#$&* course MTH 173 12/15 3:20
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Given Solution: `aSTUDENT RESPONSE: A limit is the resulting answer as x approaches some number or infinity. The limit is not that quantity exactly, but it is said to be as close as we can get without being exact. It is taken from an infinitely small interval from either side of what x approaches. INSTRUCTOR COMMENTS: That's a good expression of what it means. The formal definition, which is very necessary if we are to be sure we're on a solid foundation, is that lim{x -> a} f(x) = L if for any `epsilon, no matter how small, we can find a `delta such that whenever | x - a | < `delta, | f(x) - L | < `delta. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qGive the definition of continuity and explain what it means. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For a function f(x) over an interval, If the value of the function at x = a is equal to the value of the limit of the function as x tends to a from either sides (that is the right hand and the left hand limit), then the function is said to be continuous at x = a. If this is the case so for all the values of f over the interval, then the function f(x) is said to be continuous throughout that interval. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: For a function to be continuous, it has to have a limit. Continuity can be applied to sums, differences, products, constant multiples, and quotients where the denominator doesn`t equal zero. INSTRUCTOR CRITIQUE: ** The key is that at a given value x = a the limit of the function f(x) as we approach that a is equal to the value of the function at a--i.e., lim{x -> a} f(x) = f(a). If this is true for every x value on some interval, then the function is said to be continuous. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qGiven definition of differentiability and explain what it means. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we have a function f(x) at x = a. The left hand derivative of the function is defined by the value of the lim as h tends to -0 (f(a - h) - f(a))/-h the right derivative of the function is defined by the value of lim as h tends to 0 (f(a + h) - f(a))/h When both the right hand derivative and the left hand derivative are equal the function is differentiable at x = a. If this is true for all the x values over the interval, the function f(x) is considered differentiable for the entire interval. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: Differentiability can be found by taking the limit as x approaches some value by both sides. A function has to have a limit and therefore have a derivative to be differentiable. INSTRUCTOR CRITIQUE: When we considered differentiability of f(x) at x = a we look at the limit of the expression [ f(a + `dx) - f(a) ] / `dx, specifically lim { `dx -> 0} [ f(a + `dx) - f(a) ] / `dx. If this limit exists, then the function is differentiable at x = a. In order to exist, the limit as `dx -> 0+ (i.e., 'from above' or through positive values of `dx) must exist and must equal the limit as `dx -> 0- (i.e., 'from below' or through negative values of `dx), which of course must also exist. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery Theory 2, Differentiability and Continuity, Problem 6 (was problem 6 page 142 ) Q = C for t<0 and Ce^(-t/(RC)) for t>=0. Is the charge Q a continuous function of time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The value of the function Q as t tends to 0 from left hand side, that is tending to 0- is C The value of the function Q as t tends to 0 from right hand side, that is tending to 0+ = Ce^(0) = C Since the value of function Q is equal to C as t tends from either sides, the function is continuous for t = 0. For t<0, the function is continuous as it is a constant function. For t>0 the function is continuous as the exponential function is continuous for t > 0. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aYou know that the exponential function Ce^(-t/(RC)) is continuous, and the constant function Q = C is continuous. Therefore for t < 0 and for t > 0 the function is continuous. The question arises at the point where the two functions meet--i.e., at t = 0. Do the right-and left-hand limits exist, and are the equal? The left-hand limit is that of the constant function Q = C. No matter how close you get to t = 0, this function will equal C and its limit will therefore equal C. You should be able to state and prove this in terms of `epsilons and `deltas. The right-hand limit is that of the exponential function, which continuously and smoothly approaches its t = 0 value C. You should think this through in terms of `epsilons and `deltas also, though a rigorous algebraic proof might be difficult at this stage. Therefore both limits exist and are equal, and the function is therefore continuous at t = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qIs the current I = dQ / dt defined for all times? