Karl's Calculus Forum: March/April 2005

2x dx +

(1/2)x-1/2 dx +

(2/3)(1/x) dx

Remembering that the integral of 1/x is ln|x|, you should be able to do this using the power rule. Once you have the indefinite integral by doing this (and don't forget the constant), find the definite integral (part b) by putting 4 and 2 into the indefinite integral and subtracting the second from the first.

The second one is done by substitution. Observe that the factor, x, is a constant multiple of the derivative of the term, (x² + 1). So let u = x² + 1. Then du = 2x dx or equivalently du/2 = x dx. Now you can substitute both the (x² + 1) and the x dx. The integral becomes

(3/2)u3 du

which you can do using the power rule. Once you've done that, back substitute for u to get the indefinite integral in terms of x. Then apply the limits in the same way as in the first one to get the definite integral.
Karl <Click to Send Email to Karl>
USA - Saturday, April 30, 2005 at 16:20:22 (EDT)

2 1

x2 dx

for example, you have only one choice of line to integrate along, and that is the real number line. You can think of this example as breaking the path of integration, which is the segment of the real numbers, [1,2], into tiny vectors pointing left to right, each of them dx long. Multiply each vector times the value of the function where the vector starts, and add up all the results. If the resulting vector sum points right (which it does in the example) then the answer is positive; to the left and the result is negative.

Now what happens if you are not restricted to moving along a single line? Suppose you can move anywhere in a plane.

(0,2) (0,1)

f(x,y) ds along a specified path

Then the path from (1,0) to (2,0) is not restricted. There are an infinitude of paths you could choose from. For example the path could start at (1,0), then wander deep into any of the quadrants, neander about, and then come back and end up at (2,0). But you do the same thing. You break the path up into infinitesimal vectors, ds, each pointing tangent to the path. Multiply each of the vectors times the value of the function, f(x,y), and add up the results. You can see that this is the same idea, but in two dimensions instead of one. If the function, f(x,y) were, for a physical example, the amount of frictional force at each point on the plane, the line integral would be the total amount of work to overcome friction it would take to traverse that path.

Here's a good physical example of a surface integral. Suppose you had a flexible sheet of solar cell material. You lay it on an uneven ground, and it forms a curved surface. Because of this, the sun does not shine equally on all parts of the surface. But if you took the surface integral of the sunshine (which is directional and is thus a vector quantity) dotted with the dA vector, which is normal to the surface at every point and added all of them up, the total would be the amount of solar energy falling on the surface.
Karl <Click to Send Email to Karl>
USA - Saturday, April 30, 2005 at 16:05:17 (EDT)

On the second one, convert everything to quotients of sines and cosines. Then cancel what you can. You will see that you get something that is much easier to integrate.
Karl <Click to Send Email to Karl>
USA - Friday, April 22, 2005 at 17:43:47 (EDT)

             (x/x) - ln(x)
   f'(x)  =              
                  x2

So f'(x) = g(x)/h(x) where g(x) = 1 - ln(x) and h(x) = x². So g'(x) = -1/x and h'(x) = 2x. Applying the quotient rule again:

             -x2/x - (1 - ln(x))(2x)
   f"(x)  =                          
                        x4

            ln x
  f(x) =   ------
              x

   A_actual = s2 ± 2sε + ε2

ε2 ± 2sε × 100 s2

Replacing ε with 0.02s you get

0.0004s2 ± 0.04s2 × 100 = s2

(0.04 ± 4) % = -3.96% to 4.04%

That indicates that the possible error on the low side of predicted area is less than the possible error on the high side (by a little bit). But if, as you ask, you want only the approximate error, ±4% gets it quite nicely. That was the algebra way to do it.

The calculus way to do it is to observe that A = s², hence

   dA
       =  2s
   ds

Now use the contraint equation to solve for L in terms of W (or vice versa), then make the appropriate substitution into the cost equation (either for L or for W) to eliminate one of the independent variables. Then find the derivative of cost, C, with respect to the remaining independent variable, set that derivative to zero, and solve for the remaining independent variable. Back-substitute to find the other independent variable.
Karl <Click to Send Email to Karl>
USA - Tuesday, April 05, 2005 at 12:44:35 (EDT)

        √(a+b-c)(a+c-b)(b+c-a)(a+b+c)
  A  =                               
                      4

g = (1,1,1)

  ∂A
      =  L
  ∂a

Without going to all that trouble, the symmetry of Heron's formula makes it pretty clear that you will end up with a = b = c when you solve for all the partials. But go ahead and do it all the way through so that you see how the method works.
Karl <Click to Send Email to Karl>
USA - Monday, April 04, 2005 at 19:07:52 (EDT)

The equation of the one of the two sloped lines is y = mx, where m = h/R (where R is the radius of the cone. I will use little r as the radial coordinate). So your cylindrical limits will have z go from mr to h and r go from 0 to R. Of course θ goes from 0 to 2π.
Karl <Click to Send Email to Karl>
USA - Monday, April 04, 2005 at 18:37:08 (EDT)

dz = dt

∂z dx = + ∂x dt

∂z dy ∂y dt

is called the chain rule for partial derivatives.

