course Mth 173 z~ió͆assignment #012
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20:39:54 `qNote that there are 12 questions in this assignment. `q001. When we form the composite of two functions, we first apply one function to the variable, then we apply the other function to the result. We can for example first apply the function z = t^2 to the variable t, then we can apply the function y = e^z to our result. If we apply the functions as specified to the values t = -2, -1, -.5, 0, .5, 1, 2, what y values to we get?
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RESPONSE --> If t = -2, then z = t^2 = (-2)^2 = 4, so y = e^z = e^4 = 55, If t = -1, then z = t^2 = (-1 )^2 = 1, so y = e^z = e^1 = 2.72, If t = -.5, then z = t^2 = (-.5 )^2 = .25, so y = e^z = e^.25 = 1.28, If t = 0, then z = t^2 = ( 0 )^2 = 0, so y = e^z = e^0 = 1. If t = .5, then z = t^2 = (.5 )^2 = .25, so y = e^z = e^.25 = 1.28, If t = 1, then z = t^2 = ( 1 )^2 = 1, so y = e^z = e^1 = 2.72, If t = 2, then z = t^2 = ( 2 )^2 = 4, so y = e^z = e^4 = 55, confidence assessment: 3
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20:40:01 If t = -2, then z = t^2 = (-2)^2 = 4, so y = e^z = e^4 = 55, approx.. If t = -1, then z = t^2 = (-1 )^2 = 1, so y = e^z = e^1 = 2.72, approx.. If t = -.5, then z = t^2 = (-.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx. If t = 0, then z = t^2 = ( 0 )^2 = 0, so y = e^z = e^0 = 1. If t = .5, then z = t^2 = (.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx. If t = 1, then z = t^2 = ( 1 )^2 = 1, so y = e^z = e^1 = 2.72, approx.. If t = 2, then z = t^2 = ( 2 )^2 = 4, so y = e^z = e^4 = 55, approx..
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RESPONSE --> ok self critique assessment: 3
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20:41:08 `q002. If we evaluate the function y = e^(t^2) at t = -2, -1, -.5, 0, .5, 1, 2, what y values do we get?
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RESPONSE --> If t = -2, then y = e^(t^2) = e^(-2)^2 = e^4 = 55 If t = -1, then y = e^(t^2) = e^(-1)^2 = e^1 = 2.72 If t = -.5, then y = e^(t^2) = e^(-.5)^2 = e^.25 = 1.28 If t = 0, then y = e^(t^2) = e^(0)^2 = e^0 = 1. If t = .5, then y = e^(t^2) = e^(.5)^2 = e^.25 = 1.28 If t = 1, then y = e^(t^2) = e^(1)^2 = e^1 = 2.72 If t = 2, then y = e^(t^2) = e^(2)^2 = e^2 = 55, confidence assessment: 3
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20:41:15 If t = -2, then y = e^(t^2) = e^(-2)^2 = e^4 = 55, approx. If t = -1, then y = e^(t^2) = e^(-1)^2 = e^1 = 2.72, approx. If t = -.5, then y = e^(t^2) = e^(-.5)^2 = e^.25 = 1.28, approx. If t = 0, then y = e^(t^2) = e^(0)^2 = e^0 = 1. If t = .5, then y = e^(t^2) = e^(.5)^2 = e^.25 = 1.28, approx. If t = 1, then y = e^(t^2) = e^(1)^2 = e^1 = 2.72, approx. If t = 2, then y = e^(t^2) = e^(2)^2 = e^2 = 55, approx.
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RESPONSE --> ok self critique assessment: 3
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20:42:44 `q003. We see from the preceding two examples that that the function y = e^(t^2) results from the 'chain' of simple functions z = t^2 and y = e^z. What would be the 'chain' of simple functions for the function y = cos ( ln(x) )?
