PRACTICE FINAL. Problem 1. Find the dimensions of the isosceles triangle with largest area that can be inscribed in a circle of radius 10cm.

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1 PRACTICE FINAL Problem 1. Find the dimensions of the isosceles triangle with largest area that can be inscribed in a circle of radius 1cm. Solution. Let x be the distance between the center of the circle and the base of the isosceles triangle. Then the base of the triangle has length x 2 and the height of the triangle is 1 + x. Therefore, area of the triangle is given by We first find the critical points of A(x): A(x) = 1 x 2 (1 + x) x [ 1, 1] A (x) = 1 x 2 x(1 + x) = 1 x2 1x x 2 1 x 2 1 x 2 Thus A (x) = when 1 1x 2x 2 =. The solution to this equation is given by 1 ± 9 = 1 or 5 4 For x = ±1, we have A(x) =. For x = 5, we get A (5) = which is the maximum area of the largest isosceles triangle that can be inscribed in the circle of radius 1 cm. For x = 5 we get that the base of the isosceles triangle has length 1 3. The other sides of the triangle have length equal to = 1 3. Thus the triangle is equilateral of side length 1 3 cm. Problem 2. Find area of the region bounded by the curves x = y 2 2, y = 1, y = ln(x) and y = 1. Solution. We compute the area by integrating relative to y. The its of integral are y = 1 and y 1. The left function is x = y 2 2. The right function is x = e y (so that y = ln(x)). Thus the definite integral we need to compute is ] 1 (e y y 2 + 2) dy = [e y y y = (e e 1 23 ) Problem 3. Evaluate the following its: (a) x 3e 2x 3 tan(2x) Solution. The it is an indeterminate form /. We apply the L Hospital rule 3e 2x 3 6e 2x = x tan(2x) x 2 sec 2 (2x) = 6 2 = 3 (b) (x3 7x 2 )e x

2 2 PRACTICE FINAL Solution. The it is an indeterminate form.. We apply L Hospital rule repeatedly to compute the it as follows: (x3 7x 2 )e x = = (c) x + 7x 2 6x 14 e x = e x = 6 e x = 1 + x 1 + sin(x) Solution. We start by rationalizing the numerator 1 + x 1 + sin(x) = x sin(x) The it x + x sin(x) ( 3x 2 14x e x ) x sin(x) can be computed by L Hospital rule as follows: x sin(x) 1 cos(x) sin(x) x + = x + 3x 2 = x + 6x = cos(x) = 1 x Thus the final answer is Problem 4. x x 1 + sin(x) (a) Evaluate the integral = x + (1 + x 1/4 ) 2 dx. ( x sin(x) = = 1 12 ) ( ) x sin(x) Solution. (1 + x 1/4 ) 2 = 1 + 2x 1/4 + x 1/2. An antiderivative of this function is Thus the definite integral is x + 2 x5/4 5/4 + x3/2 3/2 = x x5/ x3/2 (1 + x 1/4 ) 2 dx = (b) Find the derivative of g(x) = Solution. calculus: [ x x5/4 + 2 ] 1 3 x3/2 = = x 4 cos(t 2 ) dt. The derivative of g(x) can be computed using the fundamental theorem of ( ) d x 4 cos(t 2 ) dt = d ( u ) ( d(x cos(t 2 4 ) ) ) dt = cos(u 2 )4 = cos(x 8 )4 dx du dx In the intermediate step, we used u = x 4.

