AP Calculus BC Unit 9 Defining and Differentiating Parametric Equations

Unit 9: Defining and Differentiating Parametric Equations ● Parametric functions: a pair of functions that allow you to find the coordinates of both x and y, usually written in terms of ‘t’ ○ Parametric equations are usually written in terms of ‘t’. ● Calculating the derivative of parametric functions: 𝑑𝑦 ○ Ex. If 𝑥 = 𝑡 2 and 𝑦 = 𝑡 3 + 1, find 𝑑𝑥 ■ 1) find dx/dt and dy/dt → dx/dt = 2t and dy/dt = 3𝑡 2 𝑑𝑦 3𝑡 2 3 ■ 2) plug the values into the formula → 𝑑𝑥 = 2𝑡 = 2 𝑡 ● Calculating the second derivative of parametric equations: ○ Using the example from above, to find the second derivative of that parametric 𝑑𝑦 function, take the derivative of 𝑑𝑥 and then divide that by 𝑑𝑥 𝑑2 𝑦 → 𝑑𝑥 2 = 𝑑𝑡 (3/2) 2𝑡 3 = 4𝑡 ● Finding the arc length of parametric curves: ○ ■ From t=𝛼 to t=𝛽 ○ Ex. (from Princeton book): Find the length of the curve defined by x=sin(t) and y=cos(t), from t=0 to t=𝜋 ■ 1) find (dx/dt) and (dy/dt) → ■ 2) plug into the formula → 𝑑𝑥 𝑑𝑡 = 𝑐𝑜𝑠(𝑡)and 𝑑𝑦 𝑑𝑡 = −𝑠𝑖𝑛(𝑡) 𝜋 L=∫0 𝜋 √(𝑐𝑜𝑠 2 𝑡) + (𝑠𝑖𝑛2 𝑡) 𝑑𝑡=∫0 𝜋 √1 𝑑𝑡=∫0 𝑑𝑡=𝜋 ● Remember that (𝑠𝑖𝑛2 𝑡) + (𝑐𝑜𝑠 2 𝑡)= 1 ● Vectors: ○ Vectors are quantities that have both a magnitude and direction ○ Vector functions are similar to parametric functions because they describe the motion of a particle on a plane ○ Vector functions have similar relationships between position, velocity, and acceleration as we’ve seen earlier. ■ If a vector function P = position of a particle, then the derivative of P gives you velocity, and the derivative of velocity gives you the particle’s acceleration ● Both x(t) and y(t) have to be differentiable for the vector function to be differentiable as well ● To find speed of a vector function, take the magnitude of the velocity vector function ○ The magnitude of a vector function = √𝑥 2 + 𝑦 2 ○ Denotation of vectors: ■ ‘i’ and ‘j’ are the unit vectors in the x and y directions ■ x(t) and y(t) are real valued functions of the variable ‘t’ ○ Ex. (from Saxon Calc BC book): ○ Integrating vector valued functions: ■ Similar to how you take the derivative of vector valued functions by differentiating x(t) and y(t), and leaving the ‘i’ and ‘j’, to find the integral of vector-valued functions, integrate the real-valued functions and leave the ‘i’ and ‘j’ ■ Integrating a velocity vector gives you the displacement of the particle over the interval of time, and integrating a speed vector gives you the total distance travelled by the particle in that interval. ■ Ex.: ● Polar curves: ○ Polar coordinates: used to describe the location of points on a polar curve/graph ■ x = rcos𝜃and y = rsin𝜃 ● *x and y are defined parametrically, in terms of ‘𝜃′ ■ Differentiating in polar form: 𝑑𝑦 ● To find 𝑑𝑥 , use the equation for finding the slope of a parametric 𝑑𝑦 𝑑𝑦/𝑑𝜃 equation → 𝑑𝑥 = 𝑑𝑥/ 𝑑𝜃 . ○ Finding the area of a single polar curve: 𝑏 ■ A = ∫𝑎 1 2 𝑟 𝑑𝜃 2 ■ Ex. Find the area of the region bounded by the polar curve r=1+cos𝜃 1 ● 2𝜋 ∫ 2 0 (1 + 𝑐𝑜𝑠𝜃)2 𝑑𝜃 → integrate that ○ *to find polar area, you will often need to know these trig identities: 1 1 1 1 ■ 𝑐𝑜𝑠 2 𝜃= 2 + 2 𝑐𝑜𝑠𝜃 ■ 𝑠𝑖𝑛2 𝜃= 2 − 2 𝑐𝑜𝑠𝜃 ○ To find the limits of the integral, find the𝜃points where r consists of that region. ■ In this case, it takes from 0 to 2𝜋for “r” to be drawn completely, which is why those are the limits ○ Finding the area of a region bounded by 2 polar curves: 𝑏 ■ A = ∫𝑎 1 2 (𝑟2 2 −𝑟 1 2 ) 𝑑𝜃 with 𝑟2 (𝜃) > 𝑟 1 (𝜃) ■ Ex. (from Princeton book): find the area inside the circle r = 4 and outside of the curve r = 4 - cos𝜃(the shaded region in the graph below) ● 1) To find the limits of the integral, set the two equations equal to each other, and solve for 𝜃→ 4 = 4 - cos𝜃 = cos𝜃 = 0 at 𝜃= -𝜋/2 and 𝜋/2 ○ You could also establish the limits of the integral as 𝜋/2 ∫0 𝑟d𝜃, and multiply the whole integral by 2, since the graphs of the polar curves are symmetrical ● 2) find which function is on the “outside”, and in this case, r = 4 is 𝜋/2 ○ 2∫0 1 (4 − 𝑐𝑜𝑠𝜃)2 ] 𝑑𝜃 ● 3) integrate 𝜋/2 [2 (4)2 − (4 − 𝑐𝑜𝑠𝜃)2 ] 𝑑𝜃 OR ∫−𝜋/2 1 [2 (4)2 − ● Connecting polar, vector, and parametric functions: ○ A function can be written in parametric, rectangular/Cartesian, or polar form. ○ Ex. (on the next page): Unit 10: Infinite Sequences and Series ● Difference between sequences and series: ○ 𝑎 𝑛= sequence ○ S = ∑∞ 𝑛=1 𝑎 𝑛= series ○ Sequences are lists of the terms, whereas series are lists of the addends of the series. ■ Series: to find the second term of the series, you ADD 𝑎 𝑎 1 and 2 together ● This is known as partial sums → 2nd partial sum = 𝑆2 = 𝑎 𝑎 1 + 2 ○ nth partial sum = the sum of the first n terms of the series ● Convergence/Divergence: ○ A series converges only if the SEQUENCE OF PARTIAL SUMS (𝑆𝑛 ) converges to a finite number. If it doesn’t converge, then it diverges. ■ 𝑙𝑖𝑚 𝑆𝑛 = finite / a real number 𝑛→∞ ● Geometric series: ○ Defined as a series with a constant ratio between each term ○ In the form of ∑∞ 𝑛=1 𝑎𝑟 𝑛 , with ‘a’ representing the 1st term of the geometric series, and ‘r’ being the ratio that’s repeatedly expressed. 𝑎 ○ If |𝑟| < 1, then the geometric series converges to 1−𝑟 ■ 𝑙𝑖𝑚 ∑∞ 𝑛=1 𝑛→∞ 𝑎 𝑎𝑟 𝑛 = 1−𝑟 1 ■ Ex. 𝑙𝑖𝑚 ∑∞ 𝑛=1 1 (2)𝑛 → r = 2, which is < 1, so the series converges. 𝑛→∞ ● To find what it converges to, plug in numbers into the formula → 1 ○ a = (2)1 (a = the first term of the series, which means n=1) 1 ○ R = 2(r = the number that has an exponent ; the ratio) ■ 𝑎 (1/2) 1 (1/2) = = = 1 → 𝑙𝑖𝑚 ∑∞ 𝑛=1 1−𝑟 1−(1/2) (1/2) (2)𝑛 = 1 𝑛→∞ ○ If |𝑟| > 1, 𝑜𝑟 = 1,then the series diverges. ■ 𝑙𝑖𝑚 ∑∞ 𝑛=1 𝑛→∞ 𝑎𝑟 𝑛 = DNE ■ Ex. 