Search
Ask a question

Answers by Richard P.

Proving this presupposes that limt -> ∞  U(t)   exists.    Since it exits, call it Q   Then consider the left hand side      lims -> 0  s ∫ U(t) exp(- st)       Interchanging the operation...

Use the triangle construction.    Draw a right triangle with opposite side of length 15 and adjacent side of length 8.   The hypotenuse is then sqrt( 152 + 82 ) = sqrt(289) = 17.    The cosine is then  adjacent over hypotenuse =   8/17...

The function f will be one to one onto if it is monotone increasing on R.  f will be monotone increasing if its slope is everywhere greater than zero.  All that is required for the slope to be everywhere greater than zero is for epsilon to be less than 1/M.    

Call  6 am   t=0,  then 9 am corresponds to t = 3   A straight line estimate of the rate as a function of t is:     R(t) = 100 + 60 t.   The volume if the water piped is the integral of R over t   with the limits: ...

The formula for volume, V, is   depth x width x length.   The depth is x.  With the squares cut out, the width is reduced to 10 - 2x  and the length is reduced to 16- 2x.   The working equation is then  137.5 = x (10- 2x) (16 - 2x)   This...

In the laser Raman effect,  the wavelength of  a small portion of the exciting optical radiation is shifted to a slightly longer wavelength.    The longer wavelength is usually also in the optical part of the spectrum,  As an example, consider using a green  Argon ion...

For the first one,   x4 - y4  factors as (x2 - y2)(X2 + y2)   the second factor cancels out leaving x2 + y2    which is zero in the limit indicated   For the second one,   change top polar coordinates:    x = r cos(θ) ...

The case b >1  is easiest to think about.    With b >1   , the argument of function f  changes more rapidly (as x increases)  than would be the case with b =1.   This means that anything that f is going to "do" happens more rapidly.  More rapid...

The anti-derivative of 1/(6 x3) is -(1/12)/ x2     So the area under the curve is (1/12) ( 1 - 1/t2 )   So  for t = 10    area =  (1/12) .99      for  t = 100   area = (1/12) (.9999)   The entire area corresponds...