Karl's Calculus Forum: Jan 2002

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im just wondering what the answer to my problem is. I need to find a graph that is symmetrical about the x-axis that is a function

kevin 





 USA - Wednesday, January 30, 2002 at 21:41:32 (EST)








I have difficulty to solve this problem:



If P pounds pe square foot is the atmospheric pressure at a height of h feet above sea level, then P=2116e^(-0.0000318h). Find the time rate of change of the atmospheric pressure outside an airplane that is 5000 ft high and rising at the rate of 160 ft/sec.



At a certain height the gauge on the airplane indicates the atmospheric pressure is 1500 lb/ft^2. approximate by differentials how much higher the airplane must rise so that the pressure will be 1480 lb/ft^2.



Could someone help me to solve this problem?



Thanks.

madie 





kansas city, MO USA - Wednesday, January 30, 2002 at 03:21:01 (EST)








Reply to Q:

There are all kinds of relationships in nature where parameter A is given

as the derivative of parameter B. If you know A, then you need integration

to find B. Some examples: Power is the time derivative of energy.

Force is the space derivative of work. Entropy is defined only in terms

of its derivative -- the derivative of entropy is the the reciprocal

of absolute temperature times the derivative of heat content. There are

plenty of other examples too.


Also for finding averages over continuous regions you need integration.

This includes finding the center of mass of an object or the moment of

inertia of an object. It also includes finding areas and volumes of

various curved shapes.


Then there are all the problems where you find a function by adding up

contributions from tiny parts everywhere. Finding the pattern that monochromatic

light makes when it passes through a narrow slit is an example of this.


Here's sample problem a physics teacher gave my class once many years

ago. Suppose you have a sphere that is 6 million meters radius (the size of

the earth). Suppose you know the material it is made out of has a small amount

of radioactive minerals in it, and he gave the rate at which the radioactive

content produced energy. He also gave the heat capacity and conductivity of

the material the sphere was made out of. Our job was to find the temperature

at the center of the sphere. It was very hot (just like the earth). But you

couldn't even approach a problem like this without using integration.



Karl <Send Email to Karl>

 USA - Monday, January 28, 2002 at 13:37:59 (EST)








Use a scalar projection to show that the distance from a point P (x,y) to the line ax + by + c=0 is ax + by + cl/(a^2+b^2)  (where the l's should be the norm, and the a squared and the b squared in the denominator should be under a square root).  Use this formula to find the distance from the point (-2,3) to the line 3x -4y + 5= 0

I know the scalar projection formula is a dot b over the norm of a.  And since the norm of a is the distance formula I know to find the distance of certain variables in this formula, but I don't understand which variables to use in the distance formula.  In the answer, however, it is obvious that a and b were used as part of the distance formula.  Why?  And how come ax + by + c remains the same?  If you could help it would be great.  Thanks.  

Scott

 USA - Saturday, January 26, 2002 at 01:04:09 (EST)








How do you do a project on integration with applications of examples and solving it/

Q

 TX USA - Thursday, January 24, 2002 at 22:01:07 (EST)








Reply to Ahmad Alzoubi:

Although 1+tan[ax]/1-tan[bx]

has an integral, I am very nearly sure it cannot be expressed as a finite

combination of elementary functions. Just because you can write a function

does not necessarily mean you can write its integral.


The integral of 1+tan[ax]/1-tan[bx] though can be

done when a and b are each integers, but even this is

way too nasty a problem for most humans to tackle without making mistakes.

The result is a VERY long expression just taking a=2

and b=3 -- much too long to post even just the

answer here, much less the method.


As for integrating (x2-1)/(x4+1),

you must factor the denominator, then apply partial fractions to the two

quadratic factors you end up with. You can see the solution to a problem

that, at one stage, has the same denominator in

Worked Example 11.8-4.

Karl <Send Email to Karl>

 USA - Thursday, January 17, 2002 at 15:43:34 (EST)








would you integrate the following function



F(x)=1+tan[a^x]/1-tan[b^x]



   thanks.....

