Karl's Calculus Forum

                Δv2
        (Vm2 + ---------- ) * m * Vm
                   2
Δe(kw) =------------------------------------
                Δs

f(x) = ax3 + bx2 + cx + d

f(x)=√(1+kcos(x))

I've tried the route of substituting x=arccos(w), x=arccos(w/k) and various mungeing ideas. I've tried expressing the resultant as logs, but no matter what twists and turns I try, I can't get it to fall out. If anyone has any clever ideas I would be very grateful to hear from you.

Many thanks,
Mark
Mark

Use the identity:

   bx = ex ln(b)

   eln(x) ln(x)  =  e(ln(x))2

   F(x3) - F(1)

f(x)=I(√(1+t3))dt from 1 to x3 

f(x)= xln(x)

I =    x
     -----
     1 + x

     x + 1 - 1  (*add and subtract 1*)
  =  --------
     1 + x

     x + 1     - 1
  =  -----    -----
     1 + x    1 + x


     1     - 1
  =        -----
           1 + x

 "Observe that x/(1+x) = 1 - 1/(x+1)"

Perhaps the material you were looking at referred to doing this integral in the domain of complex numbers. There you can choose your path of integration arbitrarily between any two points you choose as limits of the integration. So, for example, you could integrate this from -1 to 1, which would include x = 0. But in the complex domain you can detour around the troublesome point (which is called a pole of the function). Your detour can come arbitrarily close to the pole, but it must bypass it either to one side or the other. Depending upon which side you bypass it, this will add a term of ±πi, where i is the square root of -1.
Karl <Click to Send Email to Karl>
USA - Friday, July 21, 2006 at 13:04:02 (EDT)

F(x) = x2 + 2x
       ------------------
       √3x+1

Without using the Chain Rule or the Difference Quotient

I hope you can help:)

Thanks

Angie








 MN USA - Wednesday, May 24, 2006 at 16:26:13 (EDT)

d/dy(ex2)

y = x2ex2.

x2 × (d/dy(ex2) + ex2 × 2x

2xex2(x2 + 1)

x2/a2 + y2/b2 = 1

- about the x-axis
- about the y-axis

x>=0, y>=0 (I think, if I remember the problem correctly!)

(L-x)2 A = + 16

x2 4

So taking the derivative

dA 2(L-x) = + dx 16

2x = 0 4

Note that I have set this to zero in order to find the optimum. Solve for x.
Karl <Click to Send Email to Karl>
USA - Sunday, April 16, 2006 at 14:31:03 (EDT)

If you have

r = R -

θR 2π

then you can substitute the right-hand expression into the volume equation everywhere you see r. You should end up with a volume equation in which the only independent variable is θ. The symbol, R, will still be there, but it, remember, is a constant. Now find dV/dθ by taking the derivative of the volume equation with respect to θ. Whatever experssion you get for dV/dθ, set it equal to zero. Now you have an equation in which the only unknown is θ. Solve for θ and you're done.
Karl <Click to Send Email to Karl>
USA - Saturday, April 08, 2006 at 11:16:12 (EDT)

r=R-((θ R)/2π)

   R2  =  r2 + h2

   h = √R² - r²

If the wedge you cut out of the paper is angle θ radians, then the length of circumference removed is θR. That means that what's left is 2πR - θR. This will be the circumference of the base of the cone (picture in your mind what happens when you form the cone from the paper). Dividing that circumference by 2π gives the radius of the base of the cone, r.

r = R -

θR 2π

The volume of a cone is given by

V =

1 3

π r2h

Use the relationships already shown to first substitute for h using r, then to substitute for r using θ. Now take the derivative, dV/dθ, set it to zero, and solve for θ (remembering that the radius, R of the original filter paper is a constant).
Karl <Click to Send Email to Karl>
USA - Friday, April 07, 2006 at 13:25:11 (EDT)

  dV
      =  0.1 ft3/min
  dt

dV dt

= 4 π r2

dr dt

You can find r by using the volume equation replacing V with 1/2 ft³ and solving for r. You know dV/dt from the first equation. Solve for dr/dt.