As you indicated, it derives from:

δz =

∂z δx + ∂x

∂z δy ∂y

The way to understand this is to remember that in ordinary derivatives, if you have y = f(x), then

δy =

dy δx dx

This rule imagines a tangent line to y = f(x) and says that if you go δx horizontally from the point of tangency, then the curve will change by the same amount as the tangent line -- that is by δx times the slope of the tangent line.

Now lets go to the case of z = f(x,y). You can imagine this function to be a curved surface of some sort -- much like a terrain surface. You are standing at some point on the terrain. Imagine a tangent plane to where you are standing. If you move δx east, then your change in altitude along the curved surface will be the same as if you followed the tangent plane. If you resolve the slope of that plane in the easterly direction, then your change in altitude will be δx times that slope. Likewise if you moved north by δy, you would resolve the slope of the plane in the northerly direction and compute your change in altitude accordingly. These resolutions of the slope of the plane in the easterly and northerly direction are exactly what the partial derivatives, ∂z/∂x and ∂z/∂y, represent. Now what happens if you move δx to the east and δy to the north? Your change in altitude is as if you followed the tangent plane in that combined direction, and that will be the sum of the two changes resolved individually in the easterly and northerly directions. And that is the essence of the rule.
Karl <Click to Send Email to Karl>
USA - Saturday, April 02, 2005 at 09:53:15 (EST)

                       ∂z      ∂z
if z= f(x,y) then δz = --- δx + --- δy
                       ∂x      ∂y

divide both sides by δt

δz    ∂z  δx    ∂z  δy
--- = --- --- +  --- ---
δt    ∂x  δt    ∂y  δt


then if δt -> 0,


δz     dz    δx     dx   δy     dy
--- -> --,   --- -> --,  --- -> --
δt     dt    δt     dt   δt     dt

but the partial differential coefficients, which do not contain δt
remain unchanged and the result is:

dz    ∂z  dx    ∂z  dy
--- = --- --- +  --- ---
dt    ∂x  dt    ∂y  dt

∞ 0	du (1 + u2)(1 + u√2)

Now let p and q be positive integers, and replace √2 in the exponent with p/q. Then this becomes pretty straightforward to integrate. Let v^q = u and q v^q-1 dv = du. Integral becomes:

∞ 0	q vq-1 dv (1 + v2q)(1 + vp)

Both the factors in the denominator can be factored further in the complex numbers based upon the roots of unity. All the roots lie on the unit circle. The trick here is to come up with the formula for the partial fractions that result with those roots -- decomposing this thing into a sum of pole-residues. If you experiment with some small values of p and q which are easily manageable, you find that the integral always comes out to π/4. So what you need to do is prove that given the way this thing breaks into the sum of functions, C_k/(x - r_k), where the r_k's lie on the unit circle and the C_k's derive from the partial fractions, the sum of the logs that you end up with when you do the definite integral always comes out to π/4. There's some serious work involved in proving that, but I think that is the path to the solution. Once you've established that for rational p/q you always get π/4, then you can take the limit of various p and q to make p/q go to √2, and you're done.
Karl <Click to Send Email to Karl>
USA - Tuesday, March 29, 2005 at 17:54:02 (EST)

find:

Int{0,π/2} f(x) dx

where

            1
f(x)= --------------
      1 + (tan(x))√2

M = 4πK

a 0

r2 √a² - r² dr

which you can solve using  a sin(u) = r  and  a cos(u)du = dr  (remember that sin²(u)cos²(u) = sin²(2u)/4 = (1 - cos(4u))/8)

If you used the triple integral that you suggest, then f(r cos(θ), r sin(θ), z) = K √r²cos²(θ) + r²sin²(θ) = Kr. Taking

M =

2π 0

a 0

√a² - r² -√a² - r²

(Kr) r dz dr dθ

As you simplify this down to a single integral dr, you should get the same thing as I got earlier.
Karl <Click to Send Email to Karl>
USA - Saturday, March 26, 2005 at 08:39:51 (EST)

   g(h)  =  ln(h)          [outer function]

   h(x)  =  2x2 - 7x + 20  [inner function]

f'(x) =

4x - 7 2x2 - 7x + 20

For f(x) = 3x² + ln(2x + 6), this is the sum of 3x², which you can find the derivative of using the power rule, and ln(2x + 6), to which you apply the chain rule.
Karl <Click to Send Email to Karl>
USA - Tuesday, March 22, 2005 at 12:26:41 (EST)

