Karl's Calculus Forum: Oct/Nov 2003

Archived Forum Discussion for October/November 2003

Reply to Rosa: The beam will stay on a 1 ft wide person for at least 0.1 seconds provided it is not going faster than 10 ft/sec. So any x(t) such that |dx/dt| < 10 for all t will work. Of course you have to be sure that it covers the entire hall. There are many solutions to this problem. One of them would be

   x(t)  =  20 sin(t/4)

which would cover the entire hallway in 8p seconds, which is less than the 10 minutes required. See if you can think of other functions that would work. Also make sure you understand why this one works.
Karl <Click to Send Email to Karl>
USA - Saturday, November 29, 2003 at 15:59:50 (EST)

Reply to Jordan: You have the ellipse:

          ____________________
   y  =  Ö1 - (x+2)² (2.89/25)

Squaring both sides

   y²  =  1 - (x+2)² (2.89/25)

The Pythagorean distance formula tells you that

   s²  =  x² + y²

where s is the distance from the origin to a point on the ellipse. Now replacing y² with the expression for it from the ellipse equation:

   s²  =  x² + 1 - (x+2)² (2.89/25)

Taking the derivative of both sides with respect to x:

   2s s'  =  2x  -  2(x+2)(2.89/25)

Divide through by s to get s'. Now replace the s in the denominator with the expression for it suggested by the Pythagorean formula.
Karl <Click to Send Email to Karl>
USA - Saturday, November 29, 2003 at 15:49:36 (EST)

Reply to Chris Bailey: The problem was to decompose the following into partial fractions:

     2k + 1
              
   k² (k + 1)²

You could use either of the standard methods for partial fractions on this, and they are both covered in the web pages, but this one succumbs easily to inspection. Observe that if you add k² - k² to the numerator (you are adding zero so this doesn't change anything), you get

   k² + 2k + 1 - k²
                   
      k² (k + 1)²

Recall that k² + 2k + 1 = (k + 1)². See if you can finish it.
Karl <Click to Send Email to Karl>
USA - Saturday, November 29, 2003 at 15:30:43 (EST)

Reply to Julie: Take the implicit derivative of x² + xy + y² = 6. You get

   2x + xy' + y + 2yy'  =  0

Since you want horizontal tangents, you are looking for the places where y' = 0. So put zero in for y' in the implicit derivative. That gives you an equation in x and y. Use that to substitute for y into the original equation. That will allow you to solve for the points where horizontal tangents touch this curve.
Karl <Click to Send Email to Karl>
USA - Saturday, November 29, 2003 at 15:14:25 (EST)

Reply to Eric Schneider: The problem was to integrate:


    1
          dx
 1 +  Ö2x

Doing it as the book recommends, you have

                 __
    u  =  1  +  Ö2x
                 __
    du  =  (1 / Ö2x) dx

By manipulating the first equation you find that

               __
    u - 1  =  Ö2x

so you have (u - 1) du = dx. The integral becomes


 (u - 1)
         du
    u

Karl <Click to Send Email to Karl>
USA - Saturday, November 29, 2003 at 15:09:30 (EST)

Hi I'm having trouble coming up with a design of a security system. Here is the scenario. There has been an increasing number of thefts from the St. elsewhere hospital Pharmacy. My job is to design a security system??? The drug supply is kept in a storeroom whose entrance is located in the middle of a 40 foot long hallway. The entrance is 3 ft. wide door. The hospital wants monitor the entire hallway as well as the storeroom door. I must decide how to program a detector to accomplish this. The detector runs on a track and points a beam of light straight ahead on the opposite wall. The beam reaches from the floor to the ceiling. Think of the hallway to be watched as the interval [-20,20]. You need to decide what x(t) is, where x(t) represent the position of the beam at time t (in seconds). So I need to draw a graph of x versus t, assume t ranges from 0 to 10 minutes. The beam must stay on an object for at least one-tenth of a second in order to detect that object. If the width of a person is 1 ft. decide whether the graph will detect a person standing anywhere in the hallway. How long will the door be without surveillence, as long as the beam is hitting any part of the door it is consider to be under surveillence. I need to come up with a function or formula for the first 10 minutes with all the restrictions possible.
Rosa

Phoenix, AZ USA - Wednesday, November 26, 2003 at 23:21:14 (EST)

Hi, I'm looking for the antiderivative of 1/x or x^-1. Thanks! Stephanie
Stephanie

Toronto, On Can - Sunday, November 23, 2003 at 22:29:05 (EST)

If anyone could help me on an exam i have do on monday please let me know!!!
Raixa

Mayaguez, PR USA - Saturday, November 22, 2003 at 15:11:08 (EST)

Sorry, I missed something in the equation!

