Calculus Past Paper PDF

## 1. Representing a Function as a Power Series 1. Represent the function *f(x) = 10 ln(1 + 3x)* as a power series. Find its radius of convergence and its interval of convergence. - **Step 1: Expressing 1/(1-x) as a power series** * We know that the geometric series formula is: 1 / (1...

## 1. Representing a Function as a Power Series 1. Represent the function *f(x) = 10 ln(1 + 3x)* as a power series. Find its radius of convergence and its interval of convergence. - **Step 1: Expressing 1/(1-x) as a power series** * We know that the geometric series formula is: 1 / (1 - x) = ∑ (x)^n for |x| < 1 n=0 - **Step 2: Expressing 1/(1+3x) as a power series** * Substitute -3x for x in the geometric series formula: 1 / (1 + 3x) = ∑ (-3x)^n for |3x| < 1 n=0 - **Step 3: Expressing ln(1+3x) as a power series** * Integrate both sides of the equation from Step 2 with respect to *x*: ∫(1 / (1 + 3x)) dx = ∫∑ (-3x)^n dx n=0 * Evaluate the integrals: ln(1 + 3x) = ∑ (-1)^n * (3^n * x^(n+1)) / (n+1) + C for |3x| < 1 n=0 * Since the functions agree at x = 0, the constant of integration *C* is 0. - **Step 4: Expressing 10 ln(1+3x) as a power series** * Multiply both sides of the equation from Step 3 by 10: 10 ln(1 + 3x) = ∑ (-1)^n * (3^(n+1) * x^(n+1)) / (n+1) for |3x| < 1 n=0 - **Step 5: Finding the radius of convergence** * The power series converges for |3x| < 1, which simplifies to |x| < 1/3. * Therefore, the radius of convergence *R* is 1/3. - **Step 6: Finding the interval of convergence** * The interval of convergence is (-1/3, 1/3). However, we need to check the endpoints: * At x = -1/3, the series becomes: ∑ (-1)^n * (3^(n+1) * (-1/3)^(n+1)) / (n+1) = ∑ (-1)^n / (n+1) which is the alternating harmonic series and converges. * At x = 1/3, the series becomes: ∑ (-1)^n * (3^(n+1) * (1/3)^(n+1)) / (n+1) = ∑ 1/(n+1) which is the harmonic series and diverges. - **Step 7: Conclusion** * The power series representation of *f(x) = 10 ln(1 + 3x)* is: 10 ln(1 + 3x) = ∑ (-1)^n * (3^(n+1) * x^(n+1)) / (n+1) for |x| < 1/3 n=0 * The radius of convergence is *R* = 1/3. * The interval of convergence is (-1/3, 1/3]. ## 2. Finding the Maclaurin Polynomial and Estimating an Integral 2. Find the Maclaurin polynomial of degree 5 for *F(x) = ∫e^(-t^2) dt*. Use this polynomial to estimate ∫e^(-t^2) dt. Give an upper bound for your error of your estimate. - **Step 1: Finding the Maclaurin series for e^x** * The Maclaurin series for e^x is: e^x = ∑ (x^n) / n! for all x n=0 - **Step 2: Finding the Maclaurin series for e^(-t^2)** * Substitute -t^2 for x in the Maclaurin series for e^x: e^(-t^2) = ∑ (-t^2)^n / n! for all t n=0 - **Step 3: Finding the Maclaurin polynomial of degree 5 for F(x)** * Integrate the Maclaurin series for e^(-t^2) term by term: F(x) = ∫e^(-t^2) dt = ∫∑ (-1)^n * (t^(2n)) / n! dt n=0 F(x) = ∑ (-1)^n * (t^(2n+1)) / (n! * (2n+1)) + C for all x n=0 * The Maclaurin polynomial of degree 5 is obtained by taking the first 3 terms: P5(x) = x - (x^3) / 3! + (x^5) / 5! - **Step 4: Estimating ∫e^(-t^2) dt** * Substitute x = 1 into the Maclaurin polynomial: P5(1) = 1 - (1^3)/3! + (1^5)/5! = 1 - 1/6 + 1/120 = 103/120 - **Step 5: Finding an upper bound for the error** * Since ∫e^(-t^2) dt is an alternating series, we can use the Alternating Series Estimation Theorem: |∫e^(-t^2) dt - 103/120 | ≤ (1^7)/(7!) = 1/5040 - **Step 6: Conclusion** * The Maclaurin polynomial of degree 5 for *F(x) = ∫e^(-t^2) dt* is *P5(x) = x - (x^3)/3! + (x^5)/5!*. * The estimate of ∫e^(-t^2) dt using the polynomial is 103/120. * The upper bound for the error of the estimate is 1/5040. ## 3. Maclaurin Series and Derivatives 3. (a) Write down the Maclaurin series for *f(x) = cos(x^2/4)*. (b) Compute the 10th derivative of *cos(x^2/4)* at *x = 0*. - **(a) Finding the Maclaurin series for cos(x^2/4)** * The Maclaurin series for cos(x) is: cos(x) = ∑ (-1)^n * (x^(2n)) / (2n)! for all x n=0 * Substitute x^2/4 for x in the Maclaurin series for cos(x): cos(x^2/4) = ∑ (-1)^n * (x^(2n) / 4^(2n)) / (2n)! for all x n=0 - **(b) Computing the 10th derivative of cos(x^2/4) at x = 0** * The 10th derivative of cos(x^2/4) is obtained by looking at the coefficient of x^10 in the Maclaurin series. Since the Maclaurin