#$&* course mth173 017. `query 17
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Given Solution: `a** In terms of the contribution from a single increment, the rectangle has 'altitude' f(x) - g(x) so its area is [ f(x) - g(x) ] `dx. Adding up over all increments we get sum { [ f(x) - g(x) ] `dx, which by the distributive law of addition over multiplication is sum [ f(x) `dx - g(x) `ds ], which is not just a series of additions and can be rearrange to give sum (f(x) `dx) - sum (g(x) `dx). As the interval `dx shrinks to zero, the sums approach the definite integrals and we get int( f(x), x, a, b ) - int(g(x), x, a, b), where int ( function, variable, left limit, right limit) is the definite integral of 'function' with respect to 'variable' from 'left limit' to 'right limit'.** STUDENT COMMENT I'm not sure but I believe I understand. INSTRUCTOR RESPONSE: The question asks you to focus on a single increment. This means that we first divide the interval [a, b] into a large number of subintervals (i.e., we 'partition' the interval [a, b]), and consider what happens in a typical subinterval, or increment. The typical increment will have some width, which we represent by `dx, and we consider a typical x value within the increment. This idea might be clearer if expressed in terms of a trapezoidal approximation of the graph of f(x): Just like any subinterval, the chosen increment would define a thin trapezoid for the function f(x). The trapezoid would have width `dx, and would lie somewhere between x = a and x = b. x would be any point on the x axis between the left and right boundaries of the increment (i.e., any point of the x axis which lies on the boundary of the trapezoid), and f(x) would be the 'graph altitude' at x. If we sum up the areas of all the increments we get a result which is close to the actual area under the graph of f(x). The smaller the increment `dx the closer the sum of the incremental areas will tend to be to the actual area, and as `dx approaches zero, the sum of the incremental areas will approach the actual area. As long as the trapezoid is thin (i.e., as long as `dx is small), the two altitudes of the trapezoid will both be close to f(x). So f(x) will be close to the average altitude and f(x) * `dx will be close to the area of the trapezoid, which is ave. altitude * width. On the interval `dx, the function f(x) - g(x) would similarly form a trapezoid whose 'average altitude' is close to f(x) - g(x) and whose width is `dx. The area of the trapezoid would therefore be close to (f(x) - g(x)) * `dx. In the given solution the distributive law is be used to show that the sum of the (f(x) - g(x) ) * `dx contributions is equal to the sum of the f(x) * `dx minus the sum of the g(x) * `dx, so that the integral of f(x) - g(x) is equal to the integral of f(x) minus the integral of g(x). It's worth summarizing we have done here: we began with a partition of the interval [a, b] (i.e., a subdivision into small increment), then we considered one increment of the partition (i.e., one subinterval), we summed the contributions of all increments, and we drew conclusions about the integral. This procedure, where we partition some interval the consider what happens on a typical subinterval, or increment, is a crucial step in finding the integrals necessary to solve a wide variety of problems. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I can't but my understanding to words. I can see how they are the same thing, but i do not understand your explanation given. when expanded out to be explained, it makes it confusing to me. but if you just ask me if they can be, it makes sense how it could. ------------------------------------------------ Self-critique Rating: 1
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Given Solution: `a** This is also in the text, so look there for an alternative explanation and full rigor. The idea is that if f(x) > m for all x, then for any interval the contribution to the Riemann sum will be greater than m * `dx. So when all the contributions are added up the result is greater than the product of m and the sum of all `dx's. The sum of all `dx's is equal to the length b-a of the entire interval. So the Riemann sum must be greater than the product of m and this sum--i.e., greater than m * ( b - a ). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qExplain why the integral of f(x) / g(x) is not generally equal to the integral of f(x) divided by the integral of g(x). