#$&* course Mth 173 7:16 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: `a** The function is `sqrt( (x^2 * 5^x)^3 ) = (x^2 * 5^x)^(3/2). This is of form f(g(x)) with g(x) = x^2 * 5^x and f(z) = z^(3/2). Thus when you substitute you get f(g(x)) = g(x)^(3/2) = (x^2 * 5^x)^(3/2). (x^2 * 5^x) ' = (x^2)' * 5^x + x^2 * (5^x) ' = 2x * 5^x + x^2 ln 5 * 5^x = (2x + x^2 ln 5) * 5^x. `sqrt(z^3) = z^(3/2), so using w(x) = f(g(x)) with f(z) = z^(3/2) and g(x) = x^2 * 5^x we get w ' = (2x + x^2 ln 5) * 5^x * [3/2 (x^2 * 5^x)^(1/2)] = 3/2 (2x + x^2 ln 5) * | x | * 5^(1/2 x). Note that sqrt(x^2) is | x |, not just x, since the square root must be positive and x might not be. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qQuery problem 3.4.26 was 3.4.28 (3d edition 3.4.19) (was 4.4.20) derivative of 2^(5t-1). What is the derivative of the given function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This function is a composite. The inner function is g(x)=5t-1 and the outer function is f(z)=2^z. f'(z)=ln(2) * 2^z. g ' (x)=5 so (f(g(t)) ' = g ' (t)f ' (g(t))= 5 ln(2) * 2^(5t-1) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThis function is a composite. The inner function is g(x)=5t-1 and the outer function is f(z)=2^z. f'(z)=ln(2) * 2^z. g ' (x)=5 so (f(g(t)) ' = g ' (t)f ' (g(t))= 5 ln(2) * 2^(5t-1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I don’t understand ------------------------------------------------
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Given Solution: `a** We apply the Chain Rule: ( k(2x) ) ' = (2x) ' * k'(2x) = 2 k(2x). When x = 1/2 we have 2x = 1. k ' (1) = y ' (1) = 2 so when x = 1/2 ( k(2x) ) ' = 2 k(2 * 1/2) = 2 * k'(1) = 2 * 2 = 4. (k(x+1)) ' = (x+1)' k ' (x+1) = k ' (x+1) so when x = 0 we have (k(x+1) ) ' = k ' (x+1) = k ' (1) = 2 (k(x/4)) ' = (x/4)' k'(x/4) = 1/4 * k'(x/4) so when x = 4 we have (k(x/4))' = 1/4 * k'(x/4) = 1/4 k'(4/4) = 1/4 * 2 = 1/2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I really don’t understand this problem
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Given Solution: `a** We use the Chain Rule. (e^(-t/(RC)))' = (-t/(RC))' * e^(-t/(RC)) = -1/(RC) * e^(-t/(RC)). So dQ/dt = -Q0/(RC) * e^(-t/(RC)). Both functions are equal to a constant factor multiplied by e^(-t/(RC)). The time constant for both functions is therefore identical, and equal to RC. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qQuery problem 3.5.5 (unchanged since 3d edition) (formerly 4.5.6). What is the derivative of sin(3x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: g ' (x) = (3x) ' = 3 * x ' = 3 ', and f ' (z) = (sin(z) ) ' = cos(z). So the derivative is [ sin(3x) ] ' = ( f(g(x) ) ' = g ' (x) * f ' (g(x) ) = 3 * cos(3x). confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** sin(3x) is the composite of g(x) = 3x, which is the 'inner' function (the first function that operates on the variable x) and the 'outer' function f(z) = sin(z). Thus f(g(x)) = sin(g(x)) = sin(3x). The derivative is (f (g(x) ) ' = g ' (x) * f' ( g(x) ). g ' (x) = (3x) ' = 3 * x ' = 3 ', and f ' (z) = (sin(z) ) ' = cos(z). So the derivative is [ sin(3x) ] ' = ( f(g(x) ) ' = g ' (x) * f ' (g(x) ) = 3 * cos(3x). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `qQuery problem 3.5.50 was 3.5.48 (3d edition 3.5.50) (formerly 4.5.36). Give the equations of the tangent lines to graph of y = sin(x) at x = 0 and at `pi/3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a** At x = 0 we have y = 0 and y ' = cos(0) = 1. The tangent line is therefore the line with slope 1 through (0,0), so the line is y - 0 = 1 ( x - 0) or just y = x. At x = `pi/3 we have y = sin(`pi/3) = `sqrt(3) / 2 and y ' = cos(`pi/3) = .5. Thus the tangent line has slope .5 and passes thru (`pi/3,`sqrt(3)/2), so its equation is y - `sqrt(3)/2 = .5 (x - `pi/3) y = .5 x - `pi/6 + `sqrt(3)/2. Approximating: y - .87 = .5 x - .52. So y = .5 x + .25, approx. Our approximation to sin(`pi/6), based on the first tangent line: The first tangent line is y = x. So the approximation at x = `pi / 6 is y = `pi / 6 = 3.14 / 6 = .52, approximately. Our approximation to sin(`pi/6), based on the second tangent line, is: y = .5 * .52 + .34 = .60. `pi/6 is equidistant from x=0 and x=`pi/3, so we might expect the accuracy to be the same whichever point we use. The actual value of sin(`pi/6) is .5. The approximation based on the tangent line at x = 0 is .52, which is much closer to .5 than the .60 based on the tangent line at x = `pi/3. The reason for this isn't too difficult to see. The slope is changing more quickly around x = `pi/3 than around x = 0. Thus the tangent line will move more rapidly away from the actual function near x = `pi/3 than near x = 0. ** confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** At x = 0 we have y = 0 and y ' = cos(0) = 1. The tangent line is therefore the line with slope 1 through (0,0), so the line is y - 0 = 1 ( x - 0) or just y = x. At x = `pi/3 we have y = sin(`pi/3) = `sqrt(3) / 2 and y ' = cos(`pi/3) = .5. Thus the tangent line has slope .5 and passes thru (`pi/3,`sqrt(3)/2), so its equation is y - `sqrt(3)/2 = .5 (x - `pi/3) y = .5 x - `pi/6 + `sqrt(3)/2. Approximating: y - .87 = .5 x - .52. So y = .5 x + .25, approx. Our approximation to sin(`pi/6), based on the first tangent line: The first tangent line is y = x. So the approximation at x = `pi / 6 is y = `pi / 6 = 3.14 / 6 = .52, approximately. Our approximation to sin(`pi/6), based on the second tangent line, is: y = .5 * .52 + .34 = .60. `pi/6 is equidistant from x=0 and x=`pi/3, so we might expect the accuracy to be the same whichever point we use. The actual value of sin(`pi/6) is .5. The approximation based on the tangent line at x = 0 is .52, which is much closer to .5 than the .60 based on the tangent line at x = `pi/3. The reason for this isn't too difficult to see. The slope is changing more quickly around x = `pi/3 than around x = 0. Thus the tangent line will move more rapidly away from the actual function near x = `pi/3 than near x = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 3.5.34 (3d edition 3.5.40). Der of sin(sin x + cos x) What is the derivative of the given function and how did you find it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: aThe function y = sin( sin(x) + cos(x) ) is the composite of g(x) = sin(x) + cos(x) and f(z) = sin(z). The derivative of the composite is g ' (x) * f ' (g(x) ). g ' (x) = (sin(x) + cos(x) ) ' = cos(x) - sin(x). f ' (z) = sin(z) ' = cos(z). So g ' (x) * f ' (g(x)) = ( cos(x) - sin(x) ) * cos( sin(x) + cos(x) ). confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe function y = sin( sin(x) + cos(x) ) is the composite of g(x) = sin(x) + cos(x) and f(z) = sin(z). The derivative of the composite is g ' (x) * f ' (g(x) ). g ' (x) = (sin(x) + cos(x) ) ' = cos(x) - sin(x). f ' (z) = sin(z) ' = cos(z). So g ' (x) * f ' (g(x)) = ( cos(x) - sin(x) ) * cos( sin(x) + cos(x) ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!