University:
Central Michigan University
Course:
MTH 232 | Linear Algebra and Differential Equations
Academic year:

2020
Views:

292

Pages:

2
Author:

Caiden Garner

The wind chill is the temperature, in degrees Fahrenheit (°F), a human feels based on the air temperature, in degrees Fahrenheit, and the wind velocity v, in miles per hour (mph). If the air temperature is 32° F, then the wind chill is given by W( v) = 55.6 -:- 22.Jv0•16 and is valid for 5 :,; v :,; 60. (a) Find W'(20). Using correct units, explain the meaning of W'(20) in terms of the wind chill. (b) Find the average rate of change of W over the interval 5 :,; v :,; 60. Find the value of v at which the instantaneous rate of change of W is equal to the average rate of change of W over the interval 5 :,; v :,; 60. (c) Over the time interval O :,; t :,; 4 hours, the air temperature is a constant 32°F. At time t = 0, the wind velocity is v = 20 mph. If the wind velocity increases at a constant rate of 511l.ph per hour, what is the rat€: of change of the wind chill with respect to time at t = 3 �ours? Indicate units of measure. (a) W'(20) = -22.1 · 0.16 • 20-0•84 = -0.285 or -0.286 When v = 20 mph, the wind chill is decreasing at 0.286 °F /mph. (b) The average rate of change of W over the interval . W(60)-W(5) 5 :,; v :,; 60 IS = -0.253 or -0.254. 60 _ 5 W(60)-W(5) when v = 23.911. W ,(v) = 60 _ 5 (c) _ 1:value 2 { • 1 : explanation 1 : average rate of change 3 : { 1 : W'( v) = average rate of change 1: value ofv dW = dW • dv)I = W'(35) • 5 = -0.892 °F /hr I ( dt t=J dv dt t=J '1 · OR W = 55.6 - 22.1(20 + 5t)°- 16 dW = -0.892 °F /hr I dt t=l Units of °F /mph in (a) and °F /hr in (c) 1:: =5 3: 1 : uses v(3) = 35, or uses v(t) = 20 + 5t 1: answer 1 : units in (a) and (c) Let/ be a twice-differentiable function defined on the interval -1.2 < x < 3.2 with/(!)= 2. The graph of/', the derivative off, is shown above. The graph of/ ' crosses the x-axis at x = -I and x = 3 and has a horizontal tangent at x = 2. Let g bethefunction givenby g(x) = ef(x) _ -----··--·-• (a) Write an eqnation for the line tangent to the graph of g at x = I, (b) For -1.2 < x < 3.2, ·find all values of x at which g has a local maximum. Justify your answer. y 2 -1 Graph off' J. (c) The second derivative of g is g"(x) = e((x) [(f'(x)) 2 + f"(x) Is g"(-1) positive, negative, or zero? Justify your answer. (d) Find the average rate of change of g', the derivative of g, over the interval [l, 3]. (a) g(l) = ef(l) = e2 l) x = -4e2 g'(x) = ef( ) f'(x), g'(l) = ef( /'(!) The tangent line is given by y = e2 - 4e2 (x - I). ( b) g'(x) = ef(x)f'(x) ef(x) > 0 for all x So, g' changes fro111_p_o_si_ti_ve_to_negativ�only when/' changes from positive to negative. This occurs at x = -I only. Thus, g has a local maximum at x = -I. 2 (c) g"(-1) = efH) [U'(-1)) +/"(-!)] ef( -1) > 0 and/'(-!)= 0 Since /' is decreasing on a neighborhood of -1, /"(-!) < 0. Therefore, g"(-1) < 0. (d) g'(3 - g ) �l ) 3-1 ef(3)/'(3) _ ef(l) /'(!) 2 2e2 l:g'(x) 3: { I : g(l) and g'(l) I : tangent line equation : I :answer 2 { 1 : justification : I :answer 2 { I : justification 2 :{ : 1 difference quotient I : answer

The Wind Chill is the Temperature Key

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