for it. so we’re done at this step!

example 4: analyzing a de

the number y of people infected with the flu in a certain town at time t can be modeled by the differential equation, dy/dt = 0.00002y(21500-y), where y(0) = 1. determine the limit of y as t → ∞.

solution

it would be difficult to solve this de for y explicitly. instead, look closely at the equation. it has the form of a logistics equation.

that means we can answer this question with no work at all! you just have to remember that logistics equations always have solutions that tend toward the carrying capacity m as t → ∞. here, m = 21500. that’s all you need to answer the question!

the limit will be 21500.

(for more details about the logistics equation, i recommend: ap calculus bc review: logistics growth model.)

conclusion

differential equations show up only sparingly on an ap calculus exam. but it’s important to be aware of the techniques for solving them, setting them up, and analyzing them.

• remember the basic methods of integration and separation of variables.
• know how to set up a differential equation.
• look for situations in which you may avoid solving the de.

the post ap calculus review: differential equations appeared first on magoosh blog | high school.

optimization

the method of optimization uses derivatives to find maximum or minimum values. however, the functions that need to be optimized typically have more than one variable. hence, considerable work goes into transforming the problem into a single-variable function first.

let’s introduce the methods using a concrete example.

example: a widget factory

suppose you own a business that makes and sells widgets. if you sell x widgets at a price of p dollars each, then that earns your company a total of xp dollars. by the way, the quantity r = xp is called the revenue.

so why not just charge a huge price for each widget to earn the most money you can? after all, the objective of most successful companies is to maximize their revenue.

well, you would raise prices, except that you noticed in the past that every time you did so, there were fewer sales of widgets. in other words, as p increases, x decreases.

fortunately, if you know exactly what the relationship between price and number of widgets sold at that price (or demand), then you can use optimization to find the price that maximizes your revenues!

using the relation to reduce variables

next, suppose that we add one vital piece of information to this problem: the relationship between price and demand is given:

x = 400 – 40p + p², for 0 ≤ p ≤ 20

note, any equation that relates the variables of an optimization problem is a constraint equation.

now do you remember that revenue function? (in case you forgot, it was r = xp.)

let’s use the relationship between x and p replace x to create an objective function having only one variable.

r = xp = (400 – 40p + p²)p = 400p – 40p² + p³

using the derivative to find the maximum

now that we have a single-variable expression, we can use the usual techniques from calculus to find its maximum value.

first take the derivative of the objective function:

r ' = 400 – 80p + 3p²

next, set the derivative equal to zero to locate any critical points. here, we’ll have to use the quadratic formula.

400 – 80p + 3p² = 0

p = 6.67, or 20.

because we got two answers, we have to check which one (if any) gives the correct result. plugging in each one into the revenue function, we find:

r(6.67) = 400(6.67) – 40(6.67)² + (6.67)³ = 1185.18

r(20) = 400(20) – 40(20)² + (20)³ = 0

clearly p = 6.67 gives better revenue.

(by the way, since there was a restriction that 0 ≤ p ≤ 20, we do not have to worry about what the revenue function does beyond p = 20.)

solving the original problem

according to the work above, the price should be p = $6.67. then, to find the actual number of widgets that produces this amount, simply plug into the constraint.

x = 400 – 40(6.67) + (6.67)² = 177.7

rounding to the nearest widget, your company should produce and sell 178 widgets at a price of $6.67 each to achieve a maximum revenue of (178)(6.67) = $1,187 (approximately).

method of optimization

so what have we learned from the above example? even though most of the work was peculiar to this situation (revenue, price, demand, etc.), there are a number of steps that are useful in general.

1. read the problem carefully. (this should go without saying.) identify what you are trying to maximize or minimize (the objective) and what relationships exist among the variables (constraint equation).
2. write the objective as well as any constraints in terms of the variables in the problem.
3. use algebra to solve the constaint(s) and reduce the number of variables in the objective. the goal is to express your objective function in terms of only one variable.
4. use calculus to find the max or min of the objective function. this means taking a derivative, setting it equal to zero, solving, and determining which solutions give the appropriate answers for the given problem.
5. make sure you answer the original question. you might have to find the values of the other variables or quantities now.

the hardest thing about optimization problems is the setup (steps 1-2), because that changes from problem to problem.

example: rectangles and parabolas

find the largest area possible for a rectangle inscribed between the parabola y = 9 – x² and the x-axis, with its base on the x-axis.

solution

first, let’s sketch the graph of the parabola as well as a typical inscribed rectangle.

our objective is to maximize the area of the rectangle.

area of a rectangle: a = bh.

because the parabola touches the rectangle at its upper corners, we know that the height of the rectangle is the same as the y-coordinate of the upper right corner point. that is, h = y = 9 – x².

the base length of the rectangle is b = 2x (a distance of x on the left and the right side).

the two equations above are the constraints. let’s replace the variables b and h in our objective function by their equivalent expressions in terms of x.

a = bh = (2x)(9 – x²) = 18x – 2x³

now we can take a derivative and set it equal to zero.

of course, only the positive x-value makes sense in this context. plug x = √3 into the area formula (objective function) to find the maximum area.

the post ap calculus review: optimization appeared first on magoosh blog | high school.

in this review article, we will highlight the most important applications of derivatives for the ap calculus ab/bc exams.

applications of derivatives

here is a list of some of the essential derivative applications.

