Karl's Calculus Forum: Oct 2002

Archived Forum Discussion for October 2002

Reply to Kyle Struble: Think about this: A straight Riemann sum blindly adds up the value of the function at each point, paying no attention to what the function might be doing between the points. Call that a zeroth order approximation of the area. The trapezoidal method uses the assumption that what goes on between the points is a linear function. Each linear function matches its two neighbors in function value at the two boundary points for the fragment. Call that a first order approximation. Simpson's rule uses the assumption that what goes on between the points is a quadratic function. This assumption allows that the approximating function in each fragment matches its two neighbors in both function value and first derivative at the boundary points. But does it end there?? No. You can have a third order method, using cubic polynomials to approximate the behavior. You'd expect doing this you could have each fragment's approximation function match its two neighbors in function value and the first two derivatives. And you can continue this to nth order, where n is as big as you'd like. This might be Taylor's method -- I don't know if it has a name.

You can also do variations of the Riemann, trapezoid, Simpson's, and so on methods. Observe, for example, that the trapezoidal method approximates the area under a straight line perfectly. That's because the second derivative of a straight line is zero, and the error in this method is proportional to the second derivative of the function being approximated. Since you know that where the magnitude of the second derivative of f(x) is high, you will get more error, you simply adjust your fragment size inversely according to magnitude of f"(x) -- where there's greater second derivative, use smaller fragments. Think about how that reduces error at a controlled cost in terms of computation. You can apply it to the higher order methods as well, but you would use higher order derivatives to determine the fragment size. I don't know if this method has a name or not.
Karl <Click to Send Email to Karl>
USA - Thursday, October 31, 2002 at 13:03:52 (EST)

Reply to John: You can find a similar problem complete with a worked solution in the text of Karl's Calculus Tutor at Related Rates, problem 6
Karl <Click to Send Email to Karl>
USA - Thursday, October 31, 2002 at 12:42:01 (EST)

Alright, I have a major problem. I need to find different methods or theroms to compute area's under an arbitrary function. I know the regulars like Simpson/trapiziod/right bound/left bound/midpoint.... I need 4 others to complete my assignment. My professer said use the internet or any source to find some. I've done a little work and I can't seem to find any... I've come across Huen's method, Runge-Cutta's method, and Taylor's methods ( I dont even know if these are right). I need to find 4.... describe their geometry(like they use trapidoids...or parabola..ext).. and their formula.... (either normal formula or what you would put into your calculator in terms of SUM(SEQ(.........)).. Thank you so much for immediate responses!!!
Kyle Struble Vestal, NY USA - Thursday, October 31, 2002 at 12:27:49 (EST)

Im having some trouble with a calculus problem and am in need of some help, I don't even know where to start. Problem: A man 6 Ft talkk is walking away from a 15 foot lamp post at a rate of 6 ft/sec. At what rate is the end of his shadow moving away from the base of the lamp post? Any help??
John Kent, oh USA - Tuesday, October 29, 2002 at 22:44:52 (EST)

Reply to dr0dlr: You have sin(1/x), which is a composite, so it should be expected that you will have to use the chain rule to differentiate that part. But it is multiplied by x, and so you should expect to have to use the product rule to complete the problem.

First, applying the chain rule to sin(1/x). If u(v) = sin(v) and v(x) = 1/x, then u(v(x)) is this part of your function. The chain rule says that its derivative is u'(v(x))v'(x). You should already know that u'(v) = cos(v) and that v'(x) = -1/x². So put those together according to the chain rule to find the derivative of sin(1/x).

