#$&* course Mth 279 6/10 12 Question: `q001. Find the first and second derivatives of the following functions:
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. Sketch a graph of the function y = 3 sin(4 t + 2). Don't use a graphing calculator, use what you know about graphing. Make your best attempt, and describe both your thinking and your graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Multiplying the sine function by 3 will increase the amplitude of the graph so instead of bouncing between 1 and -1, the graph tavels between 3 and -3. The coefficient 4 of t inside the function will shorten the wavelenth to be one quarter the wavelenth of a standard sine graph so pi/2 instead of 2pi. The +2 shifts the graph to the left 2.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. Describe, in terms of A, omega and theta_0, the characteristics of the graph of y = A cos(omega * t + theta_0) + k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The cos wave will have an amplitude of A, a wavelength of 2pi*(1/omega), it will be shifted to the left theta_0, and shifted up k
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. Find the indefinite integral of each of the following: f(t) = e^(-3 t) x(t) = 2 sin( 4 pi t + pi/4) y(t) = 1 / (3 x + 2) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(t) = e^(-3 t) integral f(t) = (-1/3)(e^-3t) + C integral x(t) = -1/(2*pi) * cos(4*pi*t + pi/4) + C integral y(t) =(1/3)*ln(3t + 2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. Find an antiderivative of each of the following, subject to the given conditions: f(t) = e^(-3 t), subject to the condition that when t = 0 the value of the antiderivative is 2. x(t) = 2 sin( 4 pi t + pi/4), subject to the condition that when t = 1/8 the value of the antiderivative is 2 pi. y(t) = 1 / (3 t + 2), subject to the condition that the limiting value of the antiderivative, as t approaches infinity, is -1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: integral f(t) = (-1/3)(e^-3t) + C 2 = (-1/3)(e^-3t) + C at t = 0 2 = (-1/3)(e^0) + C C = 7/3 integral x(t) = -1/(2*pi)*cos(4*pi*t + pi/4) + C 2*pi = -1/(2*pi)*cos(4*pi*t + pi/4) + C at t = 1/8 2*pi = -1/(2*pi)*cos(4*pi*(1/8) + pi/4) + C C = 2*pi - sqrt(2)/(4*pi) integral y(t) = (1/3)*ln(3t + 2) + C As t approaches infinity, ln(3t + 2) approaches infinity. Therefore a value for C cannot be found. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. Use partial fractions to express (2 t + 4) / ( (t - 3) ( t + 1) ) in the form A / (t - 3) + B / (t + 1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2t + 4 = A(t+1) + B(t-3) set t=3 A = 5/2 2t + 4 = A(t+1) + B(t-3) set t=-1 B = -1/2 (5/2) / (t - 3) + (-1/2) / (t + 1) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. The graph of a function f(x) contains the point (2, 5). So the value of f(2) is 5. At the point (2, 5) the slope of the tangent line to the graph is .5. What is your best estimate, based on only this information, of the value of f(2.4)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Use y = mx +b We know the slope(m) is equal to .5 y = .5x + b use point given 5 = .5(2) + b b = 4 giving the function y = .5x + 4 at x = 2.4 y = .5(2.4) +4 y = 5.2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. The graph of a function g(t) contains the points (3, 4), (3.2, 4.4) and (3.4, 4.5). What is your best estimate of the value of g ' (3), where the ' represents the derivative with respect to t? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: slope = (y2 - y1)/(x2 - x1) slope = (4.4 - 4)/(3.2 - 3) = 2 slope = (4.5 - 4.4)/(3.4 - 3.2) = 1/2 graph is increasing at a decreasing rate, so slope at 3 should be more than 2 g'(3) = 5/2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:"