series only has even powers of *x*, the coefficient of x^10 is 0. * Therefore, the 10th derivative of *cos(x^2/4)* at *x = 0* is 0. ## 4. Line of Intersection and Distance 4. (a) Find the parametric equations for the line of intersection of the planes *x + y + z = 3* and *2x - 3y = 6*. (b) Find the distance between the point *P(0, 0, 1)* and the line of intersection. - **(a) Finding the parametric equations for the line of intersection** * **Step 1: Finding direction vector** * The normal vectors to the two planes are *n1 = <1, 1, 1>* and *n2 = <2, -3, 0>*. * The direction vector of the line of intersection is perpendicular to both normal vectors, so it can be found by taking their cross product: * **v = n1 x n2 = <3, 2, -5>** * **Step 2: Finding a point on the line** * Solve the system of equations formed by the two plane equations to find a point on the line of intersection. Let *x = 0*. * Substituting into the plane equations, we get *y = -2* and *z = 5*. * Therefore, a point on the line is *Q = (0, -2, 5)*. * **Step 3: Writing the parametric equations** * The parametric equations of the line are: * **x = 0 + 3t** * **y = -2 + 2t** * **z = 5 - 5t** - **(b) Finding the distance between P(0, 0, 1) and the line of intersection** * **Step 1: Finding the vector PQ** * **PQ = Q - P = <0, -2, 5> - <0, 0, 1> = <0, -2, 4>** * **Step 2: Finding the projection of PQ onto v** * **proj_v PQ = ((PQ · v) / ||v||^2) * v** * **(PQ · v) = <0, -2, 4> · <3, 2, -5> = -18** * **||v||^2 = 3^2 + 2^2 + (-5)^2 = 38** * **proj_v PQ = (-18/38) * <3, 2, -5> = <-27/19, -9/19, 45/19>** * **Step 3: Finding the distance** * The distance between *P* and the line is the magnitude of the vector projection of *PQ* onto *v*: * **d = ||proj_v PQ|| = √((-27/19)^2 + (-9/19)^2 + (45/19)^2) = √(1184) / √(38)** - **Conclusion** * The parametric equations for the line of intersection are *x = 0 + 3t*, *y = -2 + 2t*, and *z = 5 - 5t*. * The distance between *P(0, 0, 1)* and the line of intersection is √(1184) / √(38). ## 5. Plane Equation and Distance 5. (a) Find the equation of the plane that passes through the three points *A(1, 0, 0)*, *B(0, 2, 0)*, and *C(0, 0, 3)*. (b) Find the distance between the origin and this plane. **(a) Finding the equation of the plane** - **Step 1: Finding two vectors in the plane** * **AB = B - A = <-1, 2, 0>** * **AC = C - A = <-1, 0, 3>** - **Step 2: Finding the normal vector** * The normal vector to the plane is perpendicular to both *AB* and *AC*, so we can find it by taking their cross product: * **n = AB x AC = <6, 3, 2>** - **Step 3: Writing the equation of the plane** * The equation of the plane is: * **a(x - x0) + b(y - y0) + c(z - z0) = 0**, where (x0, y0, z0) is a point on the plane and **<a, b, c>** is the normal vector. * Using point *A(1, 0, 0)* and the normal vector *<6, 3, 2>*, the equation becomes: * **6(x - 1) + 3(y - 0) + 2(z - 0) = 0** * **6x + 3y + 2z = 6** **(b) Finding the distance between the origin and the plane** - **Step 1: Finding the unit normal vector** * **n = <6, 3, 2>** * **||n|| = √(6^2 + 3^2 + 2^2) = 7** * **n_hat = n / ||n|| = <6/7, 3/7, 2/7>** - **Step 2: Finding the distance** * The distance between the origin *O(0, 0, 0)* and the plane is given by: * **d = |OA · n_hat| = |<0, 0, 0> · <6/7, 3/7, 2/7>| = 0** * However, this result is incorrect since the origin is not on the plane. We missed the absolute value of the equation for the plane. * The correct distance is: * **d = |(6 * 0 + 3 * 0 + 2 * 0 - 6) / √(6^2 + 3^2 + 2^2)| = 6/7** - **Conclusion** * The equation of the plane passing through points *A(1, 0, 0)*, *B(0, 2, 0)*, and *C(0, 0, 3)* is *6x + 3y + 2z = 6*. * The distance between the origin and the plane is *6/7*. ## 6. Bonus Problem: Vector Operations 6. (Bonus Problem) Given nonzero vectors *u*, *v*, and *w*, use dot product and cross product notation, as appropriate, to describe the following. - **(a) The vector projection of *u* onto *v*** * **proj_v u = ((u · v) / ||v||^2) * v** - **(b) A vector orthogonal to *u* and *v*** * **u x v** - **(c) A vector orthogonal to *u x v* and *w* (you can assume that *u* is not orthogonal to *w*)** * **(u x v) x w** - **(d) The area of the parallelogram determined by *u* and *w*** * **||u x w||**