But the function you really want to find the derivative of is the product of that with x. If you let g(x) = x and h(x) = sin(1/x), then your function is f(x) = g(x)h(x). And the product rule says that f'(x) = g'(x)h(x) + h'(x)v(x). In the preceding paragraph I showed you how to take the derivative of h(x). And you should be able to find the derivative of g(x) on your own. Now see if you can put the pieces together.
Karl <Click to Send Email to Karl>
USA - Tuesday, October 29, 2002 at 18:45:38 (EST)

Find the derivative of the given function. y = x(sin1/x) Please include steps because I am lost.
dr0dlr USA - Sunday, October 27, 2002 at 19:21:44 (EST)

Reply to Amy: You didn't say what you wanted to find, so I assume you are looking to find the values of x that satisfy your polynomial equation. In general, the roots of polynomials of degree 5 or greater cannot be found using algebraic methods. Your polynomial is of degree 10. I don't see any algebraic trick to which this one succumbs. Absent that, you have to use a progressive approximation algorithm, like Newton-Raphson to find the roots, one by one. I already tried this and found 10 roots, but I have to say that they were unremarkable numbers. The lowest one, for example, was 0.997612904837836.

To do Newton-Raphson, you have to know the derivative of your polynomial. Since this one is expressed as a product, you can drop one term of the product in 10 different ways. The derivative is the sum of the 10 products that are made from each combination of 9 remaining factors.
Karl <Click to Send Email to Karl>
USA - Saturday, October 26, 2002 at 13:13:26 (EDT)

Reply to L: Your lack of understanding of calculus won't get you through your calculus course, if that's what you're taking right now. If you are lost in class, you need to make the extra effort to catch up by figuring out where in the past weeks it got away from you, and doing extra problems in that area and the areas subsequent to that until you feel confident in your understanding.

To differentiate the function you've given here, you use the quotient rule. Your numerator is N(x) = x². Your denominator is D(x) = x-2. You should be able to find N'(x) and D'(x) using the power rule. Now simply apply the quotient rule to those results to find the derivative of f(x) = N(x)/D(x).
Karl <Click to Send Email to Karl>
USA - Saturday, October 26, 2002 at 13:03:46 (EDT)

I am having difficulty on this problem: (x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)(x - 8)(x - 9)(x - 10) = (1² * 3² * 5² * 7² * 9²)/2¹⁰.
Amy
USA - Saturday, October 26, 2002 at 09:04:01 (EDT)

Your help on the previous was GREAT! Thank you. However, due to my total lack of understanding of calculus I would appreciate your help on this one. Differentiate f(x) = x^2/x-2
L
FL USA - Friday, October 25, 2002 at 16:18:21 (EDT)

Reply to Lynn: The formula for the height of an object dropped from x₀ meters high, as a function of time, is:

            1
   h(t)  =    at² + v₀t + x₀
            2

where a = -9.8 meters/sec² for an object falling under earth's gravity, v₀ is the initial velocity of the object in the upward direction (use negative for downward, and in your case, the object is dropped from rest, so your v₀ is zero), and t is time in seconds.

When the object hits the ground, its height, h(t), is zero. So use the quadratic formula to solve for t.
Karl <Click to Send Email to Karl>
USA - Friday, October 25, 2002 at 12:08:48 (EDT)

I am lost as can be on this, please help. Word problem: a water balloon is dropped from 30 meters. What is the time from release to ground?
Lynn FL USA - Thursday, October 24, 2002 at 20:42:02 (EDT)

Reply to Ben: You can get estimates of F'(x) at the midpoints of each of the sub-intervals in your table using a divided difference. That is, to find the estimate of F'(1.2), take (F(1.3) - F(1.1))/(1.3 - 1.1). Make a table of values for F'(x), similar to the one you made for F(x), except these will be at the midpoints (and your table will necessarily have one fewer entries than the original table). Your estimate for F'(1.7) will be the mean of your estimates for F'(1.6) and F'(1.8).

To do the second part, simply use the formula for a line of slope m through the point, (h,k):

   y - k  =  m(x - h)

Use the equation of the line to estimate F(1.8) by putting 1.8 in for x into the equation for the line.
Karl <Click to Send Email to Karl>
USA - Thursday, October 24, 2002 at 20:34:09 (EDT)

Please help me out.

