2021 SUMMER MATH 2100 Calculus
6/29/2021 Final Exam Instructor: Evan Haskell Course: 2021 SUMMER MATH 2100 Calculus I Student: _____________________ Date: _____________________ Assignment: Final Exam 1. Find all verticalasymptotes, x = a, of the following function. For each value ofa, evaluate lim f(x), lim f(x), and lim f(x). x →a + x →a − x →a Use ∞ or − ∞ when appropriate. 2 f(x) = x − 7x + 10 2 x − 4x + 4 Select the correct choicebelow, and fill in the answer box if necessary. A. The vertical asymptote is x = lim f(x) = . The limits at this vertical asymptote are, lim f(x) = x →a + B. The vertical asymptote is x = lim f(x) = x →a +, and lim f(x) = . x →a x →a − . The limits at this vertical asymptote are, lim f(x) =, and lim f(x) does not exist and is x →a − x →a neither ∞ nor − ∞. C. There is no vertical asymptote. 2. Determine the intervals on which the following function is continuous.f(x) = x−6 2 x − 36 The function is continuous on theinterval(s) .(Simplify your answer. Type your answer in interval notation. Use a comma to separate answers asneeded.) 3. Differentiate. g(x) = 5x − 7 + x3 6x + 5 g′(x) = https://xlitemprod.pearsoncmg.com/api/v1/print/math 1/10 6/29/2021 Final Exam 4. An object thrown vertically upward from the surface of a celestial body at a velocity of 27 m/s reaches a height of s = − 0.9t2 + 27t meters in t seconds. a. Determine the velocity v of the object after t seconds. b. When does the object reach its highestpoint? c. What is the height of the object at the highestpoint? d. When does the object strike theground? e. With what velocity does the object strike theground? f. On what intervals is the speedincreasing? a. Theobject’s velocity after t seconds is v(t) = b. It takes the object(Simplify youranswer.) (1) (2) to reach its highest point. c. The object reaches a maximum height of(Simplify youranswer.) (3) d. The object strikes the ground after(Simplify youranswer.) (4) e. The object strikes the ground with the velocity(Simplify youranswer.) (5) f. The speed is increasing on theinterval(s) .(Simplify your answer. Type your answer in interval notation. Use a comma to separate answers asneeded.) (1) meters. (2) meters per second seconds. meters meters per second. (4) meters. seconds (5) (3) meters per second. seconds. meters. meters. meters per second. meters per second. seconds. seconds. 5. Calculate the derivative of the following function. 2 y = cos 3x + 4x + 4 dy = dx 6. Calculate the derivative of the following function. y= 8− e x 8 dy = dx https://xlitemprod.pearsoncmg.com/api/v1/print/math 2/10 6/29/2021 7. Final Exam a. Use implicit differentiation to find the derivative dy . dx b. Find the slope of the curve at the given point. 4 cos (y) = 5x − 5; 1, a. π 2 dy = dx b. The slope at 1, π 2 is .(Simplify youranswer.) 8. Find the derivative of the function f(y) = tan − 1 9y2 + 2 . f′(y) = 9. A circle has an initial radius of 50 ft when the radius begins decreasing at the rate of 4 ft/min. What is the rate in the change of area at the instant that the radius is 25 ft? Write an equation relating the area of acircle, A, and the radius of thecircle, r.(Type an exactanswer, using π asneeded.) Differentiate both sides of the equation with respect to t. dA dr = dt dt(Type an exactanswer, using π as needed. Type an expression using r as thevariable.) The rate of change of the area is(Type an exact answer in terms of π.) (1) ft 2 ft 3 (1) 3 / min. ft . / min. ft. ft / min. 2 ft . 10. A stone is launched vertically upward from a cliff 160 ft above the ground at a speed of 48 ft / s. Its height above the 2 ground t seconds after the launch is given by s = − 16t + 48t + 160 for 0 ≤ t ≤ 5. When does the stone reach its maximumheight? Find the derivative of s. s′ = The stone reaches its maximum height at(Simplify youranswer.) https://xlitemprod.pearsoncmg.com/api/v1/print/math s. 3/10 6/29/2021 Final Exam 11. Determine the intervals on which the following function is concave up or concave down. Identify any inflection points. 5 4 3 g(t) = 3t − 5t − 20t + 80 Determine the intervals on which the following functions are concave up or concave down. Select the correct choice below and fill in the answerbox(es) to complete your choice.(Simplify your answer. Type your answer in interval notation. Use a comma to separate answers asneeded.) A. The function is concave up on and concave down on B. The function is concave down on . . C. The function is concave up on . Locate any inflection points of g. Select the correct choice belowand, ifnecessary, fill in the answer box to complete your choice. A. An inflection point occurs at t = .(Use a comma to separate answers asneeded.) B. There are no inflection points for g. 12. Make a complete graph of the following function. If an interval is notspecified, graph the function on its domain. Use a graphing utility to check your work. 2 f(x) = 9x ln(x) 2 What is the correct graph of f(x) = 9x ln(x)? A. B. 4 y C. 4 y 4 x -4 4 -4 D. y 4 x -4 4 -4 y x -4 4 -4 x -4 4 -4 13. A cone is constructed by cutting a sector from a circular sheet of metal with radius 14. The cut sheet is then folded up and welded. Find the radius and height of the cone with maximum volume that can be formed in this way. r 14 14 The radius is r = and the height is h =(Type exactanswers, using radicals asneeded.) h . 