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: integral of f(x) would be the entire area under f(x) integral of g(x) would be the entire area under g(x) lets say g(x) is a little bit smaller than f(x) saying integral of f(x)/g(x) would be combining the functions, and getting the area under the new function. saying integral of f(x) divided by integral of g(x) is kind of like saying, how many times can the area under g(x) fit into the area under f(x). it should be obvious when read like that, these will lead to separate values when evaluated. However thinking about it, i notice it states ""not generally equal"" then thinking, division with negatives leads to funny results sometimes, and was not thinking of negatives, so it may be possible with those values. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** For now assume that f and g are both positive functions. The integral of f(x) represents the area beneath the curve between the two limits, and the integral of g(x) represents the area beneath its curve between the same two limits. So the integral of f(x) divided by the integral of g(x) represents the first area divided by the second. The function f(x) / g(x) represents the result of dividing the value of f(x) by the value of g(x) at every x value. This gives a different curve, and the area beneath this curve has nothing to do with the quotient of the areas under the original two curves. It would for example be possible for g(x) to always be less than 1, so that f(x) / g(x) would always be greater than f(x) so that the integral of f(x) / g(x) would be greater than the integral of f(x), while the length of the interval is very long so that the area under the g(x) curve would be greater than 1. In this case the integral of f(x) divided by the integral of g(x) would be less than the integral of f(x), and would hence be less than the integral of f(x) / g(x). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qGiven a graph of f(x) and the fact that F(x) = 0, explain how to construct a graph of F(x) such that F'(x) = f(x). Then explain how, if f(x) is the rate at which some quantity changes with respect to x, this construction gives us a function representing how much that quantity has changed since x = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(x) being related to F(x), as that it is its derivative. going at intervals on f(x), having a starting point for F(x), you con figure out the rest of the values of F(x). say f(0)=1, and f(1) = 0. F(0)=0 (given) and F(1) is one increment from F(0), which means we go one increment on f(x). from f(0) to f(1), if viewed as a trapezoid, the area is .5, and with just one increment, F(1)= .5. this is essentially continues for any other value. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** To construct the graph you could think of finding areas. You could for example subdivide the graph into small trapezoids, and add the area of each trapezoid to the areas of the ones preceding it. This would give you the approximate total area up to that point. You could graph total area up to x vs. x. This would be equivalent to starting at (0,0) and drawing a graph whose slope is always equal to the value of f(x). The 'higher' the graph of f(x), the steeper the graph of F(x). If f(x) falls below the x axis, F(x) will decrease with a steepness that depends on how far f(x) is below the axis. If you can see why the two approaches described here are equivalent, and why if you could find F(x) these approaches would be equivalent to what you suggest, you will have excellent insight into the First Fundamental Theorem. ** STUDENT COMMENT So in general using trapezoid approx.. is another way to see what is going on? INSTRUCTOR RESPONSE It's very similar, in that if you first think of the trapezoids you see how the area is close to the sum of the trapezoidal areas. In each increment of the partition we choose a single value of x, and evaluate f(x) at that value. Using that value as the average 'graph altitude' we approximate the area of the trapezoid as f(x) * `dx. This has the same effect as replacing the trapezoid by a rectangle of altitude f(x) and width `dx. As long as `dx is small, f(x) is close to the actual average 'graph altitude' for the interval, and the error in the approximation is small. As `dx approaches zero, the error in the approximation approaches zero. The limiting value of our approximations is the integral, and it represents the area beneath the graph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 3.1.16 (3d edition 3.1.12 ) (formerly 4.1.13) derivative of fourth root of x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the derivative of the given function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: fourth root is x^(1/4) as long as the expression is n does not =1 (in CS notation this is n!