X  =  1.1  1.3  1.5  1.7  1.9  2.1
F(X)= 12   15   21    23   24   25

A) ESTIMATE f`(1.7)

B) WRITE AN EQUATION FOR THE TANGENT LINE TO THE GRAPH OF F AT X=1.7

C) USE YOUR ANSWER IN B) TO PREDICT THE VALUE OF F AT X=1.8

Ben Bogota, NJ USA - Thursday, October 24, 2002 at 17:32:53 (EDT)

Reply to Lynn: You could do this one either of two ways. One is to multiply out the terms (which results in a cubic polynomial), and then take the derivative of the result, term by term, using the power rule. But that is the hard way. The easy way it to recognize that this function is a product:

   f(x)  =  u(x)v(x)

where u(x) = x² + 1, and v(x) = x + 3. So you apply the product rule, which says that

   f'(x)  =  u'(x)v(x) + v'(x)u(x)

Use the power rule to find u'(x) and v'(x). Then just put the pieces together.
Karl <Click to Send Email to Karl>
USA - Thursday, October 24, 2002 at 12:05:18 (EDT)

Problem understanding: Differentiate f(x) = (x^2 + 1)(x + 3)
Lynn FL USA - Thursday, October 24, 2002 at 07:20:32 (EDT)

Reply to dr0dlr: Your function is equivalent to

   y(x)  =  x^-1/2

Step 1: Use the power rule to take the derivative and find y'(x).

Step 2: The x coordinate of the point of tangency the problem asks for is x=1. So evaluate y'(1). That will be the slope, m, of the tangent line.

Step 3: Use the formula to find the equation of the tangent line:

   y - k  =  m(x - h)

is the formula for the line that passes through the point, (h,k), and has slope of m. You want the line to go through the point, (1,1). Just put in the numbers.
Karl <Click to Send Email to Karl>
USA - Wednesday, October 23, 2002 at 19:39:21 (EDT)

Find an equation to the tangent line to the curve at the given point. y = 1 / sqrt(x), (1,1) Please show steps. Thanks.
dr0dlr USA - Wednesday, October 23, 2002 at 18:43:04 (EDT)

Reply to Nick: Your formula for curveture is

  C(t)  =  r'(t) cross r"(t) / |r'(t)|³

If r' and r" are parallel, then their cross product is zero. To find an example, just let f(t) be any scalar, differentiable function of t. Multiply f(t) by any constant vector, A, and you have such a function, which, by necessity, is a straight line. Recall that only a straight line has no curveture.

For the planes, just take the dot product and the cross product of the two vectors, A = (1,1,-1) and B = (2,-3,4). Note that these are the normal vectors to the planes. The planes can be parallel only if the cross product is zero. The planes can be perpendicular only if the dot product is zero. To get the angle between them, divide the cross product by the dot product and take the arctangent.

To find where arctan(x) has its max curveture, first apply the curveture formula:

   C(x)  =  y" / (1 + y'²)^3/2

Put in arctan(x) for y. Next find C'(x), and set it to zero. Finally, solve for x that makes that last equation work.

The last problem, you have r(t) = (f(t),g(t)). So r'(t) = (f'(t),g'(t)) and r"(t) = (f"(t),g"(t)). Just apply the formula.
Karl <Click to Send Email to Karl>
USA - Wednesday, October 23, 2002 at 00:05:39 (EDT)

1. show that if r'(t) and r"(t) are parallel at some point on the curve described by r(t), then the curvature at that point is 0. Give a (nonzero) example of a curve r(t) for which r'(t) and r"(t) are always parallel. 2. Show that the planes x+y-z=1 and 2x-3y+4z=5 are neither parallel nor perpendicular. Find, correct to the nearest degree, the angle between these planes. 3. At what point(s) does the curve y=arctan(x) have maximum curvature? 4. Use K(t)=r'(t) x r"(t)|/|r'(t)|^3 to show that the curvature of a plane parametric curve x=f(t), y=g(t) is K=|x.y..-y.x..|/ x.^2+y.^2)^(3/2), where the dots indicate the derivitives with respect to t. (the dots where above the x and y's on the sheet but I couldn't type it like this, it should read x dot y double dot and so on with the daots above the letters.)
Nick OR USA - Tuesday, October 22, 2002 at 12:24:34 (EDT)

Reply to Rafael: If limit of f as x goes to x₀ is not infinite (and exists), then f is bounded over some interval that contains x₀ as an interior point. Since g is constrained to be positive but less than f, it too must be bounded over that same interval. Hence the product of f and g must also be bounded over that same interval. So no pair of functions can exist that meet your criteria.