Ahmad Alzoubi 





Irbid, JORDAN - Wednesday, January 16, 2002 at 18:38:28 (EST)








would you please integratethe following function



F(x)=x^2-1/x^4+1

Ahmad Qaseem Alzoubi 





Irbid, JORDAN - Wednesday, January 16, 2002 at 18:30:09 (EST)








Reply to Tai Heng:

To integrate










   








 x arctan(x) dx









you have to apply integration by parts. Let




  u  =  arctan(x)


           dx
  du  =        
         1 + x2


  dv  =  x dx


  v  =  (1/2) x2





The integral becomes









   






1
 
2






 x2 arctan(x)












     1
  -   
     2






 





  x2 dx
       
 1 + x2








Now use polynomial long division to divide out the new integrand into

quotient and remainder expressions, then reform the integrand using

them and integrate (you'll still need to do trig substitution to integrate

one of the terms)







Karl <Send Email to Karl>

 USA - Wednesday, January 16, 2002 at 09:41:24 (EST)








How would I solve this problem ?

           x
          e  - 1 - x
  Lim     __________
x -> 0
              2
             x



Thanks !

CH 





 HK - Wednesday, January 16, 2002 at 05:03:26 (EST)








Hi, I'm having trouble integrating the function found below. I've gotten to the point where i've used integration by parts 4 levels down, and it keeps getting more complex.



Y=X ArcTan(X)



Any help at all would be much appreciated!

Tai Heng 





 Canada - Tuesday, January 15, 2002 at 21:02:26 (EST)








Reply to David P-T:

Your diff. eq. is




   dV        V
       =  -   
   dt       40





This equation is separable. That is you can, just by multiplying

and dividing, turn it into:




   dV       dt
       =  -   
    V       40





It's easy to integrate both sides of this to get




  ln(V)  =  -t/40 + C





From this you should be able to find V(t) in terms of t

and C. You know that V(0) = V0.

By putting t = 0 into your function, you should

be able to establish what C is in terms of V0.

See if you can finish this one based on that.

Karl <Send Email to Karl>

 USA - Tuesday, January 15, 2002 at 12:06:29 (EST)








Suppose that electricity is draining from a capacitator at a rate that is proportional to the voltage V(t) across its terminals and that, if t is measured in seconds, dV/dt=-V/40.  solve this equation for V, using Vo to denote the value when t=0.

David P-T 





Berkeley, CA USA - Tuesday, January 15, 2002 at 00:07:13 (EST)








Important! for Marty Rabens:

Your model will lead to a result that I thing will both surprise you and will

be not quite the effect you had in mind. Assuming the terrain extends outward

to infinity, you will find that the force on vehicles will be more or less

constant in an upward direction (assuming repulsion) independent of altitude.

The only exception your players will notice to this is when their altitude is

in the same ballpark as the roughness of the terrain. Near a mountain, for

example, you will feel a small side-load pushing you away from the mountain.

And in a valley that goes below the mean altitude of the terrain (that is a

valley with negative altitude, the vehicle will suddenly feel a downward

force with the same magnitude as the upward force when it was above the

mean terrain altitude. The transition of sign of the normal force will be

abrupt unless the valley is very broad.


It seems like a lot of computer power to add up all the forces from all those

triangles only to find that the force is nearly the same everywhere. But

consider what we feel on the earth. You feel very little gravitational effect

from mountains. As you increase altitude, gravity does decrease a little, but only

because the earth is curved. On a flat earth, gravity would be the same no

matter how high you are. The effect of having all the mass (or charge)

confined to an infinite surface is much the same, but with the interesting

reversal of sign when you go below sea-level (an effect you don't see even

on a flat but solid earth).


The rules of thumb are (and I remember this well from when I took E-Mag) is

that force is inversely as the square of distance from a point or spherical

source, inversely as of the first power of the distance from a source that

is an infinite line or cylinder, and is constant over a source that is an

infinite plane (provided you stay on just one side of the plane).

Karl <Send Email to Karl>

 USA - Friday, January 11, 2002 at 18:53:26 (EST)








Reply to Karl's reply to Marty Rabens


Wow, that was a quick reply!  Actually, I'm not taking any class (but I'm guessing E-mag means electromagnetism?).  I'm a game programmer, developing the physics for a racing game where the vehicles hover over the terrain through the use of "repulsors" mounted on the vehicle.  Since the terrain consists of a mesh of triangles, the physics engine needs to calculate the force each nearby terrain triangle exerts on each repulsor.  My calculus training was a long time back, so I'm pretty rusty.  I'll see if I can take it from here, with the hints you've given.  Thanks again for the speedy reply!

Marty Rabens

 USA - Thursday, January 10, 2002 at 15:25:34 (EST)








Reply to John:

Let f(x) = k cos(x) - kx2.

First find the points where f(x) is zero. This is equivalent

to solving




   cos(x) - x2  =  0





which you can do using Newton's method. When you know both solutions for x

that solve that, set up the definite integral of f(x) dx

from the lower solution to the higher solution. Evaluate that definite

integral. It will give you an expression that is a function of k.