=1, not sure how you wanted us to notate it however) expression x^n, the derivative is nx^(n-1) so (1/4)x^(-3/4) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe derivative y=x^(1/4), which is of the form y = x^n with n = 1/4, is y ' = n x^(n-1) = 1/4 x^(1/4 - 1) = 1/4 x^(-3/4). *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 3.1.39 was 3.1.27 (formerly 4.1.24) derivative of (`theta-1)/`sqrt(`theta) What is the derivative of the given function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: theta / sqrt(theta) - 1/sqrt(theta) theta * theta^1/2 - theta^1/2 theta^3/2 - theta^1/2 it would be a difference of powers 3/2 * theta^1/2 1/2 * theta^-1/2 (3/2)theta^(1/2) - (1/2) theta^ (-1/2) confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** (`theta-1) / `sqrt(`theta) = `theta / `sqrt(`theta) - 1 / `sqrt(`theta) = `sqrt(`theta) - 1 / `sqrt(`theta) = `theta^(1/2) - `theta^(-1/2). The derivative is therefore found as derivative of the sum of two power functions: you get 1/2 `theta^(-1/2) - (-1/2)`theta^(-3/2), which simplifies to 1/2 [ `theta^(-1/2) + `theta^(-3/2) ]. . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I am confused on what happened between `theta / `sqrt(`theta) - 1 / `sqrt(`theta) and `sqrt(`theta) - 1 / `sqrt(`theta) what happened to the first theta? I noticed the answers have similar values 1/2 [ `theta^(-1/2) + `theta^(-3/2) ]. (3/2)theta^(1/2) - (1/2) theta^ (-1/2) ------------------------------------------------ Self-critique Rating:1
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Given Solution: `a** The first derivative is 7 x^6 + 25 x^4 - 12 x^3. The second derivative is the derivative of this expression. We get 42 x^5 + 200 x^3 - 36 x^2. It isn't necessary to keep taking derivatives if we notice the pattern that's emerging here. If we keep going the highest power will keep shrinking but its coefficient will keep increasing until we have just 5040 x^0 = 5050 for the seventh derivative. The next derivative, the eighth, is the derivative of a constant and is therefore zero. The main idea here is that the highest power is 7, and since the power of the derivative is always 1 less than the power of the function, the 7th derivative of the 7th power must be a multiple of the 0th power, which is constant. Then the 8th derivative is the derivative of a constant and hence zero. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): this would make all derivatives past 7, 0, not just the 8th derivative, correct? ------------------------------------------------ Self-critique Rating: OK
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Given Solution: `a** The derivative of a^x is ln(a) * a^x. So the derivative of 11^x is ln(11) * 11^x. The derivative of the given function is therefore 12 e^x + ln(11) * 11^x. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 3.2.18 was 3.2.10 (3d edition 3.2.19) (formerly 4.2.20) derivative of `pi^2+`pi^x. What is the derivative of the given function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: PI is a constant. so i would treat them as a power and exponential, using the sum of them. I almost made the mistake of still taking a derivative of PI, befure noticing after squaring its still a constant, so we drop that off completely. so PI^x + ln(PI) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** `pi is a constant and so therefore is `pi^2. Its derivative is therefore zero. `pi^x is of the form a^x, which has derivative ln(a) * a^x. The derivative of `pi^x is thus ln(`pi) * `pi^x. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The first part did trick me at first, but I was glad to be tricked by it, let me know I need to pay more attention. ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 3.2.43 was 3.2.40 (3d edition 3.2.30) (formerly 4.2.34) value V(t) = 25(.85)^6, in $1000, t in years since purchase. What are the value and meaning of V(4) and ov V ' (4)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It seems like part of the function is missing. I havent looked at this problem, but I'll go off the information presented. It seems V(t) is a value, represented in thounsands, so if V(t)=1 ill assume it means $1,000. t= years. This means that V(4) represents a value, in thousands, after 4 years. with the given function, y is a constant value. V'(4) would represent how quickly the value is changing at the instant t=4 years, and could be used for a prediction of times around that. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** V(4) is the value of the automobile when it is 4 years old. V'(t) = 25 * ln(.85) * .85^t. You can easily calculate V'(4). This value represents the rate at which the value of the automobile is changing, in dollars per year, at the end of the 4th year. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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." 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: #*&!