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Calculus/Parametric equations. Second Derivative

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Expert: - 8/2/2009


Question

Second Derivative
I disagree with the books final answer.
The books answer is 2/e^t(cos t - sin t)
I belive the answer should be 2/e^t(cos t - sin t)^3

Who is correct.

If the book is correct, could you please show me the steps involved to go from (2/(cos t - sin t)^2)/e^t(cos t - sin t)  to
the final answer of 2/e^t(cos t - sin t)


Answer
Questioner: Mark
Country: United States
Category: Calculus
Private: No
Subject: Parametric curve, Second Derivative
Question:

Second DerivativeI disagree with the books final answer.
The book's answer is 2/e^t(cos t - sin t)
I believe the answer should be 2/e^t(cos t - sin t)^3

Who is correct?

If the book is correct, could you please show me the steps involved to go from (2/(cos t - sin t)^2)/e^t(cos t - sin t)  to
the final answer of 2/e^t(cos t - sin t)
...........................................
Hi, Mark,

I think you are correct -- the book probably just left out the '3'.

I worked it out like this:

x = e^t cos t,   y = e^t sin t

x' = e^t (cos t - sin t)

y' = e^t (cos t + sin t)

dy   dy/dt   (cos t + sin t)
-- = ----- = ----------------
dx   dx/dt   (cos t - sin t)

Now for  ddy    d  dy
        --- = --- ---  << I hate making those ^2's.
        dxx   dx  dx

d      d  dt     d
-- =  --- --- = --- (dy/dx) divided by dx/dt
dx     dt dx     dt

ddy      d  dy
---- =  --- --, divided by dx/dt
dxdx     dt dx

(I am guessing this is Schaum's  37.2 -- I don't have that copy.)

        2
= ------------------ divided by  e^t(cos t - sin t)
 (cos t - sin t)^2


Yes, that looks like

        2
= --------------------
 e^t(cos t - sin t)^3
.......................................
You could also do this: (more fun and games?)

dy   cos t + sin t
-- = --------------
dx   cos t - sin t

 e^t cos t + e^t sin t    x + y
= ---------------------- = -----
 e^t cos t - e^t sin t    x - y

ddy    (x - y)(1 + y') - (x + y)(1 - y')
---  = ---------------------------------
dxdx                (x - y)^2

 x - y + xy' - yy' - x - y + xy' + yy'
= ------------------------------------
              (x - y)^2
   2(- y + xy')
= ---------------
   (x - y)^2

   2(- y +  x(x+y)/(x-y))
= ------------------------  << substituting  y' as found before.
      (x - y)^2


   2(- y(x-y) +  x(x+y))
= -----------------------  << multiply all by  x-y
    (x - y)^3

   2(- xy + yy + xx + xy)
= ------------------------  << I told you I hate those ^2's
      (x - y)^3

   2(yy + xx)
= --------------
    (x - y)^3

Now you can use the original definitions of x,y:

      2(e^t)^2
= ------------------------
 (e^t)^3(cos t - sin t)^3

           2
= ------------------------
  e^t(cos t - sin t)^3

Nice.

Paul Klarreich

Expertise

All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

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I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.

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