Set that expression equal to 2, and then solve for k.

Karl <Send Email to Karl>

 USA - Thursday, January 10, 2002 at 13:13:18 (EST)








Reply to Marty Rabens:

Your taking an E-Mag class, aren't you?

You have to integrate each component of the force-vector function over the

entire area of the triangle.


First determine the equation of the plane in which the triangle lies. You'll

get something in the form of




  ax + by + cz = 1





unless the plane passes through the origin, in which case you get




  ax + by + cz = 0





Anyway, determine the expressions for a, b, and c

using linear algebra. You know that every point you have to integrate lies

along the plane given by the equation. The plane equation allows you to

eliminate one of your three variables (you pick which one, x,

y, or z), and end up doing the double integral with

just two variables. The hard part is coming up with the limits of integration.

Find the extremes of one of the remaining variables, say x. Then

for a given x, determine what the extremes are for the remaining

variable, say y. Both extremes for y will be linear functions

of x. The extremes for x become the integration limits

for x. The expressions for the extremes for y become

the integration limits for y.




You will have three of these double integrals to do, one for each of the

vector components. All three will have the same limits of integration.

See if you can take it from there.

Karl <Send Email to Karl>

 USA - Thursday, January 10, 2002 at 13:06:39 (EST)








(oops, forgot to escape the "<" character.  Let's try this post again.)



I need a formula to calculate a repelling force of a triangle against a point in 3-space.  The triangle ABC is defined by the points A, B, and C (each of which has <x,y,z> coordinates).  This triangle exerts a repelling force vector on a point at the origin.  The force it exerts is 1 / distance^2.  So if a one-unit square area within the triangle was located at <x,y,z>, the force this square exerts on the origin would be calculated as




                 -x
forcex =  -----------------
          (x2 + y2 + z2)3/2

                 -y
forcey =  -----------------
          (x2 + y2 + z2)3/2

                 -z
forcez =  -----------------
          (x2 + y2 + z2)3/2





because you're dividing each component by the distance cubed (divide by distance to normalize the vector, then divide by distance squared to get 1 / distance^2).



I need to integrate this over the area of the triangle so I end up with a sum vector of the repelling force the entire triangle exerts on the origin.  I'm not sure how to set up this integral (I know it will involve a double integral).



So, basicly, given the coordinates of A, B, and C, how do I find the repelling force vector at the origin?



Any help appreciated,





Marty Rabens

 USA - Thursday, January 10, 2002 at 12:32:42 (EST)








I need a formula to calculate a repelling force of a triangle against a point in 3-space.  The triangle ABC is defined by the points A, B, and C (each of which has  coordinates).  This triangle exerts a repelling force vector on a point at the origin.  The force it exerts is 1 / distance^2.  So if a one-unit square area within the triangle was located at , the force this square exerts on the origin would be calculated as




                 -x
forcex =  -----------------
          (x2 + y2 + z2)3/2

                 -y
forcey =  -----------------
          (x2 + y2 + z2)3/2

                 -z
forcez =  -----------------
          (x2 + y2 + z2)3/2





because you're dividing each component by the distance cubed (divide by distance to normalize the vector, then divide by distance squared to get 1 / distance^2).



I need to integrate this over the area of the triangle so I end up with a sum vector of the repelling force the entire triangle exerts on the origin.  I'm not sure how to set up this integral (I know it will involve a double integral).



So, basicly, given the coordinates of A, B, and C, how do I find the repelling force vector at the origin?



Any help appreciated,

    Marty



Marty Rabens

 USA - Thursday, January 10, 2002 at 12:29:00 (EST)








a)Find the positive value of k such that the area of the region enclosed between the graph of y = k cos x and the graph of y = kx^2 is 2.

b)Find the area of the regions enclosed by the curve: x + y^2 = 3 and 4x + y^2 = 0.



Thanks

John 





Los Angeles, CA USA - Wednesday, January 09, 2002 at 00:19:23 (EST)








a)Find the positive value of k such that the area of the region enclosed between the graph of y = k cos x and the graph of y = kx^2 is 2.

b)Find the area of the regions enclosed by the curve: x + y^2 = 3 and 4x + y^2 = 0.



Thanks

John 





Los Angeles, CA USA - Wednesday, January 09, 2002 at 00:18:28 (EST)








Reply to Allan:

To find areas and volumes the trick is to approximate the area (or volume) of

the shape in question using a series of shapes that you do know how to evaluate.

