course Mth 173 ?????K?????assignment #017s?r??????????????Calculus I
......!!!!!!!!...................................
16:52:43 Explain in terms of the contribution to the integral from a small increment `dx why the integral of f(x) - g(x) over an interval [a, b] is equal to the integral of f(x) over the interval minus the integral of g(x) over the same integral.
......!!!!!!!!...................................
RESPONSE --> When thinking in terms of the graph and the actual values of f(x) and g(x), the basic idea is that the we want to find the area between f(x) and g(x) (f(x) being higher than g(x)) over the interval [a, b]. If we take the integral of f(x) over the interval, that will give us the entire area from the axis to the line. We can then take the integral of g(x) and subtract it from the integral of f(x) to find only the area between the lines. However, there is a shortcut. The shortcut is to take the integral of the value of f(x) - g(x). This is where we take a specific point and find only the height between the two lines. We then multiply that value (the height between the lines) by the 'dx value to find the area over the interval 'dx. This then integrates to find the overall value.
.................................................
......!!!!!!!!...................................
17:03:29 07-18-2007 17:03:29 ** 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'.**
......!!!!!!!!...................................
NOTES -------> Hmm. I guess I didn't understand the question. I think I understand it now.
.......................................................!!!!!!!!...................................
17:08:05 Explain why the if f(x) > m for all x on [a,b], the integral of f(x) over this interval is greater than m (b-a).
......!!!!!!!!...................................
RESPONSE --> The areas of both the integral of f(x) and m(b-a) are basically two boxes with the same width, but different heights. If the height of f(x) is consistently greater than the height of m, then the area of the integral of f(x) will be greater than m(b-a).
.................................................
......!!!!!!!!...................................
17:08:19 ** 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 ). **
......!!!!!!!!...................................
RESPONSE --> Ok.
.................................................
......!!!!!!!!...................................
17:51:41 Explain 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).
......!!!!!!!!...................................
RESPONSE --> When dividing the integrals of both functions, each integral represents area. Each overall integral is typically different from the specific points of the result of f(x) / g(x), rendering a different curve.
.................................................
......!!!!!!!!...................................
17:57:10 ** 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). **
......!!!!!!!!...................................
RESPONSE --> Ok. Interesting.
.................................................
......!!!!!!!!...................................
18:01:18 Given 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.
......!!!!!!!!...................................
RESPONSE --> To construct a graph of a function from its rate, we would need to find on the rate function the zeroes and the periods of increasing and decreasing rates. From there we can create a function based on the zeroes of the rate and modifiy it to fit the rate function. This is basically integration in which we take f(x) (the rate) and use its graph to find the value of the function itself at a specific time. Basically we find the average value of f(x) over an interval and multiply it by the parts of the interval to find the area under the rate curve. This area represents the function.
.................................................
......!!!!!!!!...................................
18:02:16 ** 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. **
......!!!!!!!!...................................
RESPONSE --> Yes, that would be more specific than my answer.
.................................................
......!!!!!!!!...................................
18:03:09 Query problem 3.1.16 (3d edition 3.1.12 ) (formerly 4.1.13) derivative of fourth root of x. What is the derivative of the given function?
......!!!!!!!!...................................
RESPONSE --> I believe the derivative is: y' = (1/4)(x^(-3/4)
.................................................
......!!!!!!!!...................................
18:03:17 The 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). *&*&
......!!!!!!!!...................................
RESPONSE --> Ok.
.................................................
......!!!!!!!!...................................
18:07:23 Query problem 3.1.27 (formerly 4.1.24) derivative of (`theta-1)/`sqrt(`theta) What is the derivative of the given function?
......!!!!!!!!...................................
RESPONSE --> Problem 3.1.36. I believe the derivative is: y' = (1/2)('theta^(-1/2) - 'theta^(-3/2)
.................................................
......!!!!!!!!...................................
18:11:28 ** (`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) ]. . **
......!!!!!!!!...................................
RESPONSE --> Ok.
.................................................
......!!!!!!!!...................................
18:11:42 Query problem 3.1.60 (3d edition 3.1.48) (formerly 4.1.40) function x^7 + 5x^5 - 4x^3 - 7 What is the eighth derivative of the given function?
......!!!!!!!!...................................
RESPONSE --> The eighth derivative is a 0.
.................................................
......!!!!!!!!...................................
18:11:59 ** 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. **
......!!!!!!!!...................................
RESPONSE --> Ok.
.................................................
......!!!!!!!!...................................
18:13:35 Query problem 3.2.4 (3d edition 3.2.6) (formerly 4.2.6) derivative of 12 e^x + 11^x. What is the derivative of the given function?
......!!!!!!!!...................................
RESPONSE --> The derivative is: f' = 12e^x + ln(11) *11^x
.................................................
......!!!!!!!!...................................
18:13:43 ** 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. **
......!!!!!!!!...................................
RESPONSE --> Ok.
.................................................
......!!!!!!!!...................................
18:16:17 Query problem 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?
......!!!!!!!!...................................
RESPONSE --> Problem 3.2.18. The derivative is: y' = ln('pi) * 'pi^x
.................................................
......!!!!!!!!...................................
18:16:26 ** `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. **
......!!!!!!!!...................................
RESPONSE --> Ok.
.................................................
......!!!!!!!!...................................
18:18:30 Query problem 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)?
......!!!!!!!!...................................
RESPONSE --> V(4) =13.05. This means that after 4 years, the vehicle has depreciated to to a value of $13,050. V'(4) = -2.121. This means that at exactly 4 years after purchase, the vehicle is depreciating at a rate of -$2,121 per year.
.................................................
......!!!!!!!!...................................
18:18:43 ** 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. **
......!!!!!!!!...................................
RESPONSE --> Ok.
.................................................
......!!!!!!!!...................................
18:19:35 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> I need to consider the logical ramifications of the relations between functions more.
.................................................
"