𝑙𝑖𝑚 ∑∞ 𝑛=1 2𝑛 𝑛→∞ → r = 2, which is > 1, so the series diverges. ○ Ex. determine if the series converges/diverges: ∑∞ 𝑛=1 3 2𝑛 ■ *it may seem like this series converges at first, since it seems like r > 1. However, be careful when determining the value of the ratio. In this example, it may seem like r = 3/2, but if you look at the equation, 3 isn’t being raised to the nth power. ● Thus, it helps to rearrange the equation → ∑∞ 𝑛=1 1 3( )𝑛 2 ○ Now, it’s easier to see that the series does converges and find what it converges to → 𝑙𝑖𝑚 ∑∞ 𝑛=1 𝑛→∞ 1 (3/2) 3( )𝑛 = 1−(1/2)= 2 3 ● The nth term test for divergence: ○ Used to determine whether or not a function DIVERGES; it CANNOT tell you if a function converges ○ If 𝑙𝑖𝑚 𝑎𝑛 ≠0, then 𝑙𝑖𝑚 ∑∞ 𝑛=1 𝑛→∞ ■ Ex. ∑∞ 𝑛=1 𝑛→∞ 𝑎𝑛 diverges 𝑛 → 𝑙𝑖𝑚 n = ∞ ≠0 𝑛→∞ ○ If 𝑙𝑖𝑚 𝑎𝑛 = 0, then the series’ convergence cannot be determined yet. 𝑛→∞ ● Integral test for convergence: ∞ ∞ ○ Improper integrals: any integral in the form of ∫𝑐 𝑓(𝑥)dx, ∫−∞ 𝑓(𝑥) 𝑑𝑥, or 𝑏 ∫−∞ ■ If the limit of improper integrals exists, then the integral converges. If the limit DNE, then it diverges ∞ 1 ■ Ex. ∫1 1 𝑏 𝑑𝑥= 𝑙𝑖𝑚 ∫1 𝑥 𝑑𝑥= 𝑙𝑖𝑚 ln(b) - ln(1) = 𝑙𝑖𝑚 ln(b) = ∞→ 𝑥 𝑏→∞ 𝑏→∞ 𝑏→∞ integral DIVERGES because the limit of the integral isn’t FINITE ● Harmonic series and p series: 1 ○ p series: ∑∞ 𝑛=1 𝑛𝑝 , where p is a real number ■ Converges if p > 1 1 ● Ex. ∑∞ 𝑛=1 𝑛3 converges because 3 > 1 ● *unlike geometric series, you can’t determine what value a p-series converges to ■ Diverges if p ≤1 1 ● Ex. ∑∞ 𝑛=1 𝑛−3 diverges because -3 < 1 1 ○ Harmonic series: type of p series where p = 1 → ∑∞ 𝑛=1 𝑛 ■ Diverges because p is equal to 1 ■ Any multiple of the harmonic series also diverges 1 ● Ex. ∑∞ 𝑛=1 and ∑∞ 𝑛=1 2𝑛 5 𝑛 both diverge (−1)𝑛 ■ However, alternating harmonic series ( ∑∞ 𝑛=1 𝑛 ) do converge because as n → ∞,the series gets closer and closer to 0 ○ Comparison test for convergence: ■ If ∑∞ 𝑛=1 𝑎𝑛 converges, and 𝑐𝑛 < 𝑎𝑛 , then ∑∞ 𝑛=1 ● Ex. determine if the series ∑∞ 𝑛=1 ○ 1 converges or diverges 1 < 𝑛2 , and 𝑛2 converges, so by the comparison test, 𝑛2 −1 ∑∞ 𝑛=1 ■ If ∑∞ 𝑛=1 1 1 𝑛2 −1 𝑐𝑛 converges too. 1 also converges 𝑛2 −1 𝑎𝑛 diverges, and 𝑐𝑛 > 𝑎𝑛 , then ∑∞ 𝑛=1 𝑐𝑛 diverges too. ■ Limit comparison test: ● Ex. (from Rajendra Dahal’s Limit Comparison test video) 𝑛 ○ In the example above, the 𝑙𝑖𝑚 3𝑛+5= ⅓, which is a finite, 𝑛→∞ positive number, and because ∑∞ 𝑛=1 the series ∑∞ 𝑛=1 1 𝑛 diverges, so does 1 3𝑛+5 ○ Alternating series for convergence: ■ This is used to test a series that alternates from negative↔positive ● ∑∞ 𝑛=1 (−1)𝑛 𝑎𝑛 ■ Alternating series test: If you have an alternating series, and 𝑙𝑖𝑚 |𝑎𝑛 | = 𝑛→∞ 0, then ∑∞ 𝑛=1 𝑛 (−1) 𝑎𝑛 converges ● Ex. ∑∞ 𝑛=1 ○ 1 (−1)𝑛 𝑛 1 𝑙𝑖𝑚 |(−1)𝑛 | = 0, so ∑∞ 𝑛=1 𝑛→∞ 𝑛 1 (−1)𝑛 𝑛converges ○ Ratio test for convergence: ■ Let r = 𝑙𝑖𝑚 𝑎𝑛+1 𝑛→∞ 𝑎𝑛 for ∑∞ 𝑛=1 𝑎𝑛 ● If r > 1, then the series diverges ● If r < 1, then the series converges ● If r = 