Please let me know if this proof is errant somewhere.
Karl <Click to Send Email to Karl>
USA - Wednesday, October 16, 2002 at 19:10:21 (EDT)

Is it possible to find 2 functions f(x) and g(x) that fulfill

(a) 0 < g(x) < f(x) where they are defined

(b) lim(f*g)(x)=infinity
     x==>x0

and still

limf(x) not equal infinity?
x==>x0

I tried to find such f and g trying some variations of dirichlet functions to no avail. When i tried to prove that no such functions exist that is to prove that for every f and g that fulfill 1 and 2 lim(f*g)(x)=infinity i got stuck proving it with the limit definition of lim(f*g)(x) Please help
Rafael Tel Aviv, israel - Wednesday, October 16, 2002 at 17:59:52 (EDT)

Reply to Van: This is just another problem of fitting a straight line to the numbers given. Let x be the price of a ticket. Let y be the number of moviegoers per night. Then you are looking for the equation

   y  =  mx + b

and you need to find m and b. Since y, the number of moviegoers, increases by 35 with each $0.50 that, x, the ticket price decreases, the slope must be

          35
   m  =         =  -70 moviegoers per dollar
         -0.50

You know that when x = $7.50, then y = 400. Hence by the formula for passing a line of slope m through a given point, you have:

   y - 400  =  m(x - $7.50)

You can take it from there.
Karl <Click to Send Email to Karl>
USA - Wednesday, October 16, 2002 at 12:12:05 (EDT)

Problem: A movie theater has been charging $7.50 per person and selling about 400 tickets on a typical weeknight. After surveying thier customers, the heater estimates that for every 50 cents that they lower their price, the number of moviegoers will increase by 35 per night. Find the demand function and calculate the consumer surplus when the tickets are priced at $6.00.
Van USA - Tuesday, October 15, 2002 at 22:32:40 (EDT)

Reply to Richard: One way you can approach this problem is to develop a Riemann sum of the function

               1
   f(x)  =   ______
            Ö1 - x²

Divide the interval, [0,1/2], into lots of equal pieces. Find f(x) at the boundary of each piece, and sum all those f(x)'s together. Multiply the sum by the length of a piece. As the number of pieces goes to infinity, the Rieman sum will go to p/6. If you add all the interior boundaries in normally, but add only half of each of the endpoint f(x)'s, you'll have taken the trapezoidal area, which is more accurate. You can get even better accuracy using Simpson's rule (which is not yet covered in Karl's Calculus Tutor, but you can search for it on the web).

You can also develop a Maclaurin series for f(x), and find its integral term by term. Then put in 1/2 for x, and add up as many terms as you have patience to do. The more terms you add up, the closer the result will be to p/6.
Karl <Click to Send Email to Karl>
USA - Tuesday, October 15, 2002 at 17:42:04 (EDT)

Reply to Ken: If you apply the chain rule to f(-x), you have f(-x) = f(g(x)), where g(x) = -x. That means that g'(x) = -1. The chain rule requires that

   d(f(-x))
             =  f'(g(x)g'(x)  =  f'(-x)(-1)  =  -f'(-x)
      dx

Now simply apply that rule to sin(-x) and cos(-x).
Karl <Click to Send Email to Karl>
USA - Tuesday, October 15, 2002 at 17:32:30 (EDT)

How do you use the integral definition of the arcsin function to actually compute an arcsin? For example, arcsin (1/2). I know it's pi/6. How do you get there from here? Richard
Richard USA - Monday, October 14, 2002 at 17:57:15 (EDT)

Corrected my email address. Problem: Find the derivative of sin(-x) + cos (-x).
ken Richmond, CA USA - Monday, October 14, 2002 at 17:35:12 (EDT)

Find the derivative of sin(-x) + cos (-x).
ken Richmond, CA USA - Monday, October 14, 2002 at 17:29:29 (EDT)

Reply to Marv: Try SOSmath at www.sosmath.com/trig/trig.html. Karl's Calculus Tutor also has a section to review trig, which is at www.karlscalculus.org/calc7_0.html.
Karl <Click to Send Email to Karl>
USA - Wednesday, October 09, 2002 at 11:40:28 (EDT)

Reply to Mario Garza Livas: The only trick to finding the derivative of f(x) = u(x)^v(x) is to use the exponential identity:

   b^x  =  e^{x ln(b)}

for all positive values of b.

