#$&* course Mth 173 7/16 2:00 pm 025. `query 25
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Given Solution: `aSTUDENT RESPONSE: A limit is the resulting answer as x approaches some number or infinity. The limit is not that quantity exactly, but it is said to be as close as we can get without being exact. It is taken from an infinitely small interval from either side of what x approaches. INSTRUCTOR COMMENTS: That's a good expression of what it means. The formal definition, which is very necessary if we are to be sure we're on a solid foundation, is that lim{x -> a} f(x) = L if for any `epsilon, no matter how small, we can find a `delta such that whenever | x - a | < `delta, | f(x) - L | < `delta. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGive the definition of continuity and explain what it means. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A function is continuous at a point x = a, if its limit at that point exists, and that limit is equal to the value of the function at that point. This means the graph has no breaks at this point and we can find a limit for it confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I have noted that this definition must be true for every x value on some interval for the function to be continuous ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGiven definition of differentiability and explain what it means. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A function is differentiable at a point x if the limit exists at this point You cannot take a derivative of this point if it does not have a limit confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: Differentiability can be found by taking the limit as x approaches some value by both sides. A function has to have a limit and therefore have a derivative to be differentiable. INSTRUCTOR CRITIQUE: When we considered differentiability of f(x) at x = a we look at the limit of the expression [ f(a + `dx) - f(a) ] / `dx, specifically lim { `dx -> 0} [ f(a + `dx) - f(a) ] / `dx. If this limit exists, then the function is differentiable at x = a. In order to exist, the limit as `dx -> 0+ (i.e., 'from above' or through positive values of `dx) must exist and must equal the limit as `dx -> 0- (i.e., 'from below' or through negative values of `dx), which of course must also exist. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have noted that for the limit to exist the values from above and from below must exist and be equal ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qQuery Theory 2, Differentiability and Continuity, Problem 6 (was problem 6 page 142 ) Q = C for t<0 and Ce^(-t/(RC)) for t>=0. Is the charge Q a continuous function of time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes C is continuous and so is Ce^-t/RC The left had limit is Q = C, as t approaches 0 will approach C on either side so the limit exists and is always C The right hand is the exponential function, which is smoothly approach the limit C So both values are equal and continuous so the function is continuous at t = 0 confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: I don’t think it is, because the I for Q = C is 0 I for Q = Ce^-t/RC is Q = 1/ R * e^(-t/RC) and for t = 0 is not 0 The left and right limits differ so at t=0 so the derivative does not exists The limit of Q=C as it approaches 0 will now be 0 The limit of the exponential function will be: Q = 1 / R * r^(-0/RC) = 1/R which is not equal to 0 Since the limits differ for the right and left hand sides the function is not continuous and I is not defined for all times confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** For t < 0, dQ / dt is the derivative of a constant function and is therefore zero. The derivative of the exponential function at t = 0 is not zero. Therefore the left-hand limit and the right-hand limit of the derivative differ at t = 0, and the derivative does not exist. ** What is the derivative of Q = C for t < 0 and what is the derivative of C e^(-t/(RC)) for t>0? Does the derivative approach the same limit at t = 0 from the left as from the right? What does this have to do with the definition of the derivative? For t < 0 we have Q = C. The derivative of the Q = C function is at all points 0, since this is a constant function. So as we approach t = 0 from the left the limiting value of the derivative is zero. For t > 0 we have the function C e^(-t / (RC) ), which has derivative 1 / R e^(-t / (RC) ). At t -> 0 from the right this derivative, which is continuous, approaches -1 / R e^(-0 / R C) = -1 / R * 1 = -1 / R. This value is not 0. Since the derivative approaches 0 from the left and -1 / R from the right it is not continuous at t = 0. It follows that the derivative is not defined at t = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery Theory 2, Differentiability and Continuity, Problem 0 (was problem 9 page 142) g(r) = 1 + cos(`pi r / 2) -2 <= r <= 2, 0 elsewhere. Is g continuous at r = 2? Explain. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes, because as the g(r) approaches 2 from the left hand limit the limit is about 0 Since the right hand limit of g(r) = 0 for r>2 Then the limits are equal for both functions and the function is continuous at r=2 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: The derivative of g(r) = 1 + cos (pi r/ 2) = -1/2 pi sin(pi r /2 ) So at r = -2 g(r) = about 0 Derivative of g(r) = 0 is simply 0 for r < -2 Since the limits of the derivatives of these functions are equal, g is differentiable and continuous at r = -2 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: This section is proving to be pretty difficult for me