Remember, the dx represents something that is closer to zero than any real number (yet is not actually equal to zero). When you take the integral, you are adding up an infinite number of terms, but each is multiplied by the differential quantity, dx. If you divided by dx instead of multiplying, you'd be adding up an infinite number of infinite terms. There's no way you could get a finite result.
Karl <Click to Send Email to Karl>
USA - Sunday, March 20, 2005 at 18:58:14 (EST)

INT dx = x + C

    dx
INT -- = ln |x| + C
    x

    1
INT --
    dx

    1                   d
INT -- = INT {{dx}-1} = -- 1 = 0
    dx                  dx

   x2 + h2  =  502  =  2500

dx 2x + dt

dh 2h = 0 dt

Observe that the values for x and dx/dt are given in the problem. Observe that the Pythagorean equation gives you a way to find h if you know x. So you just need to solve for dh/dt.
Karl <Click to Send Email to Karl>
USA - Monday, March 14, 2005 at 18:50:14 (EST)

______ y' = 4x√4x - 3 +

(2x2 - 1)

2 √4x - 3

If you need to find the integral of this, then let u = 4x - 3. Then

  u + 3
         =  x
    4

  du
      =  dx
   4

1 4

2

u + 3 4

2 - 1

√u du

which you can multiply out and integrate term-by-term using the power rule for integrals (remember that √u = u^1/2).
Karl <Click to Send Email to Karl>
USA - Saturday, March 12, 2005 at 18:28:13 (EST)

1 V = 3

π r2 h

You have r and h changing at independent rates, dr/dt and dh/dt. So you have to use partial derivatives to find dV/dt:

   dV     ∂V dh     ∂V dr
       =         +       

   dt     ∂h dt     ∂r dt

∂V 2 = ∂r 3

π rh

and

∂V 1 = ∂h 3

π r2

The values of r, h, dr/dt and dh/dt are given in the problem. Use the partial derivative formulas above to substitute into the partial derivative equation for dV/dt. The evaluate dV/dt.
Karl <Click to Send Email to Karl>
USA - Saturday, March 12, 2005 at 18:00:49 (EST)

x4 + 2x2y2 = 8

3x3 + 4xy2 + 4x2yy'  =  0

3x3 + 4xy2  =  -4x2yy'

-(3x3 + 4xy2)
               =  y'
    4x2y

3x3 + 4xy2 + 4x2yy'  =  0

6x2 + 4y2 + 8xyy' + 8xyy' + 4x2(y')2 + 4x2yy"  =  0

Problems 2, 3, 4, and 5 are done the same way. On problem 6, they added a parametric variable, t.

y = t2 - 4       x = t + 2

dy              dx
    =  2t           =  1
dt              dt

dy dx     dy
       =    
dx dt     dt

r = 3 cos(2θ) + 1

If you need to know more, please quote the text of the problem.

Karl <Click to Send Email to Karl>
USA - Thursday, March 03, 2005 at 12:59:53 (EST)

33 = 43

27 < 64

1 2

Hence (3/4)³ⁿ < (1/2)ⁿ. And we know that (1/2)ⁿ goes to zero as n goes to infinity. But how do we know that? Because 2ⁿ grows without limit as n goes to infinity, so its reciprocal must go to zero. Since (3/4)³ⁿ < (1/2)ⁿ, it follows that (3/4)³ⁿ must also go to zero.
Karl <Click to Send Email to Karl>
USA - Thursday, March 03, 2005 at 12:47:09 (EST)

   h(t)  =  -(1/2)gt2  +  vvert t


   -(1/2)gt2  +  vvert t  +  192  =  0

_____________ vvert ± √vvert2 + 384g t = g

Since you want t to be positive, you have to take the + of the ±. The horizontal distance to impact in the gully will be x = v_horz t.

If m is the tangent of the angle from horizontal that the projectile is launched with and v is the launch velocity, then:

vvert =

mv √m2 + 1

vhorz =

v √m2 + 1

Making the substitution for v_vert and v_horz into the expression for horizontal distance traveled before impact (and doing some algebra the result):

x =	___________________ mv + √(mv)2 + 384g(m2 + 1) v g(m2 + 1)

Everything here is a constant except x and m. Take the derivative, ^dx/_dm, and set it to zero. Then solve for m. I grant you that solving it is nasty, but it is doable. Back substitute to find x. Use arctan(m) for the angle.

And do check my work here for mistakes.
Karl <Click to Send Email to Karl>
USA - Wednesday, March 02, 2005 at 19:46:33 (EST)