So your problem becomes:

   f(x)  =  e^{v(x) ln(u(x))}

to which you can apply the chain rule and product rule:

   f'(x)  =  (e^{v(x) ln(u(x))})(v'(x) ln(u(x))  +  v(x)u'(x)/u(x))

You can then use the identity in reverse to convert this to

   f'(x)  =  (u(x)^v(x))(v'(x) ln(u(x))  +  v(x)u'(x)/u(x))

Karl <Click to Send Email to Karl>
USA - Wednesday, October 09, 2002 at 11:19:56 (EDT)

I want to know the rule used to differentiate this: f(x)=u(x)^v(x) In other worsd I want d u(x)^v(x) ___________ d x Note: Consider that f(x), u(x) and v(x) are all functions of x. Please send me the rule and also the proof... Whats the use of learning it without understanding it... The proof is more important for me. Thats all, I´ll wait for your response pasted in your forum or sent to my mail. Thanks and forgive my english... I´m Mexican.
Mario Garza Livas Monterrey, NL Mexico - Wednesday, October 09, 2002 at 10:16:25 (EDT)

I am falling behind quite tremendously in a Calc 1 course due to Trig shortcomings. It has been about 13 years since Trig. Can anyone suggest a good online resource for a review of the Calc specific Trig functions and how to use them?
marv Orlando, FL USA - Wednesday, October 09, 2002 at 00:42:15 (EDT)

Reply to Van: Substitute tan(u) = x and (1 + tan²(u))du = dx. Then use trig identities to convert the result into a simple quotient of sines and cosines of u. Finally, substitute v = sin(u) and dv = cos(u)du. Then you can integrate and substitute back.
Karl <Click to Send Email to Karl>
USA - Tuesday, October 08, 2002 at 18:17:41 (EDT)

Can you help me on this one. I'm not sure what to do with the square roots. Evaluate the integral.

    dx
-----------
   ________
x²Ö1 + x²

Van USA - Tuesday, October 08, 2002 at 17:47:39 (EDT)

Reply to Van: You can easily use the comparison test to show your integral is convergent. Just show that for sufficiently large x, it is always the case that

    1     ln(x)
       >       
   x²      x³

Since dx/x² is convergent, your integral must be convergent as well.

As for evaluating this integral, use the substitution:

   u  =  ln(x)

   e^u = x

   du = dx/x

   e^u du = dx

Find the indefinite integral and evaluate it from 1 to infinity.
Karl <Click to Send Email to Karl>
USA - Monday, October 07, 2002 at 18:25:20 (EDT)

Reply to Cheron: I'm not sure how you are suppose to plot this thing. It has two traces, one along y=0, and the other along y=x. But both traces are made up of infinitely many completely disconnected dots, packed infinitely densely together.

To answer the question about continuity of f(x) at x=0, use the delta-epsilon test. Since zero is rational, it is clear that f(0) = 0. So does the limit as x approaches zero of f(x) indeed go to zero? So, by the definition of a limit, can you show that for any epsilon greater than zero, no matter how small, that you can always find a delta such that:

   |f(x)|  <  e

whenever |x| < d? How far can f(x) be from zero if |x| < 0.01? What if |x| < 0.000001? Follow that lead to its conclusion.
Karl <Click to Send Email to Karl>
USA - Monday, October 07, 2002 at 18:18:15 (EDT)

I need help on the following problems. Please explain in detail if possible. Determine whether each integral is convergent or divergent. Evaluate those that are convergent. integrate from 1 to infinte

 ln x
----- dx
 x^3

Van Upper Darby, PA USA - Monday, October 07, 2002 at 14:31:14 (EDT)

Reply to James: In the case of f+g, use a delta-epsilon proof. According to the premise of the problem, f is continuous, so the limit as x goes to a of f(x) must be equal to f(a). That is the definition of continuity.