14. Use a linear approximation to estimate the following quantity. Choose a value of a to produce a small error. e e 0.04 0.04 ≈(Simplify youranswer.) https://xlitemprod.pearsoncmg.com/api/v1/print/math 4/10 6/29/2021 15. Final Exam dT, which is the rate at which the temperature in a snowpack T dh changes with respect to its depth h. A large temperature gradient may lead to a weak layer in the snowpack. When these dT weak layerscollapse, avalanches occur. Avalanche forecasters use the rule of thumb that if exceeds 10°C / m dh anywhere in thesnowpack, conditions are favorable forweak-layer formation, and the risk of avalanche increases. Assume the temperature function is continuous and differentiable. Complete parts(a) through(d) below. Avalanche forecasters measure the temperature gradient a. An avalanche forecaster digs a snow pit and takes two temperature measurements. At the surface (h = 0), the temperature is − 13°C. At a depth of 1.9 m, the temperature is − 3°C. Using the Mean ValueTheorem, what can he conclude about the temperaturegradient? The average temperature gradient from h = 0 to h = 1.9 is(Simplify your answer. Round to the nearest tenth asneeded.) °C / m . Is the formation of a weak layerlikely? Select the correct answer below. No Yes b. One mileaway, a skier finds that the temperature at a depth of 1.1 m is − 2°C, and at the surface it is − 15°C. What can be concluded about the temperaturegradient? The average temperature gradient from h = 0 to h = 1.1 is(Simplify your answer. Round to the nearest tenth asneeded.) °C / m . Is the formation of a weak layerlikely? Select the correct answer below. Yes No c. Because snow is an excellentinsulator, the temperature ofsnow-covered ground is near 0°C. Furthermore, the surface temperature of snow in a particular area does not vary much from one location to the next. Explain why a weak layer is more likely to form in places where the snowpack is not too deep. Choose the correct answer below. A. A weak layer is more likely to formbecause, as the snowpack getsthinner, the temperature gradient gets larger. B. A weak layer is more likely to formbecause, as the change in temperature getslarger, the temperature gradient gets larger. C. A weak layer is more likely to formbecause, as the snowpack getsdeeper, the temperature gradient gets smaller. D. A weak layer is more likely to formbecause, as the change in temperature getssmaller, the temperature gradient gets smaller. d. The term isothermal is used to describe the situation where all layers of the snowpack are at the same temperature(typically near the freezingpoint). Is a weak layer likely to form in isothermalsnow? A. No. If all layers of snowpack are the sametemperature, then the change in temperature is zero and the temperature gradient is zero. B. No. If all layers of snowpack are the sametemperature, then the change in temperature is double and the temperature gradient is large. C. Yes. If all layers of snowpack are the sametemperature, then the change in temperature is zero and the temperature gradient is zero. D. Yes. If all layers of snowpack are the sametemperature, then the change in temperature is double and the temperature gradient is large. https://xlitemprod.pearsoncmg.com/api/v1/print/math 5/10 6/29/2021 Final Exam 16. Evaluate the following limit. Usel’Hôpital’s Rule when it is convenient and applicable. lim (x + π)cot x x→ − π lim (x + π)cot x =(Type an exactanswer.) x→ − π 17. Find all the antiderivatives of the following function. Check your work by taking the derivative.f(x) = 6 sin x − 5 The antiderivatives of f(x) = 6 sin x − 5 are F(x) = . 18. For the following functionf, find the antiderivative F that satisfies the given condition.f(u) = 4 e u + 5; F(0) = − 2 The antiderivative that satisfies the given condition is F(u) = . 19. The velocity of an object is given by the following function defined on a specified interval. Approximate the displacement of the object on this interval by subdividing the interval into the indicated number of subintervals. Use the left endpoint of each subinterval to compute the height of the rectangles. v = 1/(3t + 3) (m/s) for 0 ≤ t ≤ 8; n = 4 The approximate displacement of the object is(Round to two decimal places asneeded.) https://xlitemprod.pearsoncmg.com/api/v1/print/math m. 6/10 6/29/2021 Final Exam 20. Consider two functions f and g on [3,7] such that 7 7 7 5 3 3 5 3 ∫ f(x)dx = 10, ∫ g(x)dx = 6, ∫ f(x)dx = 5, and ∫ g(x)dx = 2. Evaluate the following integrals. 5 a. ∫ 2f(x)dx =(Simplify youranswer.) 3 7 b. ∫ (f(x) − g(x))dx =(Simplify youranswer.) 3 5 c. ∫ (f(x) − g(x))dx =(Simplify youranswer.) 3 7 d. ∫ (g(x) − f(x))dx =(Simplify youranswer.) 5 7 e. ∫ 8g(x)dx =(Simplify youranswer.) 5 3 f. ∫ 3f(x)dx =(Simplify youranswer.) 5 https://xlitemprod.pearsoncmg.com/api/v1/print/math 7/10
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