1, then the convergence/divergence of the series is inconclusive ■ Ex. determine if the series ∑∞ 𝑛=1 𝑛2 2𝑛 converges or diverges 1 ○ Because 𝑟 = 2 < 1, the series converges ○ Absolute convergence: ■ If an alternating series converges after taking the absolute value of the series function, then that series is said to converge absolutely ● Basically, if ∑∞ 𝑛=1 |𝑎𝑛 |converges, then ∑∞ 𝑛=1 𝑎𝑛 converges absolutely (−1)𝑛 ■ Ex. ∑∞ 𝑛=1 ∑∞ 𝑛=1 𝑛3 (−1)𝑛 𝑛3 → ∑∞ 𝑛=1 (−1)𝑛 | 𝑛3 |converges because of p-series, so is said to converge absolutely ○ Conditional convergence: ■ If an alternating series converges, and does NOT converge absolutely, then it’s said to converge conditionally ■ An alternating series that conditionally converges must follow these conditions: ● 𝑎𝑛 is positive for all n ● 𝑎𝑛+1 ≤ 𝑎𝑛 for all n ● 𝑙𝑖𝑚 𝑎𝑛 = 0 𝑛→∞ ○ Alternating Series Error Bound: ■ |𝑆 − 𝑆𝑛 | ≤ |𝑎𝑛+1 | ● S = the actual sum of the series ● 𝑆𝑛 =the partial sum of a convergent alternating series ● 𝑎𝑛+1 = the first unused term 1 ■ Ex. (from Princeton book): Find the error bound of ∑7𝑛=1 (−1)𝑛+1 𝑛2 ● Take the absolute value of the first unused term ○ In this case, to find 𝑆𝑛 , you added up the terms up until n=7, so the first unused term is when n=8 1 ■ |𝑆 − 𝑆7 | ≤ |𝑎7+1 |→ (−1)8 2= 1/64 = error bound 8 ● Taylor polynomials: ○ In the form of: ∑∞ 𝑛=0 𝑓′′′(𝑎) 3! 𝑓 (𝑛) (𝑎) 𝑛! (𝑥 − 𝑎)𝑛 = 𝑓(𝑎) 0! + 𝑓′(𝑎) 1! (𝑥 − 𝑎) + 𝑓′′(𝑎) 2! (𝑥 − 𝑎)2 + (𝑥 − 𝑎)3 . .. ■ Special case: when a=0, this is known as a Maclaurin series ■ *𝑓 𝑛 just represents the degree of the derivative ● For ex., 𝑓 5 stands for the fifth derivative of the function ○ Ex. Find the Taylor series of sinx expanded around x=𝜋 ■ 1) find the derivatives of sinx, and plug in their values at x=𝜋 n 𝑓𝑛 𝑓 𝑛 (𝜋) 0 sin(x) 0 1 cos(x) -1 2 -sin(x) 0 3 -cos(x) 1 4 sin(x) 0 0 ● 2) plug in the values into the Taylor Series formula → Sinx = 0! (𝑥 − 1 0 1 𝜋)0 − 1! (𝑥 − 𝜋)1 + 2! (𝑥 − 𝜋)2 + 3! (𝑥 − 𝜋)3 +. . = - ( x -𝜋) + (𝑥−5)5 5! ... (𝑥−𝜋)3 3! - ■ 3) combine the terms into one series equation → ∑∞ 𝑛=0 (−1)𝑛+1 (𝑥−𝜋)2𝑛+1 (2𝑛+1)! ○ Ex. Find the Taylor polynomial for 𝑒 𝑥 expanded around x=0 ● *IMPORTANT - memorize the taylor polynomial for 𝑒 𝑥 !! ○ Approximations: ■ To approximate the value of a function at a particular x value, just plug in the given x value into your equation ■ Ex. use the 4th degree Taylor polynomial to estimate 𝑒 3 ● Now that you know the Taylor polynomial for 𝑒 𝑥 ,all you do to find 𝑒 3 is to plug in x=3 into the first 4 terms of the series → 𝑒 3 = 1 + 3 + 32 33 + 3! = 17.5 2! ● Lagrange Error Bound ○ Lagrange’s Remainder: 𝑅𝑛 = 𝑓 (𝑛+1) (𝑐) (𝑥−𝑎)𝑛+1 (𝑛+1)! , where ‘c’ is in between ‘x’ and ‘a’ ■ 𝑅𝑛 represents the remainder (error) AFTER the nth degree partial sum ○ Ex. approximate e with an error less than 0.0001 ■ ∑∞ 𝑛=0 𝑥𝑛 𝑛! ≤0.001 ● For this problem, you need to first