• (instantaneous) rates of change
  • velocity
  • acceleration
• slope and tangent lines
• analysis of graphs
  • increase/decrease
  • relative extreme points
  • concavity
  • inflection points
• related rates
• optimization

rates of change

perhaps the most important application of the derivative is in finding rates of change. in fact, this is why newton (and leibniz independently) invented calculus in the first place!

suppose y = f(x) is a function that models some real-world behavior. here, the input is x and output is y. a natural question in the sciences is: how does the output change as one changes the input? in other words, what is the change in y resulting from a change in x?

for most purposes, it makes sense to consider the ratio of the changes. for example, if y changes twice as quickly as x, then that would be a rate of change of 2.

so this leads to the concept of average rate of change, Δy/Δx. (see ap calculus review: average rate of change for more about this.)

finally, by allowing the change in input to approach 0 (by taking a limit), we get the definition of derivative:

of course after you learn the definition, then you get to learn all of the shortcut rules for derivatives that let you avoid taking limits at all!

example: rates of change

suppose the population of a small city can be modeled by p = 40532 e^0.03t, where p is total population and t is time in years after the year 2000. how quickly was the population increasing in 2002?

solution

the phrase “how quickly” means “at what rate”? so we use a derivative to find the rate of change.

p ‘ = 40532 e^0.03t (0.03) = 1215.96 e^0.03t

the year 2002 corresponds to t = 2.

p ‘(2) = 1215.96 e^0.03(2) = 1291.15

thus the city was growing at a rate of about 1291 people per year.

velocity and acceleration

by definition, velocity is the rate of change of position. so, if you have a position function s(t), then the velocity function is nothing more than the derivative of s(t).

analogously, acceleration is the rate of change of velocity – hence the second derivative of position.

slope and tangent lines

the derivative value f ‘(a) measures the slope of the curve y = f(x) at the point x = a. by extension, then the derivative also gives you the slope of the tangent line there.

here is another reference for tangent lines and slopes of curves: how to find the slope of a line tangent to a curve.

example: tangent lines

find the equation of the tangent line to the curve y = (1/3)x³ – (1/2)x² + 3x – 5 at x = 6.

solution

slope = derivative. here, dy/dx = (1/3)3x² – (1/2)2x + 3 = x² – x + 3.

therefore, the slope at x = 6 is:

m = (6)² – (6) + 3 = 33.

we also need the y-coordinate of the point. so plug x = 6 into the original function as well.

(1/3)(6)³ – (1/2)(6)² + 3(6) – 5 = 67.

finally, using point-slope form, we construct the equation of the tangent line.

y – 67 = 33(x – 6) = 33x – 198, or

y = 33x – 131

analysis of graphs

there’s quite a lot that calculus can tell us about the shape of a graph. we’ll focus on what the derivative and second derivative can say about a graph.

for a detailed review of these topics, you can also check out: ap calculus exam review: analysis of graphs

first derivatives — increase, decrease, and relative extrema

the first derivative measures the slope of the curve. so just as slopes of lines can tell you whether the line increases, decreases, or is horizontal, the derivative can be used to locate intervals of increase and decrease, as well as any turnaround points.

at a critical point, the graph may turn around (but it’s not a guarantee). then you can use the first derivative test to decide whether each critical point is a relative maximum, minimum, or neither.

first derivative test: suppose x = a is a critical point for a function f.