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RESPONSE --> The first function encountered by the variable x is the ln(x) function, so we will say that z = ln(x). The result of this calculation is then entered into the cosine function, so we say that y = cos(z). Thus I have y = cos(z) = cos( ln(x) ). the function y(x) is the composite of the functions cosine and natural log functions, i.e., the composite of y = cos(z) and z = ln(x). confidence assessment: 3
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20:42:59 The first function encountered by the variable x is the ln(x) function, so we will say that z = ln(x). The result of this calculation is then entered into the cosine function, so we say that y = cos(z). Thus we have y = cos(z) = cos( ln(x) ). We also say that the function y(x) is the composite of the functions cosine and natural log functions, i.e., the composite of y = cos(z) and z = ln(x).
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RESPONSE --> ok self critique assessment: 3
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20:44:02 `q004. What would be the chain of functions for y = ( ln(t) ) ^ 2?
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RESPONSE --> first function encountered by the variable t is ln(t), so we say that z = ln(t). This value is then squared so we say that y = z^2. we have y = z^2 = (ln(t))^2. we have the composite of the squaring function and the natural log function confidence assessment: 3
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20:44:08 The first function encountered by the variable t is ln(t), so we say that z = ln(t). This value is then squared so we say that y = z^2. Thus we have y = z^2 = (ln(t))^2. We also say that we have here the composite of the squaring function and the natural log function, i.e., the composite of y = z^2 and z = ln(t).
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RESPONSE --> ok self critique assessment: 3
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20:45:31 `q005. What would be the chain of functions for y = ln ( cos(x) )?
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RESPONSE --> first function encountered by the variable is cos(x), so we say that z = cos(x). then take the natural log of this function, so y = ln(z). y = ln(z) = ln(cos(x)). This function is the composite of the natural log and cosine function confidence assessment: 3
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20:45:36 The first function encountered by the variable is cos(x), so we say that z = cos(x). We then take the natural log of this function, so we say that y = ln(z). Thus we have y = ln(z) = ln(cos(x)). This function is the composite of the natural log and cosine function, i.e., the composite of y = ln(z) and z = cos(x).
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RESPONSE --> ok self critique assessment: 3
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20:47:14 `q006. The rule for the derivative of a chain of functions is as follows: The derivative of the function y = f ( g ( x) ) is y ' = g' ( x ) * f ' ( g ( x) ). For example if y = cos ( x^2 ) then we see that the f function is f(z) = cos(z) and the g function is the x^2 function, so that f ( g ( x) ) = f ( x^2 ) = cos ( x^2 ) . By the chain rule the derivative of this function will be (cos(x^2)) ' = g ' ( x) * f ' ( g ( x) ) . g(x) = x^2 so g'(x) = 2 x. f ( z ) = cos ( z) so f ' ( z ) = - sin( z ), so f ' ( g ( x ) ) = - sin ( g ( x ) ) = - sin ( x^2). Thus we obtain the derivative (cos(x^2)) ' = g ' ( x ) * f ' ( g ( x ) ) = 2 x * ( - sin ( x^2 ) ) = - 2 x sin ( x^2). Apply the rule to find the derivative of y = sin ( ln ( x ) ) .
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RESPONSE --> y = sin ( ln(x) ) is the composite f(g(x)) of f(z) = sin(z) and g(x) = ln(x). The derivative of this composite is g ' (x) * f ' ( g(x) ). Since g(x) = ln(x), g ' (x) = ( ln(x) ) ' = 1/x. Since f(z) = sin(z) f ' (z) = cos(z). the derivative of y = sin( ln (x) ) is y ' = g ' (x) * f ' (g(x)) = 1 / x * cos ( g(x) ) =1 / x * cos( ln(x) ). confidence assessment: 3
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20:47:20 We see that y = sin ( ln(x) ) is the composite f(g(x)) of f(z) = sin(z) and g(x) = ln(x). The derivative of this composite is g ' (x) * f ' ( g(x) ). Since g(x) = ln(x), we have g ' (x) = ( ln(x) ) ' = 1/x. Since f(z) = sin(z) we have f ' (z) = cos(z). Thus the derivative of y = sin( ln (x) ) is y ' = g ' (x) * f ' (g(x)) = 1 / x * cos ( g(x) ) = 1 / x * cos( ln(x) ). Note how the derivative of the 'inner function' g(x) = ln(x) appears 'out in front' of the derivative of the 'outer' function; that derivative is in this case the sine function.