For example, the area of a circle can be approximated by breaking it up into

some number of pie-slices and pretending that each slice is a triangle

(Archimedes was the first to observe this 2300 years ago). The more slices

you have, the closer each slice is to actually being a triangle. So you take

the limit as the number of slices goes to infinity, and that yields the area

of the circle. All area and volume problems are done in a manner similar to

this, except the shape of the slices varies. The process of adding up the

slices and taking the limit as the number of them goes to infinity is called

integration. Integration is covered in online pages starting at

unit 10.

Karl <Send Email to Karl>

 USA - Tuesday, January 08, 2002 at 12:28:51 (EST)








Reply to Chery Dady:

If the rotor goes from 400 RPM to 280 RPM in 2.4 minutes, then its rate

of change is




   dr     280 - 400
       =            RPM/minute  =  - 50 RPM/minute
   dt        2.4





To find out how long it takes the rotor to spin down to zero RPM,

divide its initial rate (400 RPM) by dr/dt. The equation

for the rate, r(t), is




                  dr
   r(t)  =  400 +    t
                  dr





Now integrate r(t) dt from zero to the time it takes for

the rotor to spin down to zero. That will give you the total number

of revolutions.

Karl <Send Email to Karl>

 USA - Tuesday, January 08, 2002 at 12:16:52 (EST)








OK bear with me this is the first time I am doing this...so here it goes. I am having trouble with this problem:  "The figure below shows a lamp located 3 units to the right of the y-axis abd a shadow created by the eliptical region


 x2 + 4y2 less than or equal to 5 
 . If the point (-5,0) is on the edge of the shadow, how far above the x-axis is the lamp located?



T.

 USA - Sunday, January 06, 2002 at 19:54:14 (EST)








OK bear with me this is the first time I am doing this...so here it goes. I am having trouble with this problem:  "The figure below shows a lamp located 3 units to the right of the y-axis abd a shadow created by the eliptical region


 x2 + 4y2 
 . If the point (-5,0) is on the edge of the shadow, how far above the x-axis is the lamp located?



T.

 USA - Sunday, January 06, 2002 at 19:52:49 (EST)








hi,

how to calculate areas of triangle, circle, pyramids and cone using calculus?

allan 





las vegas , nv USA - Saturday, January 05, 2002 at 14:15:28 (EST)








Reply to Ali:  Your derivative should look something like:






dy/dx = [ (x + 1/2) / sqrt(x^2+x) ] - 1








...if you get something that is not equivalent, recall that:





d/dx[u^n] = nu^(n-1)(du/dx)








There's no particular need to factor this in order to set it equal to 0 and solve.  Consider:







[ (x + 1/2) / sqrt(x^2+x) ] - 1 = 0





[ (x + 1/2) / sqrt(x^2+x) ] = 1





sqrt(x^2 + x) = x + 1/2        ...cross-multiplied





x^2 + x = x^2 + x + 1/4        ...squared both sides





0 <> 1/4                       ...subtracted x^2 + x from both sides








Obviously 0 does not equal 1/4, but no need to panic.  This just means the derivative never equals 0 because there was no solution to the equation (dy/dx)=0.  BUT don't forget when you're looking for critical points, there are two rocks you need to look under:





1st rock.  Where the derivative is zero





2nd rock.  Where the derivative is undefined (but the original function is defined.)






The derivative is undefined where:





x^2 + x =< 0   ...can you see why?






Algebra review---to find where this occurs, first solve the equation:





x^2 + x = 0





x = -1, 0





These numbers divide a number line into three intervals:





<---------(-1)-------(0)---------->








Test a number in each interval to see if x^2+x is <0 there:





1.  (-2)^2 + (-2) > 0   ...so this interval is not part of the solution





2. (-1/2)^2 + (-1/2) < 0 ...so this interval is part of the solution





3.  (1)^2 + 1 > 0     ...so this interval is not part of the solution






So the derivative is undefined where:





-1 =< x =< 0





However...the original function is also undefined on the open interval:





-1 < x < 0





So anything in the "open" interval are not critical values.  BUT, the original function is defined where x=-1 and x=0 (the endpoints of that interval) and the derivative is undefined there.  Therefore, -1 and 0 are critical values for this function.  To find the critical *points*, plug in -1 and 0 into the original function to get the y-coordinates.  

Darrell Ryan 





 USA - Saturday, January 05, 2002 at 06:46:56 (EST)








When I differentiate this equation y=(((x^2)+x)^0.5)-x I get an unfactored form, the trouble is that, how do I factor it to get the critical points on the graph?