2) The volume of a cylinder is V = π r²h. The radius, is given as 20 feet. Again dV/dt is given. Taking the derivative of the volume equation, assuming fixed radius:

dV dT

= pi r2

dh dt

Solve for dh/dt.

3) The volume of the cylinder is, again, V = π r²h. Note that V is given, so you can use that to find h in terms of r. The cost function is (two circles of radius, r, at $10 per square meter plus a rectangle that is 2πr by h at $8 mer square meter).

  C  =  $10 2π r2  +  $8 2π rh

  2p/q  =  7

  2p  =  7q

   y = √r² - x²

   V  =  2π x2 √r² - x²

   0  = 4π x √r² - x²  -  2π x3 / √r² - x²

   0  = 2x(r2 - x2) - x3

2int(f(x)*f(x)) between 0 and 2, instead of (x*f(x))

Still need assistance with the second one however.
Tony

http://apcentral.collegeboard.com/repository/05836apcoursdesccalc0_4313.pdf

First is #22. The base of a solid S is a semicircular region enclosed by the graph of y=sqrt(4 - x²) and the x-axis. If the corss sections of S perpendiculr to the x-axis are squares, then the volume os S is?

I set this up as 2*int(x*f(x)dx) between 0 and 2. I keep getting half of what they get.

Next is Question 24.

                pi*ex
If f'(x) = sin( ---- ) and f(0)=1, then f(2) =
                  2

             dx
       --------------
         [A-x][B-x]

    1         [B - x][A]
--------- ln -------------
[B] - [A]     [A - x][B]

    [A - x]
ln -------------
    [B - x]

             1  dx
integrate -----------
          (x2 + a2)3/2

             1  dx
integrate -----------
          (x2 + a2)3/2

             1  dx
integrate -----------
          (x2 + a2)3/2

V = 24π inches3 = πr2h

From this you can easily get h in terms of r:

h =

V πr2

Your cost function is proportional to (and we'll just say it's equal to) the area of the sides (2πrh) plus three times the area of the bottom (3πr²).

C = π (2rh + 3r2)

Now replace h with its equivalent from the constraint equation:

C = π

2rV + πr2

3r2

=

2V + r

3πr2

Now you find dC/dr of this, set it to zero, and solve for r.

dC = 0 = - dr

2V + r2

6πr

Substituting 24π inches³ for V (as that's what volume is given in the problem), applying the commutative law, and multiplying through by r²:

0 = 6πr3 - 48π inches3

which you can solve easily for r. Back-substitute the solution for r into the constraint equation to get h.
Karl <Click to Send Email to Karl>
USA - Tuesday, January 10, 2006 at 12:43:47 (EST)

Couldn't get that mad email link to work so here is a temp mail of mine: bromfieldcourtpathway@hotmail.com

Cheers!
Mason
London, UK - Tuesday, January 03, 2006 at 19:13:03 (EST)

             Q2
            ----
             10

             Q
            ----
             5

Best Regards, Mason
Mason
London, UK - Tuesday, January 03, 2006 at 18:57:33 (EST)

(H4+3H2+1) / (4H4 + 2H2 +-1)

I multiplied it by H^-4, and I then found the standard part to be 1/4. Am I correct?
Bill Awesome

2π

r ds

In your problem, r = f(y) (because you are rotating about the y-axis), and ds = √1 + (f'(y))² dy. Since you have f(y) = sin(y), it means that (f'(y))² = cos²(y). So the integral is

2π

π 0

sin(y) √1 + cos²(y) dy

Let u = cos(y). Then du = -sin(y) dy, and the integral becomes

-2π

-1 1

√1 + u² du

which can be integrated by substituting tan(v) = u, or by hyperbolic subsitution.
Karl <Click to Send Email to Karl>
USA - Sunday, December 18, 2005 at 18:25:06 (EST)

  s2  =  (x1 - x2)2 + (y1 - y2)2

         _______________________
  s  =  √(x1 - x2)2 + (y1 - y2)2

lim x → ∞

3x - 1 x - 2

by the rule I indicated you should get a limit of 3. You can see why if you substitute 1/u = x and take the limit as u goes to zero.