           /---------------------
          /  /    (x+2)²  \
f(x) =   /  ( 1 - -------  ) *  2.89
       \/    \      25    /

Jordan

Chilliwack, BC Canada - Thursday, November 20, 2003 at 03:19:05 (EST)

I need help with this problem... OK, so I have half an ellipse, equation is:



           /---------------------
          /    (x+2)²
f(x) =   / 1 - -------  *  2.89
       \/        25

I need find an equation that produces the derivative of the distance between the origin and the ellipse at any point. I've worked on this problem for hours and I still can't find a solution. Any help would be very much appreciated!
Jordan

Chilliwack, BC Canada - Thursday, November 20, 2003 at 03:13:22 (EST)

I need help on how to do the Partial Fractions on the following Infinite series:

        2k + 1
f(x) = ------------

        k²(k+1)²

Chris Bailey

Tampa, Fl USA - Wednesday, November 19, 2003 at 20:25:49 (EST)

Help please x² + xy + y² = 6 find the points of the horizontal tangent lines
Julie

baltimore, MD USA - Monday, November 17, 2003 at 21:50:44 (EST)

I need a little help on a problem from the section in my text on the natural logarithmic function in integration. Find the indefinite integral of:

               1
    f(x) = ------------
           1 + sqrt(2x)

The text gives a hint to "Let u be the denominator of the integrand." Thank you
Eric Schneider

Tucson, AZ USA - Sunday, November 16, 2003 at 18:23:09 (EST)

I need to find a numeric method that solves differential equations and uses an adaptable step. It must be of the quality of Runge-Kutta fouth order. If you could post an explanation, or you know any book or URL where i could find that information i will be gratefull to know. César Villa.
César Villa

Monterrey, NL Mexico - Saturday, November 15, 2003 at 20:51:03 (EST)

Reply to Nick: If you have a vector, (u(t), v(t)), then use


   tan(q)  =  lim
              t->0

   v(t)
       
   u(t)

To find the angle to the x axis with which it approaches the origin. You can do that with both your vectors. To find the angle between them, use the identity

   tan(a-b)  =

  tan(a) - tan(b)
                  
  1 + tan(a)tan(b)

Karl <Click to Send Email to Karl>
USA - Saturday, November 15, 2003 at 15:58:09 (EST)

Reply to lavontestewart: To find the

        e^2x - 1
   lim         
   x->0    x

Use L'Hopital's rule. Take the derivative of the numerator and derivative of the denominator and find the limit of the quotient of the results.
Karl <Click to Send Email to Karl>
USA - Saturday, November 15, 2003 at 15:27:30 (EST)

Reply to mona: You didn't put any equal sign into your statement of the problem. If you had xy + y² = 0 then I get

   xy' + y + 2yy'  =  0

   xy' + 2yy'  =  -y

           -y
   y'  =        
          x + 2y

To find y", take the implicit derivative again of xy' + 2yy' = -y. I get

   xy" + y' + 2yy" + 2y'²  =  -y'

Now solve for y" in terms of x, y, and y'. When you are done, replace y' with the expression in x and y that you solved for earlier.
Karl <Click to Send Email to Karl>
USA - Saturday, November 15, 2003 at 15:22:36 (EST)

Reply to mona: You need to find the limit of f(x) as x approaches zero. That limit is what k needs to be. Look at the definition of continuity to see why.

Note that sine of anything is always between -1 and 1 inclusive. So as x goes to zero, then x sin(1/x) must also go to zero, in accordance with the squeeze theorem.
Karl <Click to Send Email to Karl>
USA - Saturday, November 15, 2003 at 15:12:32 (EST)

not sure why the first vector description didn't show up, it did in the preview, but the first vector was < t , t^2 > Thanks.
Nick

Pgh, pa USA - Thursday, November 13, 2003 at 09:11:15 (EST)

I have a problem involving vectors. Given two vectors and < sin t, 1/2 sin 2t >, intersection occurs when t = 0. find the angle of their intersection. I cannot remember the rest of the question, but it involved using tangent vectors to find the solution. Any help would be appreciated. Nick
Nick

Pittsburgh, pa USA - Thursday, November 13, 2003 at 09:04:51 (EST)

lim _x->infinity e^2x - 1
               ----
                x

lavontestewart







chicago , Il USA - Wednesday, November 12, 2003 at 23:06:14 (EST)

dear sir, i want to ask that

dy/dx of xy+y² is -y/x+y
so what is the d²y/dx² of same equation is it y/x+y.