The existence of that limit means that for any epsilon, no matter how small, you can find delta such that

   |f(x) - f(a)|  <  e

whenever |x - a| < d. On the other hand, g(x) is discontinuous at x = a. That means that there is some epsilon that you can never get under no matter how small your delta is:

   |g(x) - g(a)| ³ e

for some epsilon, no matter how small |x - a| is. Now, considering that such an epsilon exists for g, see if you can figure out what the implications are for f+g.

As far as f × g goes, there is a very easy counterexample. Suppose that f(x) = 0 for all x. That is a continuous function. What about its product with the discontinuous function, g(x)? And if you stipulate that g(x) is nonzero everywhere, what about the quotient function, f(x)/g(x)?
Karl <Click to Send Email to Karl>
USA - Monday, October 07, 2002 at 13:02:44 (EDT)

I have a question on a graph called a salt and pepper graph. f(X) = 0, when x is irrational and f(X)= x, when x is rational. Sketch this graph. Explain why the limit when x approaches 0 eguals 0 if x takes on only irrational values. Explain why the limit when x approaches 0 eguals 0 if x takes on only rational values. Does the limit as x approaches 0 of f(x) exist?
Cheron HazelCrest, il USA - Sunday, October 06, 2002 at 21:33:16 (EDT)

I am just beginning proofs in Calculus and I am having a very hard time on how to start them. An example he gave us in class to try was the following: 1. Suppose that the function f is continuous at the point a, and the function g is discontinuous at a. Prove that the following are either true or false (i) f+g can never be continuous at a. (ii) f x g can never be continuous at a. (iii) f/g can never be continuous at a.
James Ottawa, ON Canada - Sunday, October 06, 2002 at 15:39:42 (EDT)

Reply to Lara: If s is supply, and p is presumably price, you have

        1
  s  =     -  0.55p
        3

This is nothing more than the equation for a line (just replace s with y and p with x, and you'll see what I mean). So the rate of change is simply the slope of that line. For the interval, (1,3), simply find the average slope (which, for a line, is the same as the slope anywhere on the line) by taking the rise over the run:

         s(3) - s(1)
   m  =             
            3 - 1

Karl <Click to Send Email to Karl>
USA - Thursday, October 03, 2002 at 20:21:25 (EDT)

Reply to Lara: I assume that "using technology" means using your calculator, and "estimate" means take the slope of progressively smaller secant lines. Your function is

   -0.5t³ + 1.4t² - 3.8t + 6

Take your calculator and evaluate this function at t = 2.65. Call the result, s₀.

Now evaluate it again at the following values and write them down, naming the values accordingly:

at t = 2.65 + 0.1        call it s₁

at t = 2.65 + 0.01       call it s₂

at t = 2.65 + 0.001      call it s₃

Now evaluate the following slope-of-the-secant for each of these points by

   m₁  =  (s₁ - s₀) / 0.1

   m₂  =  (s₂ - s₀) / 0.01

   m₃  =  (s₃ - s₀) / 0.001

Base your "estimate" on the trend.
Karl <Click to Send Email to Karl>
USA - Thursday, October 03, 2002 at 20:10:23 (EDT)

HELP!! The calc. lab is closed and I can't figuire these problems out.... #1.Using technology and the limit concept estimate the slope of the tangent line to the curve s(t)= -0.5t raised to the third +1.4t squared - 3.8t + 6 at t=2.65 #2. You have been selling sodas at sporting events and have developed the following model for the supply: 1/3-0.55p a. Estimate the rate of change of the quantity when the price is 1, 1.50, 3 b. Find the avg. rate of change at interval (1,3) c. How do your answers in a compare to b
Lara Murfreesboro, Tn USA - Wednesday, October 02, 2002 at 20:12:24 (EDT)

hello :)
cat20000
USA - Tuesday, October 01, 2002 at 16:31:02 (EDT)

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