lim u → 0

3/u - 1 1/u - 2

Now multiply top and bottom by u.

lim u → 0

3 - u 1 - 2u

You can see that as u goes to zero, this limit goes to 3. In the other problem where you have the limit as x goes to infinity of a ratio of two fourth degree polynomials, try the same thing. Replace x with 1/u and take the limit as u goes to zero. In this case you will multiply top and bottom by u^{4 to simplify the limit.
Karl <Click to Send Email to Karl>
USA - Tuesday, December 13, 2005 at 18:59:07 (EST)}

Your heat equation is looking for a function, T(x,t), which makes it an equation in time and in one spacial dimension. So it could be a staight bar, or as you indicate, a slab where temperature is uniform over the areas of each of the two surfaces.

∂T = A ∂t

∂2T ∂x2

First thing -- before applying boundary conditions -- is to get a general solution to this. We do this by imagining that the solution is the product of two functions, one purely a function of time, the other purely a function of space. So we let T(x,t) = F(x)G(t), and apply the partial differential equation to the product.

  F(x)G'(t)  =  A F"(x)G(t)

  G'(t)  =  -Aμ G(t)

  -μF(x)  =  F"(x)

The rest of the problem is to apply the initial condition and boundary conditions to establish the eigenvalues that work for ω and the weights, M and N for each eigenvalue. With a heat equation we do the latter by applying the boundary conditions to establish the steady-state solution, T(x, ∞). The steady state solution has the property of ∂T/∂t = 0, which implies that ∂²T/∂x² = 0. This means that T(x, ∞) must be a linear function of x.

   T(x, ∞)  =  mx + b

m =

K 1-K

Now we find the eigenvalues, eigenfunctions, and their weights. We do this by taking the difference between the initial condition and the steady state solution and extracting a Fourier series from it.

T(x, 0) - T(x, ∞) =

K x K-1

The left-hand boundary condition requires that we use only sin(ωx) terms to ensure that the eigenfunctions are zero at the left boundary. The right-hand boundary condition requires that ω be odd multiples of π/2 to ensure that the derivative of the eigenfunctions are zero at the right boundary. So ω_i = (2i+1)π/2 are the eigenvalues, with i being integers from zero to infinity. The eigenfunctions are

  Ei(x)  =  sin(((2i+1)π/2)x)

The full solution is

T(x, t) = T(x, ∞) +

∞ ∑ i=0

Ni e-Aωi2t sin(ωix)

Karl <Click to Send Email to Karl>
USA - Friday, December 02, 2005 at 09:13:04 (EST)

  dT       d2T
 ---- = A  ----
  dt       dx2

T=1 at t=0 for all x

T=1 at x=0 for t>0

  dT
 ---- = KT    at x=1  for t>0
  dx

lim x → 0+

x = |sin(x)|

lim x → 0

x sin(x)

but from below with x < 0

lim x → 0-

x = |sin(x)|

lim x → 0

x -sin(x)

Karl <Click to Send Email to Karl>
USA - Wednesday, November 30, 2005 at 17:58:17 (EST)

lim(x--->0) x/abs(sin(x))

x/sin(x)

Your function can be simplified using identities and algebra to make it much easier to differentiate or integrate. First identity is ln(√b) = (1/2)ln(b). So

ln(√x² - 1) =

1 2

ln(x2 - 1)

You can also factor x² - 1 = (x + 1)(x - 1). And then you can apply ln(ab) = ln(a) + ln(b). So

1 2

ln(x2 - 1) =

1 2

ln(x + 1) +

1 2

ln(x - 1)

I'll leave it to you to integrate or differentiate this on your own because with the function in this form, either operation is easy.
Karl <Click to Send Email to Karl>
USA - Wednesday, November 30, 2005 at 08:56:14 (EST)

If you want to find the integral

x3(5x2 + 1)-2/3 dx

then substitute u = 5x² + 1. Hence du = 10x dx or equivalently du/10 = x dx. Also you have (u-1)/5 = x². Making the substitution, the integral becomes: