Expert: Paul Klarreich - 10/14/2007

Question
differentiation of implicit functions are simple no? but what happens when a
function x = (1 t)^1/2 , y= (1-t)^1/2 is differentiated to - x/y. i don't get it.
also, isn't the differentiated form of a parametric equation x = at^2, y = at^3
going to be a3t^2/a2t ? or did i go wrong somewhere along the line cause the
answer is 3bt/2a. help, please? please, i need to understand for midterms

Answer
Questioner:   liz
Category:  Calculus
Private:  No
 
Subject:  calculus
Question:  differentiation of implicit functions are simple no? but what happens when a function x = (1 t)^1/2 , y= (1-t)^1/2 is differentiated to - x/y. i don't get it.

also, isn't the differentiated form of a parametric equation x = at^2, y = at^3
going to be a3t^2/a2t ? or did i go wrong somewhere along the line cause the
answer is 3bt/2a. help, please? please, i need to understand for midterms
..............................................
Hi, Liz,

In my profile, I have a short list of instructions to questioners, which includes a request that you PROOFREAD YOUR QUESTION before submitting it.  Your submission has a couple of omissions/errors that make the questions either meaningless or incorrect.  Perhaps that is the reason for your difficulty.

.......................

In your first example

x = (1 t)^1/2 , y= (1-t)^1/2 is differentiated to - x/y

did you mean:

x = (1 + t)^1/2 , y= (1 - t)^1/2 is differentiated to - x/y

[the plus sign didn't come through.]

I assume your question asks:

If  x = (1 + t)^1/2 , y= (1 - t)^1/2,  then what is dy/dx?

You use the chain rule in this form:

dy   dy/dt
-- = -----
dx   dx/dt

dy/dt = (1/2)(1 - t)^-1/2(-1) =
             -1
dy/dt = -------------
       2(1 - t)^1/2

        -1
dy/dt = --- << remember, y = (1 - t)^1/2
        2y
. . . . . . . . . .

dx/dt = (1/2)(1 + t)^-1/2(1) =
            1
dx/dt = -------------
       2(1 + t)^1/2

        1
dx/dt = --- << remember, x = (1 + t)^1/2
        2x

. . . . . . . . . .

dy   -1/2y
-- = ----- = -x/y
dx    1/2x
.................................

You wrote:

x = at^2, y = at^3

I think you meant to write:

x = at^2, y = bt^3

Now

dy/dt = 3bt^2
dx/dt = 2at

and

dy    3bt^2   3bt
-- =  ----- = ---
dx     2at    2a