Question Description

1.

Find the particular solution to y ‘ = sin(x) given the general solution is y = C – cos(x) and the initial condition y of pi over 2 equals 1 . (5 points)

-cos(x)

2 – cos(x)

-1 – cos(x)

1 – cos(x)

2.

The slope of the tangent to a curve at any point (x, y) on the curve is negative 1 times x divided by y . Find the equation of the curve if the point (2, -2) is on the curve. (5 points)

x + y = 0

x2 – y2 = -2

x2 + y2 = 16

x2 + y2 = 8

3.

The rate of decay in the mass, M, of a radioactive substance is given by the differential equation dM dt equals negative 1 times k times M , where k is a positive constant. If the initial mass was 100g, then find the expression for the mass, M, at any time t. (5 points)

M = e-kt

M = 100 e-kt

M = 100 ekt

M = 100ln(kt)

4.

The temperature of a pot of coffee varies according to Newton’s Law of Cooling: dT dt equals negative k times the quantity T minus A , where T is the temperature of the coffee, A is the room temperature, and k is a positive constant. If the water cools from 90°C to 85°C in 1 minute at a room temperature of 30°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes. (5 points)

72

42

68

81

5.

The differential equation dy dx equals the quotient of the quantity y minus 2 and y plus 1 (5 points)

I. produces a slope field with horizontal tangents at y = 2
II. produces a slope field with vertical tangents at y = -1
III. produces a slope field with columns of parallel segments

I only

II only

I and II

III only

6.

Which of the following differential equations is consistent with the following slope field?

In quadrant one, all slopes are positive. Greater values of y have slopes approaching horizontal and y less than 1 have slopes approaching vertical. For quadrant two slopes are negative with slopes approaching horizontal for greater values of y. Quadrant three also has all negative slopes. Larger negative values of y have slopes approaching horizontal. All slopes are positive in quadrant four. (5 points)

dy over dx equals x divided by y squared

dy over dx equals x divided by y

dy over dx equals x squared divided by y

dy over dx equals x squared divided by y squared

7.

The general solution of the differential equation dy – 0.2x dx = 0 is a family of curves. These curves are all (5 points)

lines

hyperbolas

parabolas

ellipses

8.

Estimate the value of the integral from negative 1 to 3 of x cubed, dx by using the Trapezoidal Rule with n = 4. (5 points)

8

19

22

36

9.

The table below gives selected values for the function f(x). With 5 rectangles, using the midpoint of each rectangle to evaluate the height of each rectangle, estimate the value of the integral from 1 to 2 of f of x, dx . (5 points)

x 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0
f(x) 1 0.909 0.833 0.769 0.714 0.667 0.625 0.588 0.556 0.526 0.500

0.7456

0.6456

0.6919

0.6932

10.

Given f(x) > 0 with f ′(x) < 0, and f ′′(x) < 0 for all x in the interval [0, 1] with f(0) = 1 and f(1) = 0.3, the left, right, trapezoidal, and midpoint rule approximations were used to estimate the integral from 0 to 1 of f of x, dx. The estimates were 0.7915, 0.8405, 0.8410, 0.8421 and 0.8895, and the same number of subintervals were used in each case. Match the rule to its estimate. (5 points)

abcde 1. trapezoidal
abcde 2.

right endpoint

abcde 3.

actual area

abcde 4.

midpoint

abcde 5.

left endpoint

a. 0.8895
b.

0.7915

c.

0.8421

d.

0.8405

e.

0.8410

1.Find the particular solution to y ‘ = sin(x) given the general solution is y = C – cos(x) and the initial condition . (5 points)-cos(x)2 – cos(x)-1 – cos(x)1 – cos(x)2.The slope of the tangent to a curve at any point (x, y) on the curve is . Find the equation of the curve if the point (2, -2) is on the curve. (5 points)x + y = 0x2 – y2 = -2×2 + y2 = 16×2 + y2 = 83.The rate of decay in the mass, M, of a radioactive substance is given by the differential equation , where k is a positive constant. If the initial mass was 100g, then find the expression for the mass, M, at any time t. (5 points)M = e-ktM = 100 e-ktM = 100 ektM = 100ln(kt)4.The temperature of a pot of coffee varies according to Newton’s Law of Cooling: , where T is the temperature of the coffee, A is the room temperature, and k is a positive constant. If the water cools from 90°C to 85°C in 1 minute at a room temperature of 30°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes. (5 points)724268815.The differential equation (5 points)I. produces a slope field with horizontal tangents at y = 2 II. produces a slope field with vertical tangents at y = -1 III. produces a slope field with columns of parallel segmentsI onlyII onlyI and IIIII only6.Which of the following differential equations is consistent with the following slope field? (5 points)7.The general solution of the differential equation dy – 0.2x dx = 0 is a family of curves. These curves are all (5 points)lineshyperbolasparabolasellipses8.Estimate the value of by using the Trapezoidal Rule with n = 4. (5 points)81922369.The table below gives selected values for the function f(x). With 5 rectangles, using the midpoint of each rectangle to evaluate the height of each rectangle, estimate the value of . (5 points) x11.11.21.31.41.51.61.71.81.92.0f(x)10.9090.8330.7690.7140.6670.6250.5880.5560.5260.5000.74560.64560.69190.693210.Given f(x) > 0 with f ′(x) < 0, and f ′′(x) < 0 for all x in the interval [0, 1] with f(0) = 1 and f(1) = 0.3, the left, right, trapezoidal, and midpoint rule approximations were used to estimate . The estimates were 0.7915, 0.8405, 0.8410, 0.8421 and 0.8895, and the same number of subintervals were used in each case. Match the rule to its estimate. (5 points)abcde1.trapezoidalabcde2.right endpointabcde3.actual areaabcde4.midpointabcde5.left endpointa.0.8895b.0.7915c.0.8421d.0.8405e.0.8410