• if the sign of the derivative f ‘ changes from positive to negative at a, then there is a relative (local) maximum at x = a.
• if the sign of the derivative f ‘ changes from negative to positive at a, then there is a relative (local) minimum at x = a.
• lastly, if the sign of the derivative f ‘ does not change at a, then there is no relative extreme point there.

second derivatives — concavity and inflection points

second derivatives give you information about concavity (direction of curving), and inflection points (transitions between different concavities).

for more details, check out: ap calculus review: inflection points.

related rates

sometimes the behaviors of two quantities are tied together. for example, raising the temperature of a gas in a fixed container will raise its pressure as well.

the method of related rates can be used to understand how rates of change are related based on an algebraic or geometric relation among the original variables.

because the word “rate” appears in this topic, it should be no surprise that derivatives will be used!

example: related rates

a spherical balloon is filled with helium at a constant rate of 5 cm³ per second. how quickly is the radius increasing when the radius is 16 cm?

solution

first we need to remember the volume formula for the sphere. this relates the two variables mentioned in the problem: volume (v) and radius (r).

first, we take time derivatives of both sides of the volume formula to produce a related rate.

next, we plug in the known information, dv/dt = 5, and r = 16.

thus, the radius is increasing at 0.00155 cm per second.

for more practice, try this link: ap calculus review: related rates.

optimization

optimization is simply the process of finding a maximum or minimum value. however, in most optimization problems, there is more than one input variable and certain constraints on the variables. so it takes quite a bit of algebra to set up the function before you can take its derivative.

understanding optimization problems really takes seeing a fair number of them worked out. you can find a detailed review of optimization here: ap calculus review: optimization.

summary

now that you’ve seen the various ways in which applications of derivatives can appear on the ap calculus exams, you can be more confident on test day! just keep in mind a few key concepts from which all of these applications arise.

• derivatives measure rates of change.
• derivative values are slopes.

the post ap calculus review: applications of derivatives appeared first on magoosh blog | high school.

what are inverse functions?

basically, an inverse function is a function that “reverses” what the original function did.

for example, consider f(x) = 3x – 6. what does f do to its input x? using correct order of operations, f has the following effect:

1. first multiply by 3.
2. then subtract 6 from the result.

so, in order to reverse what f does, we have to follow the steps backwards. this is very much like reversing your driving directions to return home from an unfamiliar place.

1. first add 6 (to undo the subtraction).
2. then divide by 3 (to undo the multiplication).

therefore, the inverse function, which we’ll call g(x) for right now, has the formula,

g(x) = (x + 6)/3

the notation for the inverse function of f is f ^-1. so we could write:

f ^-1(x) = (x + 6)/3

our purpose here is not to be able to solve to find inverse functions in all cases. in fact, the main theorem for finding their derivatives does not require solving for f ^-1(x) explicitly.

finding the derivative of an inverse function

computing the derivative of an inverse function is not too much more difficult than computing derivatives in general.

first, here’s a quick review of the basic derivative rules: calculus review: derivative rules.

the main theorem for inverses

suppose that f is a function that has a well-defined inverse f ^-1, and suppose that (a, b) is a point on the graph of y = f(x). then

however, you might see a different version of this rule. another way to say that (a, b) is a point on the graph of y = f(x) is to say that b = f(a). moreover, by properties of the inverse, then we can say that a = f ^-1(b).

finally, replacing b by x, we discover the second version of the main theorem:

the two versions are useful in different contexts, which we shall see in the examples.

it’s also good to know that the condition of “having a well-defined inverse” is satisfied whenever the function is one-to-one.

example 1

suppose that g(x) is the inverse function for f(x) = 3x⁵ + 6x³ + 4. find the value of g '(13).

solution

it won’t do us any good to try to solve for the inverse function algebraically. this polynomial is just too complicated for that. instead, we must rely on the main theorem.

first, find the derivative of the original function.

f '(x) = 15x⁴ + 18x²

the other piece of the puzzle is the value to plug in. identify b = 13 from the problem statement. but don’t plug 13 into anything! instead, the formula requires a value a such that f(a) = b; that is, f(a) = 13.

so it looks like we might have to solve:

3x⁵ + 6x³ + 4 = 13

however, that polynomial is still way to difficult to solve algebraically. fortunately, it doesn’t take long to guess and check a value for x that would make this equation true.

just look at the coefficients: they add up to 13… thus, if we just plugged in x = 1, then we’d get the correct result.

3(1)⁵ + 6(1)³ + 4 = 3 + 6 + 4 = 13

this means we should use a = 1 in the formula for the derivative.

finally, put it all together using the formula from the main theorem:

g '(b) = 1/f '(a)

g '(13) = 1/f '(1) = 1/(15(1)⁴ + 18(1)²) = 1/33.

common inverses and their derivatives

some functions are so important that their inverses get special names. we’ll take a look at two such cases:

• logarithms, and
• inverse trig functions

logarithmic functions

logarithms are nothing more than the inverses of exponentiation. for a quick review, check out: ap calculus review: properties of exponents and logarithms.

for our purposes in this article, it suffices to understand the following inverse relationship.

• if f(x) = a^x, then f ^-1(x) = log_a x

so in particular, when the base is e, the natural base, then we get a relationship between e^x and the natural logarithm, ln x.

• if f(x) = e^x, then f ^-1(x) = ln x

let’ use the main theorem to prove that the derivative of ln x is indeed equal to 1/x. remember, if f(x) = e^x, then f '(x) = e^x as well.

you could do a similar thing to calculate the derivative of log_a x. but it’s easier just to memorize the end result.

here are the derivatives of the logarithmic functions:

trigonometric functions

each of the six trig functions has its own inverse function. (each of them requires a certain restriction on the domain of the original function to ensure that it’s one-to-one. but we won’t get into the technical details here.)

the inverse for sin x is either written    sin ^-1 x   or   arcsin x.

similar notation applies for the other five.

you can use the main theorem for inverse functions to work out their derivatives, but it’s better just to have these memorized.

example 2

find the slope of the tangent line to y = arctan 5x at x = 1/5.

solution

we know that arctan x is the inverse function for tan x, but instead of using the main theorem, let’s just assume we have the derivative memorized already. (you can cheat and look at the above table for now… i won’t tell anyone.)

when you take the derivative, keep in mind that there will be a chain rule, with inside function u = 5x and outside function arctan u.

then plug in x = 1/5 to compute the slope.

the post ap calculus review: derivatives of inverse functions appeared first on magoosh blog | high school.

solids of revolution and the shell method

briefly, a solid of revolution is the solid formed by revolving a plane region around a fixed axis.

we defined solids of revolution in a previous article, ap calculus review: disk and washer methods. so you might want to read up before continuing.

shells, shells, and more shells…

suppose you need to find the volume of a solid of revolution. first we have to decide how to slice the solid. if you wanted to slice perpendicular to the axis of revolution, then you would get slabs that look like thin cylinders (disks) or cylinders with circles removed (washers). however, the shell method requires a different kind of slicing.

imagine that your solid is made of cookie dough. and you have a set of circular cookie cutters of various sizes. starting with the smallest cookie cutter and progressing to larger ones, let’s slice through the dough in concentric rings.

making sure to slice in the same direction as the axis of revolution, you will get a clump of nested shells, or thin hollow cylindrical objects.

approximating the volume

now let’s take a closer look at a single shell.

.

as long as the thickness is small enough, the volume of the shell can be approximated by the formula:

v = 2πrhw

note that the volume is simply the circumference (2πr) times the height (h) times the thickness (w). in fact, you can think of cutting the shell along its height and “unrolling” it to produce a thin rectangular slab. then the volume is simply length × height × width as in any rectangular solid.

now suppose we have a solid of revolution with generating region being the area under a function y = f(x) between x = a and x = b. and suppose that the y-axis as its axis of symmetry. (this is the easiest case).

finally, after taking the limit as n → ∞ (so that we have infinitely many shells to fill out the solid), we get the exact formula.

example 1

find the volume of the solid generated by revolving the region under f(x) = x² + 1, where 2 ≤ x ≤ 6, around the y-axis.

solution

it might help to sketch a figure. fortunately, this is exactly what’s pictured in the figure above.

first identify the dimensions of a typical shell.

• r = x
• h = f(x) = x² + 1
• thickness = dx

in addition, we use a = 2 and b = 6 because we have 2 ≤ x ≤ 6.

now set up the shell method integral and evaluate to find the volume.

thus the volume is equal to 672π cubic units.

more than one function

when the generating region is defined as the area between two functions, then we have to modify the formula somewhat.

consider the region between two curves, y = f(x) on top and y = g(x) on bottom, between x = a and x = b.

the height at a typical sample x-value is equal to the difference of the two function values. that is,

h = y₂ – y₁ = f(x) – g(x)

this observation leads directly to the following version of the shell method formula:

revolving around different axes

there are also variations of the formula to cover cases in which the axis of revolution is not the y-axis. in all cases though, the axis must not be interior to the region itself.

be careful! the horizontal axis cases require that the functions be solved for x rather than y.

• vertical axis x = h entirely to the left of the region bounded above by y = f(x) and below by y = g(x):

  

• horizontal axis y = k entirely below the region bounded on the right by x = f(y) and on the left by x = g(y):

  

example 2

let r be the region bounded by y = x⁴ and y = 3x³. find the volume of the solid generated by revolving r around the line x = -2.

solution

let’s make a sketch. there is only a tiny sliver of area between the two curves.

to find the bounds of integration, we need to set the two functions equal.

so we get x = 0 and x = 3. these two values will be our a and b in the integral.

then identify the radius, height, and thickness of the typical shell.

• r = x – (-2) = x + 2
• h = f(x) – g(x) = 3x³ – x⁴
• thickness = dx

finally, put it all together and evaluate the definite integral.

the post ap calculus review: shell method appeared first on magoosh blog | high school.