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RESPONSE --> ok self critique assessment: 3
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20:48:34 `q007. Find the derivative of y = ln ( 5 x^7 ) .
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RESPONSE --> f(z) = ln(z) and g(x) = 5 x^7. f ' (z) = 1 / z and g ' (x) = 35 x^6. We see that f ' (g(x)) = 1 / g(x) = 1 / ( 5 x^7). derivative of y = ln( 5 x^7) = y ' = g ' (x) * f ' (g(x)) = 35 x^6 * [ 1 / ( 5 x^7 ) ] confidence assessment: 3
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20:48:41 For this composite we have f(z) = ln(z) and g(x) = 5 x^7. Thus f ' (z) = 1 / z and g ' (x) = 35 x^6. We see that f ' (g(x)) = 1 / g(x) = 1 / ( 5 x^7). So the derivative of y = ln( 5 x^7) is y ' = g ' (x) * f ' (g(x)) = 35 x^6 * [ 1 / ( 5 x^7 ) ]. Note again how the derivative of the 'inner function' g(x) = 5 x^7 appears 'out in front' of the derivative of the 'outer' function; that derivative is in this case the reciprocal or 1 / z function.
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RESPONSE --> ok self critique assessment: 3
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20:49:41 `q008. Find the derivative of y = e ^ ( t ^ 2 ).
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RESPONSE --> f(z) = e^z and g(t) = t^2. f ' (z) = e^z and g ' (t) = 2t. the derivative of y = e^(t^2) is y ' = g ' (t) * f ' (g(t)) = 2 t * e^(t^2). confidence assessment: 3
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20:49:48 This function is the composite of f(z) = e^z and g(t) = t^2. We see right away that f ' (z) = e^z and g ' (t) = 2t. Thus the derivative of y = e^(t^2) is y ' = g ' (t) * f ' (g(t)) = 2 t * e^(t^2). Note once more how the derivative of the 'inner function' g(t) = t^2 appears 'out in front' of the derivative of the 'outer' function.
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RESPONSE --> ok self critique assessment: 3
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20:50:22 `q009. Find the derivative of y = cos ( e^t ).
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RESPONSE --> the derivative of y = cos ( e^t ) is y ' = g ' (t) * f ' (g(t)) = e^t * -sin( e^t) = - e^t sin ( e^t). confidence assessment: 3
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20:50:28 We have the composite of f(z) = cos(z) and z(t) = e^t, with derivatives f ' (z) = -sin(z) and g ' (t) = e^t. Thus the derivative of y = cos ( e^t ) is y ' = g ' (t) * f ' (g(t)) = e^t * -sin( e^t) = - e^t sin ( e^t). Note how the 'inner function' is unchanged, as it has been in previous examples, and how its derivative appears in front of the derivative of the 'outer' function.
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RESPONSE --> ok self critique assessment: 3
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20:51:03 `q010. Find the derivative of y = ( ln ( t ) ) ^ 9.
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RESPONSE --> We have y = f(g(t)) with f(z) = z^9 and g(t) = ln(t). f ' (z) = 9 z^8 and g ' (t) = 1 / t. So, y ' = g ' (t) * f ' (g(t)) = 1/t * 9 ( ln(t) )^8. confidence assessment: 3
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20:51:13 We have y = f(g(t)) with f(z) = z^9 and g(t) = ln(t). f ' (z) = 9 z^8 and g ' (t) = 1 / t. Thus y ' = g ' (t) * f ' (g(t)) = 1/t * 9 ( ln(t) )^8.
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RESPONSE --> right self critique assessment: 3
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20:52:16 `q011. Find the derivative of y = sin^4 ( x ).