mona

abu dhabi, emirates - Wednesday, November 12, 2003 at 10:50:25 (EST)

sir, plz help me to solve following questions

Question  # 1

(a )  If f(x)={xsin1/x   ,x is not equal to 0 and k=0}is continuous at x =0 then what is the value of  k.

mona

abu dhabi, emirates - Tuesday, November 11, 2003 at 14:31:39 (EST)

Reply to mark hutton: Your equation of velocity is

   v = 3 t² - 3t

Solving this for v = 0, you have

   0  =  3 t² - 3t  =  3t(t - 1)  :  t = 0  and  t = 1

Integrating the velocity equation to get position you find:

   x(t)  =  t³ - (3/2)t² + C

The boundary conditions require that x(0) = 4. So that means C = 4. Put the solutions for when the body is at rest into the position equation to find where it is at rest. Finally take the derivative of the velocity equation:

   a(t)  =  6t - 3

to find acceleration. Put t = 2 to get the last part.
Karl <Click to Send Email to Karl>
USA - Saturday, November 08, 2003 at 13:15:31 (EST)

Could someone please help me with this question? A particle P is moving along the x-axis with velocity v=3tsquared-3t. When t=0 x=4 find: a. the times when P is at rest. b. the position of P each time it is at rest. c. the acceleration when t=2. your help will be greatly appreciated.
mark hutton

nuneaton, UK - Wednesday, November 05, 2003 at 08:48:06 (EST)

*Note: this is calculus pertaining to Quantitative Methods for Business I need some hints as to how to structure this question, and how I can answer it, preferrably using Microsoft Excel spreadsheets with Solver function (I think graphing the problem should help):

A mining company wishes to connect four points lying on the same elevation. Places are referenced by a grid system where (a,b) is a point located "a" metres east and "b" metres north of a standard point. The four points are located at (100,400), (200,100), (500,400) and (600,200). The company wishes to minimize the total distance of the tunnels needed to connect the four locations.

Formulate this problem, and solve it using a spreadsheet Solver. Use the spreadsheet to calculate the ending coordinates of each segment of the tunnel system, and give the length of each segment, as well as the total length.

Thanks, Chuck
Chuck Smith

St. John's, NL Can - Monday, November 03, 2003 at 00:48:33 (EST)

Reply to Vincent Talent: These exercises are asking you to apply the various rules for combining derivatives. The first one is asking you to use the sum rule, which simply states that the derivative of a sum is the sum of the derivatives. The second is asking you to apply the product rule. The third is asking you to apply the quotient rule. Follow the links to learn more about these rules for combining derivatives, and then you should be able to do the examples you list yourself.

On the "what's wrong with..." example: What's wrong with the equation is that it is untrue for x = 1. At that x the left side of the equation is undefined because its denominator is zero. Look at the online pages about limits starting here to get a feel for what limits are all about.
Karl <Click to Send Email to Karl>
USA - Saturday, November 01, 2003 at 10:02:04 (EST)

sir, please send me the solution of following question plz as soon as possible.

·Express the following function in piecewise form without using absolute values.
f(x)=3|x-2|-|x+1|

mona

abu dhabi, united arab emirates - Friday, October 31, 2003 at 05:53:32 (EST)

I've been trying to do these concept-based questions, (but I think my concept isn't that sound).

"Suppose f'(2)=4, g'(2)=3, f(2)=-1 and g(2)=1. Find the derivative at 2 of each of the following functions
a. s(x)=f(x)+g(x)
b. p(x)=f(x)g(x)
c. q(x)=f(x)/g(x)"
I began doing this, without reading the find the derivative part. What order would I exactly solve it in? Or does it work straight in by plugging in the derivatives? (too simple, so must not be it)

"If f(x)=x, find f'(137)"
This is a pure concept question I'm sure...

"Explain what is wrong with the equation (x^2-1)/(x-1)=x+1, and why lim(x^2)/(x-1)=lim(x+1) both x->1"
The top factors out and supposedly cancels, though I'm not sure why I can't do that.

Vincent Talent

Athens, GA USA - Thursday, October 30, 2003 at 16:42:37 (EST)

sir, plz i am asking you the following maths notations so please solve them and send to me by mail.i shall be very thankful to you.

Q1:
Given a function y = 9 + cos(x+7) + square root of (x+7)
 , explain the graph of which (basic) function is translated horizontally
and vertically to obtain this (given) function and how much translation
in each direction?

mona

abu dhabi, united arab emirates - Thursday, October 30, 2003 at 05:52:45 (EST)

Reply to Daniel Morgan: There are lots of solutions to this. I'm going to suggest just one, but after you understand my approach, see if you can find a different one using the same method.