Ali Akhtar 





Etobicoke, On Canada - Friday, January 04, 2002 at 21:22:40 (EST)








A lelicopter rotor at 400 revolutions per minute and begins to slow down at a constant rate.After 2.4 minutes it is rotating at 280 revolutions per minute.How many revolutions does the rotor make from it is rotating at 400 revolutions per minute to when it stops?



 Help me to resolve this exercice

Chery Dady 





Brooklyn, NY USA - Friday, January 04, 2002 at 19:24:16 (EST)








1)The Seattle ferris wheel is 50 feet in diameter and has a ground level loading platform that raises the wheel 10 feet above the ground .It makes one revolution every 15 seconds in the counterclockwise direction. assume that you start at the platform level(6 O'clock position)at time t =0.Write your height above ground,h(t),as a trigonometric function of time.

2) The value in dollars of a U.S. Treasury bond is a complicated function of its yield.The derivative of the bond value with respect to yield is called dollar-duration.Today you buy bond  for $945,000.Suppose its dollar-duration is -500,000.if tommorrow the yield drops by 0.10 and you sell the bond ,will you have lost money or made money? Approximately how much?





Please can you help me to resolve those exercices

Chery Dady 





Brooklyn, NY USA - Friday, January 04, 2002 at 19:15:43 (EST)








My first entry didn't make it correctly, so I will try ONE more time (sorry).

I am having difficulty with the following motion problem:

Given:

        acceleration = t - 1 + 6

                               -

                               t



find: (a) a formula for v when v(1) = 1.5 and (b) max value for v on the

interval [1,9]. For (a) I have no difficulty using the indefinite integral

of acceleration followed by using v = 1.5 to give the velocity formula:



           (t2  - t  + 6 ln |2| + 2

            ------------

                 2



but I can't find part (b) the max velocity over the given interval. I

put acceleration = 0 and have tried unsuccessfully to solve for t. TX.

Ralph Moskowitz 





w, CA USA - Friday, January 04, 2002 at 18:04:21 (EST)








I am having difficulty with the following motion problem: given



         acceleration  =  t - 1 +  6

                                  ----

                                   t



find: (a) a formula for v when  v(1) =  1.5



      (b) max value for v on interval [1,9].



For (a), I have no difficulty using the indefinite integral of acceleration



followed by using v = 1.5 to obtain the formula for velocity:



        

                 (t2)  - t + 6 ln |2| + 2
                 ---------------
                       2



but I can't find part (b) the max velocity over the given interval.  I have



put acceleration = 0 and have tried to solve for t unsuccessfully.  Comments?



                

Ralph Moskowitz 





Walnut Creek, CA USA - Friday, January 04, 2002 at 17:44:58 (EST)








Reply to Jim Quigley:
To integrate





   1
   
   0




 lnn(x) xs dx





you have to apply integration by parts. Let


  u = lnn(x)

  du = n (lnn-1(x) / x) dx

  dv = xs dx

  v = (1/(s+1))xs+1



The integral becomes






   





 1
   
s+1




 lnn(x) xs+1




1

0




      n
  -     
     s+1




1

0




 lnn-1(x) xs dx





Notice the similarity between the integrand here and the original integrand.
Notice in particular that you still have xs, but
the exponent of the log is reduced by one. Clearly you have to apply
integration by parts again and again until the exponent of the log
gets down to 1. Then you apply it one more time to get the final solution.
You will end up with a sum of terms taken from x=0 to x=1. They will all
have a xs+1 factor, which makes them all zero when
evaluated at x=0 (note that you have to apply L'Hopital's rule
n times to establish that the limit as x goes to
zero of xs+1 lnn(x) is zero
whenever n is a positive integer and s is positive).
All of the terms, except the last,
will have a lnk(x)
term in them, for some positive integer, k. That means that
all of them except the last will be zero when x=1. So you only
have to worry about the last term evaluated at x=1. Once you have
established what that is, you will easily be able to prove the recurrence
identity asked for in the problem.

Just so that you will see the right path as you develop the sum in
question here, observe that the signs of the terms in the sum will
alternate. For the kth term, you should see that it has
a constant factor of
n!/(n+1-k)!. Can you see also that the kth term will
also have a factor of (s+1)-k?


Karl <Send Email to Karl>

 USA - Wednesday, January 02, 2002 at 17:48:16 (EST)








My question is this: If J(sub n)= the integral from 0 to 1 of (ln X)^nx^sdx, where s>0 and n is a non negative integer, show that (s+1)J(sub n) + nJ(sub n-1)= 0, and hence find J(sub n).

Thanks

Jim Quigley 





toronto, on Canada - Sunday, December 23, 2001 at 17:26:30 (EST)





    










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