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RESPONSE --> The composite here is y = f(g(x)) with f(z) = z^4 and g(x) = sin(x). t f ' (z) = 4 z^3 and g ' (x) = cos(x). Thus, y ' = g ' (x) * f ' (g(x)) = cos(x) * 4 ( sin(x) ) ^ 3 = 4 cos(x) sin^3 (x). confidence assessment: 3
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20:52:22 The composite here is y = f(g(x)) with f(z) = z^4 and g(x) = sin(x). Note that the notation sin^4 means to raise the value of the sine function to the fourth power. We see that f ' (z) = 4 z^3 and g ' (x) = cos(x). Thus y ' = g ' (x) * f ' (g(x)) = cos(x) * 4 ( sin(x) ) ^ 3 = 4 cos(x) sin^3 (x).
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RESPONSE --> ok self critique assessment: 3
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20:52:45 `q012. Find the derivative of y = cos ( 3x ).
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RESPONSE --> y ' = g ' (x) * f ' (g(x)) = 3 * (-sin (3x) ) = - 3 sin(3x). confidence assessment: 3
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20:52:53 This is the composite y = f(g(x)) with f(z) = cos(z) and g(x) = 3x. We obtain f ' (z) = - sin(z) and g ' (x) = 3. Thus y ' = g ' (x) * f ' (g(x)) = 3 * (-sin (3x) ) = - 3 sin(3x).
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RESPONSE --> ok self critique assessment: 3
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ސJ~ny assignment #012 zKÑQߔ Calculus I 10-06-2008
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20:54:39 What is the seventh power of (x + `dx) (use the Binomial Theorem)?
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RESPONSE --> x^7 +7x^6'dx +21x^6'dx^2+35x^4'dx^3+35x^3'dx^4 +21x^2'dx^5 +7x'dx^6+'dx^7
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20:54:44 ** Using the binomial Theorem: x^7+7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7
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RESPONSE --> ok
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20:55:26 What is ( (x + `dx)^7 - x^7 ) / `dx and what does the answer have to do with the derivative of x^7?
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RESPONSE --> x + `dx)^7 - x^7 = x^7+7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7 - x^7 = 7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7, [ (x + `dx)^7 - x^7 ] / `dx = (7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7)/`dx = 7x^6+21x^6'dx+35x^5'dx^2+35x^3'dx^3+21x^2'dx^4+7x'dx^5+'dx^6.
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20:55:33 ** (x + `dx)^7 - x^7 = x^7+7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7 - x^7 = 7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7, so [ (x + `dx)^7 - x^7 ] / `dx = (7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7)/`dx = 7x^6+21x^6'dx+35x^5'dx^2+35x^3'dx^3+21x^2'dx^4+7x'dx^5+'dx^6. As `dx -> 0, every term with factor `dx approaches 0 and the quotient approaches 7 x^6. **
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RESPONSE --> ok
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20:57:02 Query problem 2.1.16 (prev edition 2.1.19 (was 2.1.8)) sketch position fn s=f(t) is vAve between t=2 and t=6 is same as vel at t = 5 Describe your graph and explain how you are sure that the velocity at t = 5 is the same as the average velocity between t=2 and t= 6.
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RESPONSE --> The slope of the tangent line at t = 5 is the instantaneous velocity, and the average slope (rise / run between t = 2 and t = 6 points) is the average velocity. If the function has constant curvature the function is increasing it must be increasing at a decreasing rate (i.e., increasing and concave down), and if the function is decreasing it must be decreasing at a decreasing rate (i.e., decreasing and concave up)
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20:57:07 ** The slope of the tangent line at t = 5 is the instantaneous velocity, and the average slope (rise / run between t = 2 and t = 6 points) is the average velocity. The slope of the tangent line at t = 5 should be the same as the slope between the t = 2 and t = 6 points of the graph. If the function has constant curvature then if the function is increasing it must be increasing at a decreasing rate (i.e., increasing and concave down), and if the function is decreasing it must be decreasing at a decreasing rate (i.e., decreasing and concave up). **
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RESPONSE --> ok
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20:57:47 What aspect of the graph represents the average velocity?