3 (c) If f(x) is a continuous function and PRACTICE FINAL 3 2 f(x) dx = 6, then compute π/2 f(2 sin(θ)) cos(θ) dθ. Solution. By the substitution rule, setting x = 2 sin(θ) gives dx = 2 cos(θ)dθ and we get π/2 f(2 sin(θ)) cos(θ) dθ = 1 f(x) dx = 3 2 Problem 5. Consider the function f(x) = ln(x 2 3x + 2). (a) Find the domain and range of f(x). Solution. f(x) = ln(x 2 3x + 2) = ln((x 1)(x 2)). Since ln is only defined for positive input values, the domain of f(x) is (, 1) (2, ). (b) Find the intervals of increase and decrease of f(x). Find the local max. and min. points of f(x). Solution. f (x) = 2x 3 x 2. Since the denominator is positive on the domain, we have 3x + 2 f (x) < for x < 3/2 f (x) > for x > 3/2. Thus f (x) = for x = 3/2. But this point is not in the domain of f(x), so f(x) has no critical points. (c) Find the intervals of concavity and inflection points of f(x). Solution. 2 f (x) = (x2 3x + 2)(2) (2x 3) 2 (x 2 3x + 2) 2 = 2x2 + 6x 5 (x 2 3x + 2) 2 f (x) is always negative, so the function is concave down everywhere. (d) Sketch the graph of f(x). We can compute the vertical asymptotes of f(x) as: f(x) = = f(x) x 2 + x 1 Moreover as x ±, the input x 2 3x + 2 for ln goes to and hence ln(x 2 3x + 2) goes to infinity.

4 4 PRACTICE FINAL Problem 6. Two people start walking from the same point. One walks east at 3 miles/hr and the other walks north at 2 miles/hr. How fast is the distance between them increasing after 15 minutes? Solution. Let x be the distance travelled by eastbound person and let y be the distance travelled by northbound person. We are given the information dx dy = 3 and = 2 (in miles per hour). The dt dt distance between them is given by s 2 = x 2 + y 2 Taking derivative of this equation relative to t gives 2s ds dt = 2xdx dy + 2y dt dt After 15 minutes, we have x = 3/4 and y = 2/4. At that time we have s = 13/4. Substituting these values in the equation above, we get ( 13/4) ds dt Thus ds dt = 13 miles per hour. = (3/4)(3) + (2/4)(2) Problem 7. Consider the region bounded between the curves y = x 2 and y = x. Find the volume of the solid obtained by rotating this region around y axis. Solution. The curves y = x 2 and y = x meet at (, ) and (1, 1). We find the volume by integrating relative to y. A typical y cross-section is the region bound between two concentric circles, with the radius of the outer circle y and the radius of the inner circle y 2. Thus the area of a typical y cross-section is Thus the volume of the solid is given by A(y) = π(y y 4 ) [ ] y Volume = π(y y ( ) dy = π 2 y5 1 = π ) = π Problem 8. Show that the sum of the x and y intercepts of any tangent line to the curve x + y = c is equal to c. Solution. Let us compute the equation of the tangent line to the given curve at a point (x, y ). Taking the derivative of x + y = c with respect to x gives 1 2 x dy y dx = dy y dx = x Thus the slope of the tangent line is y / x. The equation is y y y = (x x ) x The x intercept of this line is when y = and the y intercept is when x = : x intercept = x + x y y intercept = y + x y

5 The sum of these intercepts can be computed as: Problem 9. Compute the following: PRACTICE FINAL 5 x + 2 x y + y = ( x + y ) 2 = ( c) 2 = c (a) d dx ((cos(x))x ) Solution. Let y = cos(x) x. Therefore ln(y) = x ln(cos(x)). Taking derivative with respect to x gives: Hence y = cos(x) x (ln(cos(x)) x tan(x)) y y = ln(cos(x)) + x sin(x) cos(x) (b) 2 t 2 t + 3 dt Solution. Using the substitution x = 2 t + 3 we get dx = 2 t (ln(2))dt and hence 2 t 2 t + 3 dt = 1 dx ln(2) x = 1 ln(2) ln( x ) = 1 ln(2) ln(2t + 3) Problem 1. The acceleration of a car is given by a(t) = 3t 5m/s 2. Find the distance traveled by the car from t = to t = 3 s, if the initial velocity is 8 3 m/s. Since a(t) = 3t 5, we get that the velocity is given by v(t) = a(t)dt = 3t 2 /2 5t + constant The value of constant is determined by v() = constant = 8/3. Thus v(t) = 3t2 2 5t = 1 ( 9t 2 3t + 16 ) = 1 ((3t 8)(3t 2)). 6 6 This function changes sign at t = 8/3 and t = 2/3. Namely we have v(t) > for < t < 2/3 and 8/3 < t < 3. v(t) < for 2/3 < t < 8/3. Therefore the distance travelled is given by Distance = 2/3 v(t)dt 8/3 2/3 3 v(t)dt + v(t)dt 8/3 An antiderivative of v(t) = 3/2t 2 5t + 8/3 is given by s(t) = t 3 /2 5t 2 /2 + 8t/3. The distance travelled is given by Distance = s(2/3) s() s(8/3) + s(2/3) + s(3) s(8/3) = s(3) 2s(8/3) + 2s(2/3) s() The value of the function s(t) at these points is s(3) = 27/2 45/2 + 8 = 1 s(8/3) = 256/27 16/9 + 64/9 = 32/27 s(2/3) = 4/27 1/9 + 16/9 = 22/27

