Karl's Calculus Forum: Summer/Fall 2005

(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:

1 50

(u - 1)u-2/3 du

Make sure you see how the substitution of the original leads to this.

Multiply the integrand out using the distributive law and integrate it term-by-term using the power rule. Then back substitute and you're done.
Karl <Click to Send Email to Karl>
USA - Sunday, November 27, 2005 at 16:13:10 (EST)

√1 + x + x² - √1 - x + x²

√1 + x + x² + √1 - x + x² = √1 + x + x² + √1 - x + x²

1 + x + x2 - 1 + x - x2 = √1 + x + x² + √1 - x + x²

2x √1 + x + x² + √1 - x + x²

In the denominator as x gets negatively huge, the x² dominates the other terms that are under the radical. So as x goes to minus infinity, the denominator looks more and more like

2x = √x² + √x²

2x = -1 for x < 0 |2x|

Karl <Click to Send Email to Karl>
USA - Sunday, November 20, 2005 at 20:00:49 (EST)

            
   lim(x-->-infinity)  (√(1+x+x2)-√(1-x+x2))
             

y' =

1 2√x

Let x_t be the point of tangency. Since you know that the y intercept of the tangent line is 1, then the equation of the line is y = mx + 1. From the fact that the slope of the tangent must match the derivative of the function at the point of tangency, you know that

m =

1 2√xt

In the equation of the tangent line we replace m with the expression above for m, x with x_t, and y with the expression for y(x_t). This way we solve for the point of tangency, x_t.

__ √xt =

1 xt + 1 2√xt

By simplifying the above and solving, you find that x_t = 4. Back-substitute x_t to find m.
Karl <Click to Send Email to Karl>
USA - Thursday, November 17, 2005 at 16:16:45 (EST)

   y - k  =  -(4/3)(x - h)

tan(π/2 - x) =

sin(π/2 - x) = cos(π/2 - x)

cos(x) = sin(x)

1 tan(x)

So if you if you know that y = tan(x), then 1/y = tan(π/2 - x). Taking the arctangent of both of those equations: arctan(y) = x and arctan(1/y) = π/2 - x. What happens if you add the two equations together?
Karl <Click to Send Email to Karl>
USA - Wednesday, November 09, 2005 at 13:06:49 (EST)

By proving that  


f(x) = tan -1 x + tan -1 (1/x), x > 0, 

is a constant function, find the value of the constant C such that 

C = tan -1 x + tan -1 (1/x)

Joseph








Winnipeg, mb canada - Tuesday, November 08, 2005 at 13:45:19 (EST)

  y" + y(y')3  =  0

1 2	y2 + C dy = dt

you should accrue an additional constant and get

1 6	y3 + Cy + D = t

Finally to your question about getting the solution as the inverse function to a cubic. If you really need y as a function of t, you can subtract t from both sides of the solution, then use the cubic formula to solve for y as a function of t. But if all you need to do is verify your solution, then take the derivative of the solution above:

1 2	y2y' + Cy' = 1

Now divide both sides by y' and take the derivative again (note that we divide both sides by y' because that will eliminate the remaining integration constant, C, when we differentiate the second time -- you need to eliminate both constants in the verification process):

1 2

y2 + C =

1 y'

Differentiating:

              y"
  yy'  =  -      
            (y')2

             dy
let v = y' = --
             dt
                dv
from that y'' = --
                dt

therefore:

dv
-- + y(v3) = 0
dt

by the chain rule:

dy     dv    dv      dv
--  ×  -- =  -- =  v --
dt     dy    dt      dy     


so the problem becomes:

  dv
v --  + y(v3) = 0
  dy

dv
--  + y(v2) = 0 {divide across by v}
dy

dv
--  = - y(v2) {move y(v2) to RHS}
dy


int v(-2) dv  = - int y dy {seperate and get ready to integrate}

-v(-1) = - 0.5 y2 + C

v = (0.5 y2 + C)-1 {multiply each side by -1 and invert}

dy
-- = (0.5 y2 + C)-1 {from letting v = dy/dt earlier}
dt

int (0.5 y2 + C) dy = int dt {seperate and get ready to integrate}

1 
- y3 + Cy = t
6

Solved for t but not for y and I can not see how to find what is asked for

m =	y - 50 = y'(x) = 0.02x x + 100

This is because the tangent line must pass through both the point of tangency, (x,y), and through the point (-100, 50). Since the derivative of the parabola at the tangent point must be equal to the slope of the line, you have m = y'(x) = 0.02x.