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RESPONSE --> The straight line through two points has a rise representing the change in position and a run representing the change in clock time so the slope represents change in position / change in clock time = average rate of change of position = average velocity
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20:57:52 ** The straight line through two points has a rise representing the change in position and a run representing the change in clock time, so that the slope represents change in position / change in clock time = average rate of change of position = average velocity **
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RESPONSE --> ok
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20:58:43 What aspect of the graph represents the instantaneous veocity at t = 5?
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RESPONSE --> The slope of the tangent line at the t = 5 graph point represents the instantaneous velocity at t = 5. According to the conditions of the problem this slope must equal the slope between the t = 2 and t = 6 point
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20:58:47 ** The slope of the tangent line at the t = 5 graph point represents the instantaneous velocity at t = 5. According to the conditions of the problem this slope must equal the slope between the t = 2 and t = 6 points **
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RESPONSE --> ok
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21:00:20 Query problem 2.1.14 (3d edition 2.1.16) graph increasing concave down thru origin, A, B, C in order left to right; origin to B on line y = x; put in order slopes at A, B, C, slope of AB, 0 and 1.What is the order of your slopes.
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RESPONSE --> 0 will be the first of the ordered quantities since all slopes are positive. C is the rightmost point and since the graph is concave down will have the next-smallest slope. The slope of the line from the origin to B is 1. The slope of the tangent line at B is less than the slope of AB and the slope of the tangent line at A is greater than the slope of AB. So slope at A is the greatest of the quantities, 1 is next, Then B, then slope of AB, then slope at A and finall 0 (in descending order).
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21:00:26 ** The graph is increasing so every slope is positive. The downward concavity means that the slopes are decreasing. 0 will be the first of the ordered quantities since all slopes are positive. C is the rightmost point and since the graph is concave down will have the next-smallest slope. The slope of the line from the origin to B is 1. The slope of the tangent line at B is less than the slope of AB and the slope of the tangent line at A is greater than the slope of AB. So slope at A is the greatest of the quantities, 1 is next, followed by slope at B, then slope of AB, then slope at A and finall 0 (in descending order). **
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RESPONSE --> ok
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21:01:40 Query problem 2.2.8 (was 2.1.16) f(x) = sin(3x)/x. Give your f(x) values at x = -.1, -.01, -.001, -.0001 and at .1, .01, .001, .0001 and tell what you think the desired limit should be.
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RESPONSE --> my values for f(x): -.1, 2.9552 -.01, 2.9996 -.001, 3 -.0001, 3 .1, 2.9552 .01, 2.9996 .001, 3 .0001, 3 . limiting value is 3
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21:01:50 COMMON ERROR: Here are my values for f(x): -.1, 2.9552 -.01, 2.9996 -.001, 3 -.0001, 3 .1, 2.9552 .01, 2.9996 .001, 3 .0001, 3 . So the limiting value is 3. INSTRUCTOR COMMENT: Good results and your answer is correct. However none the values you quote should be exactly 3. You need to give enough significant figures that you can see the changes in the expressions. The values for .1, .01, .001 and .0001 are 2.955202066, 2.999550020, 2.999995500, 2.999999954. Of course your calculator might not give you that much precision, but you can see the pattern to these values. The limit in any case is indeed 3.
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RESPONSE --> oh ok I get it
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21:04:57 Describe your graph.
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RESPONSE --> the graph passes horizontally through the y axes at (0,3), then as x increases it decreases an increasing rate -- i.e., it is concave downward--for a time, but gradually straightens out then decreases at a decreasing rate as it passes through the x axis, etc..