find the value of ‘n’ that gives you a number that’s ≤ 0.001 ○ The only way to do this is to plug in values ○ For ex. When n=3, 𝑎3 = 13 3! ≈ 0.1667 ■ *you know that x=1 because the problem asks you to approximate 𝑒 1 ■ 0.1667 > 0.001, so you have to keep on plugging in numbers for ‘n’, until you get a value ≤ 0.001 ■ In this case, when n=8, 𝑎8 = 18 8! ≈0.0000248, which is ≤ 0.001, and thus, to approximate e with an error less than 0.0001, find ∑8𝑛=0 1𝑛 𝑛! ● Radius and interval of convergence of power series: ○ *remember: a power series: ∑∞ 𝑛=0 𝑎𝑛 (𝑥 − 𝑟) ○ If a power series converges, it either converges over an interval, or to a single point. ■ Use the ratio test to determine interval of convergence: ● Ex. (from Princeton) Find the interval of convergence for : ∑∞ 𝑛=0 𝑥𝑛 1+𝑛2 ○ You have to check the endpoints of the interval to see if the function converges at the endpoints! If it does converge at an endpoint, place the interval in brackets, and if not, place the interval in parentheses. ■ If it converges at 1 endpoint and not the other, denote it like this: [a,b) or (a,b], with the bracket indicating convergence at that endpoint ○ Radius of convergence: ■ Similar to how the radius of a circle describes the length from the outer edge of the circle to the centerpoint, the radius of convergence describes the distance between the midpoint and the endpoint of the interval of convergence. ■ In the example above, the interval of convergence was [-1,1]. Thus, the radius of convergence is 1, since the midpoint of that interval is 0, and the distance between 0 (the midpoint) and 1 (the endpoint) is 1. ● Representing functions as Power series: ○ Differentiation of power series ■ 1) write out the first few terms of the power series ● Ex. f(x) = 𝑒 𝑥 = 1 + x + 𝑥2 + 2! 𝑥3 + 3! 𝑥4 4! +. .. ■ 2) differentiate the written terms as you normally would 𝑥2 ● Ex. f’(x) = 𝑒 𝑥 = 1 +x + 2! + 𝑥3 3! + …. ■ 3) rewrite the new Taylor polynomial (if needed) ● Ex. f’(x) = ∑∞ 𝑛=0 (𝑥)𝑛+1 (𝑛+1)! ○ Integration of power series: ■ Same steps as differentiating power series goes for integration, except for you’re going to integrate the function instead of differentiating it ● Ex. h(x) = 1 + x + 𝑥 2 + 𝑥 3 + 𝑥 4 +. .. (Maclaurin series for -ln(1-x)). If g’(x) = h(x), then find the power series for g(x) ○ ∫ (1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 +. ..)dx = x + =∑∞ 𝑛=0 ○ Substitution of power series: 𝑥 𝑛+1 𝑛+1 𝑥2 + 2 𝑥3 + 3 𝑥4 4 +. .. ■ If f(u) = ∑∞ 𝑛=0 ∑∞ 𝑛=0 𝑎𝑛 𝑢𝑛 , and u = g(x), then f(u) = f(g(x)) = 𝑎𝑛 (𝑔(𝑥))𝑛 ● Ex. the power series for 𝑒 𝑥 = 1 + x + 3 The power series for 𝑒 𝑥 =1 + 𝑥 3 + 𝑥6 2! 𝑥2 + 2! + 𝑥9 3! 𝑥3 + 3! + 𝑥4 𝑥 12 4! +. .. = ∑∞ 𝑛=0 4! +. .. = ∑∞ 𝑛=0 𝑥𝑛 𝑛! 𝑥 3𝑛 𝑛! 3 ○ Notice how to find the power series for 𝑒 𝑥 , you just plug in (𝑥 3 ) into the power series for the original function, 𝑒 𝑥