You also have from the original equation, y = 0.01 x², which you can solve simultaneously with the above to find the coordinates of the point of tangency, x and y. Do the same for the other endpoint of the wall as well. You will have solved for two points of tangency. Call them, (x₁,y₁) and (x₂,y₂).

You are interested in the distance the car travels between those two points. To find that use the length integral:

s =

x2 x1

√1 + (y'(x))² dx =

x2 x1

√1 + 0.0004x² dx

Now suppose I look at all the solutions I can get with x(0) = 0. They will be in the form of x = A(pf(t) + qg(t)), where A is an arbitrary constant, but the ratio of p to q is established by the constraint that x(0) = 0. Likewise you can come up with the solution set for v(0) = 0 of x = B(rf(t) + sg(t)), where B is arbitrary and the ration of r to s is established by v(0) = 0.

With all that in place, you can now think of the solution to some boundary condition problem that requires a specific value (nonzero) for x(0) and some specific nonzero value for v(0) in this way: The amount that the v(0) boundary condition must be is in proportion to how much of the x = A(pf(t) + qg(t)) solution I need -- that is it establishes A. That is your v(0) contributes A amount of this solution. Likewise the x(0) boundary condition contributes B amount of the solution, x = B(rf(t) + sg(t)).

So nature isn't solving the linear combination, it is simply contributing the two functions -- one that is characteristic of one boundary condition and the other that is charateristic of the other boundary condition -- in proportion to how much of each boundary condition you have.
Karl <Click to Send Email to Karl>
USA - Monday, October 10, 2005 at 11:48:20 (EDT)

   p''(t) + u*p'(t) + K*p(t) = 0

   p''(t) + 5*p'(t) = 0

p1(t) = C1*exp(-5*t) and p2(t) = C2.  Good.

p1(t) = C1*exp(-4.8*t) and p2(t) = C2*exp(-0.2*t).

  f(x)  =  Ax2 + Bx + C

  A  +  B  +  C  =  -1

  A  -  B  +  C  =   3

  9A + 3B  +  C  =   3

   p''(t) = -K/p(t)2

   p''(t)p'(t)  =  -Kp'(t)/p(t)2

1 2

p'(t)2 =

K + C p(t)

Choose C to satisfy your initial condition for p'(t).

Multiply through by 2, take the square root of both sides, then solve by separation of variables.
Karl <Click to Send Email to Karl>
USA - Monday, September 26, 2005 at 12:30:00 (EDT)

p''(t) = -K/p(t)2

v =

ds dt

Now multiply that equation by

dv = a dt

You get

v

dv = dt

a

ds dt

Now multiply through by dt and you get the differential result shown in your dynamics book.
Karl <Click to Send Email to Karl>
USA - Monday, September 05, 2005 at 10:49:36 (EDT)

v dv = a ds

    dv   d2s
a = -- = --- 
    dt   dt2

    ds
v = --
    dt

   Za + Zb  =  a + c

   Za - a  =  c - Zb

   a(Z - 1)  =  c - Zb

a =	c - Zb Z - 1

You can solve for b or c in a similar way.

Note that Z will be independent of a only when b = c, in which case Z is always equal to 1.
Karl <Click to Send Email to Karl>
USA - Wednesday, August 24, 2005 at 12:22:33 (EDT)

    (a + c)
Z = -------
    (a + b)

dy = y2 - 1

dx x

Integrating both sides gives

   arctanh(y)  =  ln(x) + C

   y  =  tanh(ln(x) + C)

y =	Ax - 1/(Ax) Ax + 1/(Ax)

or equivalently

y =	A2x2 - 1 A2x2 + 1

Put your boundary conditions into this and I think you'll find that you can solve for A. The solution is disappointing, but it does work in the original equation and does solve the boundary condition.
Karl <Click to Send Email to Karl>
USA - Friday, August 19, 2005 at 13:21:30 (EDT)

   dy    y2-1
   --- = -----
   dx     x