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21:05:01 ** the graph passes horizontally through the y axes at (0,3), then as x increases it decreases an increasing rate -- i.e., it is concave downward--for a time, but gradually straightens out then decreases at a decreasing rate as it passes through the x axis, etc.. However the important behavior for this graph is near x = 0, where the graph reaches a maximum of 3 at x = 0, and approaches this value as a limit. **
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RESPONSE --> ok
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21:06:32 Find an interval such that the difference between f(x) and your limit is less than .01.
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RESPONSE --> As the numbers quoted earlier show, f(x) is within .01 of the limit 3 for -.01 < x < .01. f(x) = 3 - .01, i.e., f(x) = 2.99, for x you obtain solutions x = -.047 and x = .047 (approx). The maximum interval is therefore approximately -.047 < x < .047.
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21:06:39 ** As the numbers quoted earlier show, f(x) is within .01 of the limit 3 for -.01 < x < .01. This interval is a good answer to the question. Note that you could find the largest possible interval over which f(x) is within .01 of 3. If you solve f(x) = 3 - .01, i.e., f(x) = 2.99, for x you obtain solutions x = -.047 and x = .047 (approx). The maximum interval is therefore approximately -.047 < x < .047. However in such a situation we usually aren't interested in the maximum interval. We just want to find an interval to show that the function value can indeed be confined to within .01 of the limit. In general we wish to find an interval to show that the function value can be confined to within a number usually symbolized by `delta (Greek lower-case letter) of the limit. **
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RESPONSE --> ok I get it
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21:07:20 Query problem 2.2.17 (3d edition 2.3.26 was 2.2.10) f(x) is cost so f(x) / x is cost per unit. Describe the line whose slope is f(4) / 4
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RESPONSE --> A line from (0, 0) to (4, f(4) ) has rise f(4) and run 4 so the slope of this line is rise / run = f(4) / 4. Similarly the slope of the line from the origin to the x=3 point has slope f(3) / 3. If the graph is concave down then the line from the origin to the x = 4 point is less than that of the line to the x = 3 point and we conclude that f(4) / 4 is smaller than f(3) / 3; average cost goes down as the number of units rises.
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21:07:24 ** A line from (0, 0) to (4, f(4) ) has rise f(4) and run 4 so the slope of this line is rise / run = f(4) / 4. Similarly the slope of the line from the origin to the x=3 point has slope f(3) / 3. If the graph is concave down then the line from the origin to the x = 4 point is less than that of the line to the x = 3 point and we conclude that f(4) / 4 is smaller than f(3) / 3; average cost goes down as the number of units rises. **
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RESPONSE --> ok
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21:08:45 Query problem 2.2.23 (3d edition 2.3.32 was 2.2.28) approximate rate of change of ln(cos x) at x = 1 and at x = `pi/4. What is your approximation at x = 1 and how did you obtain it?
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RESPONSE --> The values of ln(cos(x)) at x = .99, 1.00 and 1.01 are -0.6002219140, -0.6156264703 and -0.6313736258. The changes in the value of ln(cos(x)) are -.0154 and -.0157, giving average rates of change -.0154 / .01 = -1.54 and -.0157 / .01 = -1.57. The average of these two rates is about -1.56 The values of ln(cos(x)) at x = pi/4 - .01, pi/4 and pi/4 + .01 are -0.3366729302, -0.3465735902 and -0.3566742636. The changes in the value of ln(cos(x)) are -.009 and 0.0101, giving average rates of change -.0099 / .01 = -.99 and -.0101 / .01 = -1.01. The average of these two rates is about -1