6 6 PRACTICE FINAL Thus the distance travelled is = = 3 m Problem 11. Use linear approximation to find the approximate value of sec(.1). Solution. We linearize the function f(x) = sec(x) at x =. f (x) = sec(x) tan(x) f() = 1 f () = Thus the linearization is given by L(x) = f() + f ()(x ) = 1. Thus sec(.1) 1. Problem 12. Find the absolute minimum of f(x) = x2 + 1 over x > and use it to prove that x x for every x >. x Solution. The derivative of f(x) is computed as follows. ( d x 2 ) + 1 = d ( x + 1 ) = 1 1 dx x dx x x 2 Over the domain (, ) there is only one critical point of f(x), namely x = 1. For < x < 1 the derivative is negative and for x > 1 the derivative is positive. Hence the function takes absolute minimum value at x = 1, namely f(1) = 2. Therefore for any x > we have f(x) f(1) = 2

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Solutions to old Exam 1 problems Solutions to old Exam 1 problems Hi students! I am putting this old version of my review for the first midterm review, place and time to be announced. Check for updates on the web site as to which sections

x 2 y 2 +3xy ] = d dx dx [10y] dy dx = 2xy2 +3y MA7 - Calculus I for thelife Sciences Final Exam Solutions Spring -May-. Consider the function defined implicitly near (,) byx y +xy =y. (a) [7 points] Use implicit differentiation to find the derivative

CALCULUS 2. 0 Repetition. tutorials 2015/ Find limits of the following sequences or prove that they are divergent. CALCULUS tutorials 5/6 Repetition. Find limits of the following sequences or prove that they are divergent. a n = n( ) n, a n = n 3 7 n 5 n +, a n = ( n n 4n + 7 ), a n = n3 5n + 3 4n 7 3n, 3 ( ) 3n 6n

TOPIC 3: CONTINUITY OF FUNCTIONS TOPIC 3: CONTINUITY OF FUNCTIONS. Absolute value We work in the field of real numbers, R. For the study of the properties of functions we need the concept of absolute value of a number. Definition.. Let

AP Calculus AB 1998 Scoring Guidelines AP Calculus AB 1998 Scoring Guidelines These materials are intended for non-commercial use by AP teachers for course and exam preparation; permission for any other use must be sought from the Advanced

AP Calculus AB 2006 Free-Response Questions AP Calculus AB 2006 Free-Response Questions The College Board: Connecting Students to College Success The College Board is a not-for-profit membership association whose mission is to connect students to

The graphs of f and g intersect at (0, 0) and one other point. Find that point: f(y) = g(y) y 2 4y 2y 2 6y = = 2y y 2. 2y(y 3) = 0 . Compute the area between the curves x y 4y and x y y. Let f(y) y 4y y(y 4). f(y) when y or y 4. Let g(y) y y y( y). g(y) when y or y. x 3 y? The graphs of f and g intersect at (, ) and one other point.

www.mathsbox.org.uk ab = c a If the coefficients a,b and c are real then either α and β are real or α and β are complex conjugates Further Pure Summary Notes. Roots of Quadratic Equations For a quadratic equation ax + bx + c = 0 with roots α and β Sum of the roots Product of roots a + b = b a ab = c a If the coefficients a,b and c