Since it must be piecewise, I'm going to choose as one piece the line that connects (1,20) with (2,4). That will be the part of the function when x < 2. For the other branch of the curve, I will do a set up that will find a polynomial that meets the critera given. Since it must have three inflection points, it must be a polynomial of at least 5th degree. That means we can have six constraints on it. Here they are. First it must pass through (2,4) to ensure continuity with the line. Second, its derivative at (2,4) must be equal to the slope of the line to ensure differentiability. Third through fifth, it must pass through the points (5,3), (6,2), and (10,1). Finally we must ensure that is has a critical point that is not one of the points given. I'll choose that it should have that critical point at x = 11. So if the polynomial is

   P(x)  =  Ax⁵ + Bx⁴ + Cx³ + Dx² + Ex + F

   P'(x) =  5Ax⁴ + 4Bx³ + 3Cx² + 2Dx + E

and we have the constraints (in the order listed above):

   P(2)  =  4

   P'(2) =  (20-4)/(1-2)  =  -16

   P(5)  =  3

   P(6)  =  2

   P(10) =  1

   P'(11) = 0

Expand the above equations using the numbers and the definitions given for P(x) and P'(x). This translates into six linear equations in the six unknowns, A through F. Solve for those coefficients using Gaussian elimination. Finally check the result to make sure that you have met the requirements of having three inflection points. If you have, you will nearly certainly also have at least one max and one min. If you don't have three inflection points, try a different choice for the outside critical point and do the problem again. You'll hit on a solution eventually.
Karl <Click to Send Email to Karl>
USA - Monday, October 27, 2003 at 19:23:12 (EST)

Hello to all, I have a question about our current lesson in my high school calculus class. We are given the following five points that lie on a function: (1,20); (2,4); (5,3); (6,2); and (10,1). I must find an equation that passes through these points and has the following features: There are three inflection points; there are at least one local max, and one local min; there is at least one critical point that is not a given point; the curve is continuous and differentiable throughout; and the equation is not a single polynomial, but must be a piecewise-defined function. I need to know how to do this because the teacher is evidently wanting us to find it on our own but we got nothing to go on and there were no examples given in class so I am searching for everything. thanks in advance for the help, Daniel Morgan
Daniel

douglasville, ga USA - Monday, October 27, 2003 at 17:58:26 (EST)

Hi Ben, you were correct in using partial fractions for the integral. Once you have the expanded fraction 1 + 1/(1 + e^x) - 4/(2 + e^x), then you can use the common integral solution for

  /
 |      1              ln(a + b e^x)      x
 |  ---------- dx = - --------------  + ---
 |   a + b e^x                 a          a
/

Hi Stephanie, I am not really sure of your question. E.g., if you mean x^(2*4/3) by "root3 over 4(x^2)", so I'll hesitate to answer anything since I'm not clear on your question. If you still need help, feel free to e-mail me.
Ian

Waterloo, ON CA - Friday, October 17, 2003 at 23:09:54 (EDT)

k this one's getting to me, i've seen some strange intergrals but this one's a toughy. Please help me out! we are suppose to use techniques of rationalizing substitutions (partial fractions, integration by parts, trig subs) The integral


              e^2x
 f (x)   =  --------
         e^2x + 3e^x + 2

I tried factoring the denominator into (e^x + 2)(e^x + 1) then using the rules of partial fractions but i couldn't get it to come out correct the answer is ln[(e^x + 2)^2/(e^x + 1)] + C Where are they getting the first squared term? I'm sure it has an easy solution im just not seeing it.
Ben
USA - Thursday, October 16, 2003 at 22:58:31 (EDT)

Correction...should read the root of 3 over 4
stephanie
KT, ON Canada - Monday, October 13, 2003 at 23:08:05 (EDT)

I have a problem...I am part way through a problem (maxima/minima) and have come to a road block...simple math/derivatives. I have the following and don't know how to solve for it... A= root3 over 4(x^2) + x(300x-3x^2/2) I need to simplify for A and then take the derivative dA/dx Please help show me the simplification steps...I'd appreciate it!
Stephanie

KT, ON Canada - Monday, October 13, 2003 at 22:25:17 (EDT)

Reply to Stephanie: I posted something last night but it had a mistake and I had to take it down. I don't get square root of -1/3, but when I solve for second derivative equal to zero, I do get the square root of a negative number. After taking the second derivative (and checking it with Mathematica), I found its numerator to be (after cancelling common factors):

   24(1 + 10x² + 5x⁴)

which does not have any positive solutions for x². So you have to conclude that there are no inflection points.
Karl <Click to Send Email to Karl>
USA - Wednesday, October 01, 2003 at 13:09:20 (EDT)

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