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The value of the derivative of the Q function is 0 for t < 0. Thus for t < 0, dQ/dt = 0 Thus the value of the limit as h tends to 0- (f(0 - h) - f(0))/-h = 0 For t > 0 Q = Ce^(-t/(RC)) Using the rules of differentiation we get dQ/dt = Ce^(-t/(RC)) * (-1/(RC)) Thus for t > 0 I = (-1/R)e^(-t/RC) The value of the derivative as t tends to 0 will be I = -1/R Thus lim h tending to 0+ (f(0 + h) - f(0))/h = -1/R Since lim h tending to 0- (f(0 - h) - f(0))/h is not equal to lim h tending to 0+ (f(0 + h) - f(0))/h the function is not differentiable at t = 0. Thus the current function I is defined for all t values except t = 0. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** For t < 0, dQ / dt is the derivative of a constant function and is therefore zero. The derivative of the exponential function at t = 0 is not zero. Therefore the left-hand limit and the right-hand limit of the derivative differ at t = 0, and the derivative does not exist. ** What is the derivative of Q = C for t < 0 and what is the derivative of C e^(-t/(RC)) for t>0? Does the derivative approach the same limit at t = 0 from the left as from the right? What does this have to do with the definition of the derivative? For t < 0 we have Q = C. The derivative of the Q = C function is at all points 0, since this is a constant function. So as we approach t = 0 from the left the limiting value of the derivative is zero. For t > 0 we have the function C e^(-t / (RC) ), which has derivative 1 / R e^(-t / (RC) ). At t -> 0 from the right this derivative, which is continuous, approaches -1 / R e^(-0 / R C) = -1 / R * 1 = -1 / R. This value is not 0. Since the derivative approaches 0 from the left and -1 / R from the right it is not continuous at t = 0. It follows that the derivative is not defined at t = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery Theory 2, Differentiability and Continuity, Problem 0 (was problem 9 page 142) g(r) = 1 + cos(`pi r / 2) -2 <= r <= 2, 0 elsewhere. Is g continuous at r = 2? Explain. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function g(r) = 1 + cos(pi*r/2), for -.2 <= r <= 2 The value of the function g(r) = 0 for r > 2. Thus the function g(r) tends to 0 as r tends to 2 from right hand side. The value of the function at x = 2 will be g(2) = 1 + cos(pi) = 0 Thus the value of the function tends to 0 as r tends to 2 from the left hand side. Since the value of the function tends to 0 as r tends to 2 from either sides the function g(r) =is continuous at r = 2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** g(2) = 0, as is easily seen by substituting. Since the cosine function is continuous for r < 2, the limiting value of the function as r -> 2 form the left is 0. Since g(r) = 0 for r > 2 the limit of g(r) as r -> 2 from the right is 0. Since the limiting values are identical as we approach r = 2 from the right and left, the function is continuous at r = 2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qIs g differentiable at r = -2? Explain. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since the value of the function is 0 for all value of r < -2 thus the value of the derivative of the function will be 0 for r values less than -2. Thus the value of the limit h tending to -0 (f(-2-h) - f(-2))/-h is 0 For r >= -2 and<= 2 g(r) = 1 + cos(pi r / 2) g’(r) = 0 - (pi/2)sin(pir/2) g’(-2) = 0 Thus the value of the limit as h tends to 0+ (f(-2 + h) - f(-2))/h = 0 Since the value of the limit as h tends to 0+ (f(-2 + h) - f(-2))/h is equal to the value of the limit as h tends to 0- (f(-2-h) - f(-2))/-h. Thus the function is differentiable at r = -2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The derivative of g(r) is g ' (r) = 1 + cos(`pi r / 2) is `pi / 2 * sin(`pi r / 2). g(-2) = 0 and this derivative function is continuous, so as r -> -2 from the left this has limit zero. The derivative of y = 0, which is the function for r < -2, is 0. Since the limiting value of the derivative as we approach r = -2 is 0 from both sides, and the same is true at at r = 2, the derivative is continuous. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aI dealt better with this focus on theory than the other in the chapter. This shows a lot about how to prove what you have already known, and it shows you why things work in a particular way. Again there is more relevance and meaning added to the tangent approximation, and the factor of error. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!