Catholic Schools Trial Examination 2004 Mathematics 0 Catholic Trial HSC Examination Mathematics Page Catholic Schools Trial Examination 0 Mathematics a If x 5 = 5000, find x correct to significant figures. b Express 0. + 0.. in the form b a, where a and

7.3 Volumes Calculus 7. VOLUMES Just like in the last section where we found the area of one arbitrary rectangular strip and used an integral to add up the areas of an infinite number of infinitely thin rectangles, we are

Limit processes are the basis of calculus. For example, the derivative. f f (x + h) f (x) SEC. 4.1 TAYLOR SERIES AND CALCULATION OF FUNCTIONS 187 Taylor Series 4.1 Taylor Series and Calculation of Functions Limit processes are the basis of calculus. For example, the derivative f f (x + h) f

D f = (2, ) (x + 1)(x 3) (b) g(x) = x 1 solution: We need the thing inside the root to be greater than or equal to 0. So we set up a sign table. . Find the domains of the following functions: (a) f(x) = ln(x ) We need x > 0, or x >. Thus D f = (, ) (x + )(x 3) (b) g(x) = x We need the thing inside the root to be greater than or equal to 0. So we

Microeconomic Theory: Basic Math Concepts Microeconomic Theory: Basic Math Concepts Matt Van Essen University of Alabama Van Essen (U of A) Basic Math Concepts 1 / 66 Basic Math Concepts In this lecture we will review some basic mathematical concepts

AP Calculus AB 2004 Scoring Guidelines AP Calculus AB 4 Scoring Guidelines The materials included in these files are intended for noncommercial use by AP teachers for course and eam preparation; permission for any other use must be sought from

AP Calculus AB First Semester Final Exam Practice Test Content covers chapters 1-3 Name: Date: Period: AP Calculus AB First Semester Final Eam Practice Test Content covers chapters 1- Name: Date: Period: This is a big tamale review for the final eam. Of the 69 questions on this review, questions will be

Calculus with Parametric Curves Calculus with Parametric Curves Suppose f and g are differentiable functions and we want to find the tangent line at a point on the parametric curve x f(t), y g(t) where y is also a differentiable function

Extra Problems for Midterm 2 Extra Problems for Midterm Sudesh Kalyanswamy Exercise (Surfaces). Find the equation of, and classify, the surface S consisting of all points equidistant from (0,, 0) and (,, ). Solution. Let P (x, y,

Mark Howell Gonzaga High School, Washington, D.C. Be Prepared for the Calculus Exam Mark Howell Gonzaga High School, Washington, D.C. Martha Montgomery Fremont City Schools, Fremont, Ohio Practice exam contributors: Benita Albert Oak Ridge High School,

Solutions to Study Guide for Test 3. Part 1 No Study Guide, No Calculator Solutions to Study Guide for Test 3 Part 1 No Study Guide, No Calculator 1. State the definition of the derivative of a function. Solution: The derivative of a function f with respect to x is the function

Items related to expected use of graphing technology appear in bold italics. - 1 - Items related to expected use of graphing technology appear in bold italics. Investigating the Graphs of Polynomial Functions determine, through investigation, using graphing calculators or graphing

f(x) = lim 2) = 2 2 = 0 (c) Provide a rough sketch of f(x). Be sure to include your scale, intercepts and label your axis. Math 16 - Final Exam Solutions - Fall 211 - Jaimos F Skriletz 1 Answer each of the following questions to the best of your ability. To receive full credit, answers must be supported by a sufficient amount

Analyzing Piecewise Functions Connecting Geometry to Advanced Placement* Mathematics A Resource and Strategy Guide Updated: 04/9/09 Analyzing Piecewise Functions Objective: Students will analyze attributes of a piecewise function including

(b)using the left hand end points of the subintervals ( lower sums ) we get the aprroximation (1) Consider the function y = f(x) =e x on the interval [, 1]. (a) Find the area under the graph of this function over this interval using the Fundamental Theorem of Calculus. (b) Subdivide the interval 21-114: Calculus for Architecture Homework #1 Solutions November 9, 2004 Mike Picollelli 1.1 #26. Find the domain of g(u) = u + 4 u. Solution: We solve this by considering the terms in the sum separately: Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x