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21:08:49 ** At this point the text wants you to approximate the value. The values of ln(cos(x)) at x = .99, 1.00 and 1.01 are -0.6002219140, -0.6156264703 and -0.6313736258. The changes in the value of ln(cos(x)) are -.0154 and -.0157, giving average rates of change -.0154 / .01 = -1.54 and -.0157 / .01 = -1.57. The average of these two rates is about -1.56; from the small difference in the two average rates we conclude that this is a pretty good approximation, very likely within .01 of the actual instantaneous rate. The values of ln(cos(x)) at x = pi/4 - .01, pi/4 and pi/4 + .01 are -0.3366729302, -0.3465735902 and -0.3566742636. The changes in the value of ln(cos(x)) are -.009 and 0.0101, giving average rates of change -.0099 / .01 = -.99 and -.0101 / .01 = -1.01. The average of these two rates is about -1; from the small difference in the two average rates we conclude that this is a pretty good approximation, very likely within .01 of the actual instantaneous rate. **
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RESPONSE --> ok
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21:09:05 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> ok
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21:09:12 STUDENT QUESTION: I did have another opportunity to go back and look at Pascal`s triangle. I always had a problem with it in earlier calculus courses. I am still uncertain when I use it to get results, but I think it is a matter of becoming more comfortable with the process. INSTRUCTOR RESPONSE: If the first row in Pascal's Triangle is taken to be row number 0, and if the first number in a row is taken to be at the 0th position in the row, then the number in row n, position r represents the number of ways to get r heads on n flips of a fair coin, or equivalently as the number of ways to select a set of r objects from a total of n objects. For example if you have 26 tiles representing the letters of the alphabet then the number of ways to select a set of 6 tiles would be the number in the 26th row at position 6. The six tiles selected would be considered to be dumped into a pile, not arranged into a word. After being selected it turns out there would be 6! = 720 ways to arrange those six tiles into a word, but that has nothing to do with the row 26 position 6 number of the triangle. The number of ways to obtain 4 Heads on 10 flips of a coin is the number in row 10 at position 4. The two interpretations are equivalent. For example you could lay the tiles in a straight line and select 6 of them by flipping a coin once for each tile, pushing a tile slightly forward if the coin comes up 'heads'. If at the end exactly six tiles are pushed forward you select those six and you are done. Otherwise you line the tiles up and try again. So you manage to select 6 tiles exactly when you manage to get six Heads. It should therefore be clear that the number of ways to select 6 tiles from the group is identical to the number of ways to get six Heads. When expanding a binomial like (a + b) ^ 3, we think of writing out (a+b)(a+b)(a+b). When we multiply the first two factors we get a*a + a*b + b*a + b*b. When we then multiply this result by the third (a+b) factor we get a*a*a + a*a*b + a*b*a + a*b*b + b*a*a + b*a*b + b*b*a + b*b*b. Each term is obtained by selecting the letter a or the letter b from each of the three factors in turn, and every possible selection is represented. We could get any one of these 8 terms by flipping a coin for each factor (a+b) to determine whether we choose a or b. We would have 3 flips, and the number of ways of getting, say, two a's and one b would be the same as the number of ways of getting two Heads on three flips. As we can see from Pascal's triangle there are 3 ways to do this. These three ways match the terms a*a*b, a*b*a and b*a*a in the expansion. Since all three terms can be simplified to a^2 b, we have [ 3 * a^2 b ] in our expansion. Using this line of reasoning we see that the expansion a^3 + 3 a^2 b + 3 a b^2 + b^3 of (a+b)^3 has coefficients that match the n=3 row of Pascal's Triangle. This generalizes: the expansion of (a + b) ^ n has as its coefficients the nth row of Pascal's Triangle. The number in position r of row n is designated C(n,r), the number of combinations of r elements chosen from a set of n elements. C(n,r) = n! / [ r! * (n-r)! ]. This formula can to be proven by mathematical induction, or it can be reasoned out as follows: In choosing r elements out of n there are n choices for the first element, n - 1 choices for the second, n-2 for the third, ..., n - r + 1 choices for the rth element, so there are n (n-1)(n-2) ... (n-r+1) ways of choosing r elements in order. There are r! Possible orders for the chosen elements, so the number of combinations, in which order doesn't matter, is n (n-1)(n-2) ... (n-r+1) / r!. This is the same as n! / [ r! (n-r)! ], since n! / (n-r)! = n(n-1) ... (n-r+1). **
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RESPONSE --> No Comments
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