Calculus I Practice Exam 2B

Transcription

1 This practice exam emphasizes conceptual connections and understanding to a greater degree than the exams that are usually administered in introductory single-variable calculus courses. It is designed to guide students who are taking such courses to a deeper mastery of the material. While a number of questions here are fairly typical for actual examinations, you should not infer from the expression practice exam that exams encountered in introductory singlevariable calculus courses will ask the same types of questions. Multiple choice 1. Use linear approximation to approximate 401 and round to 6 decimal places. You will not get credit for a decimal approximation of the exact value. a. 20 b c d Find the tangent slope of the curve xy 2 + x 2 y = 2 at (1,1). a. dy dx = 0 b. dy dx = 1 3 c. dy dx = 1 d. dy dx = 1 3. A geologist finds that a stalagmite increases in volume by approximately 1 cubic centimeter every century. Material is deposited on it in such a way that it maintains an approximately cylindrical shape with a constant ratio between the height and the radius of 10:1. The stalagmite has already reached a height of 1 meter. At which rate is its height currently increasing, in centimeters per century? Round your answer to three decimal places. a b c d

2 4. Given the graphs of f and g on the right, and if H = f g, evaluate H (1). a. 0 b. 1 c. 1 d. Does not exist because f is not differentiable at x = Given the graphs of f and g on the right, and if H = g/f, evaluate H (1.5). a. 2 3 b. 2 3 c. 1 d. Does not exist. 6. The Solar and Heliospheric Observatory (SOHO) found the radius of the sun to be 696,342 ± 65km. The sun is approximately a sphere. What is the uncertainty in its volume that corresponds to the uncertainty in its radius? Use differentials. You will not get credit for the exact answer computed as the difference of two spheres. a million km 3 b million km 3 c trillion km 3 d trillion km 3 7. Based on the table of values for the differentiable, invertible function f and its derivative, evaluate (f 1 ) (2). a. 1 x f(x) f (x) b. 1 7 c. 1 4 d. Cannot be determined based on the given information.

3 8. Evaluate the it: arctan( ln x) a. 1 b. π 2 c d. 9. If f has the tangent y = 2 x at x = 1 and g has the tangent y = 2x + 1 at x = 1, what is the tangent of fg at x = 1? a. y = (2 x) (2x + 1) b. y = 4 x c. y = 3 4x d. y = 2 x 10. Find and simplify the derivative of f(x) = sin 1 ( 1 x ). a. sin 2 ( 1 x 3) b. sin 2 ( 1 x 3) c. 1 x 4 x 2 1 d. x 4 x 2 Free response 1. Find the tangent line L(x) to the curve f(x) = sin x at x = 0. Then approximate sin 0.1 by evaluating L(x) at a suitable x value. Compare to the calculator value. 2. Apply L Hospital s Rule to evaluate the it exactly: ( 1 e x 1 1 x ) x 3. Use L Hospital s Rule to evaluate the it exactly: xsin

4 Answers: Multiple Choice: 1B 2B 3B 4D 5A 6C 7A 8B 9B 10C Free Response: 1. The equation of the tangent line of a function f at x = a is L(x) = f(a) + f (a)(x a) In this case, a = 0 and f (x) = cos x, so f (a) = 1. Since f(a) = sin 0 = 0, the tangent is L(x) = x We know that the tangent approximates the function value near the point of attachment. Therefore, sin 0.1 L(0.1) = 0.1 This agrees well with the calculator value sin 0.1 = Since the given it is an indeterminate form of type, we have to turn the difference into a suitable quotient before we can apply L Hospital s rule. We do this by creating a common denominator and combining the two fractions. ( 1 e x 1 1 x ) = x (e x 1) (e x 1)x The new it is an indeterminate form of type 0, so L Hospital s rule applies: 0 x (e x 1) (e x 1)x = 1 e x e x (x + 1) 1

5 This it is again an indeterminate form of type 0, so we use L Hospital s rule again. After that, 0 we can find the it by evaluation. 1 e x e x (x + 1) 1 = e x e x (x + 2) = 1 (x + 2) = Since the given it is an indeterminate form of type 0 0, we have to use the technique we learned in class to be able to apply L Hospital s rule: xsin x x sin x = eln We must now evaluate the it of the exponent: ln x sin x. This is an indeterminate form of type 0. We rewrite it as a quotient of type : ln x sin x = ln x = (sin x) 1 1/x (sin x) 2 cos x To simplify this it, we factor the fraction into a product of two fractions whose its we know and then 1/x (sin x) 2 cos x = sin x x sin x cos x = sin x x Therefore, the original it we were asked to evaluate is e 0 = 1. tan x = 0 1 = 0. Note: while this guide is being made freely available to ASU students and the general public for personal use, it is not to be uploaded to third-party websites, especially not ones that profit from such content. If you found this document on a third-party website such as Course Hero or Chegg, the document is being served to you in violation of copyright law.