Expert: Paul Klarreich - 7/16/2007

Question
Hi,

I am having trouble trying to figure out how approach this problem;

How would you go about solving for the derivative of y with respect to x? What is the slope of the function y(x) if t=1?

y(t)=t^3-t
x(t)=1-t^2
dy/dx(vertical line) t-1 =?

Thanks in advance!


Answer
Questioner:   kerel
Category:  Calculus
Question:  Hi,

I am having trouble trying to figure out how approach this problem;

How would you go about solving for the derivative of y with respect to x? What is the slope of the function y(x) if t=1?

>> You mean the slope of the GRAPH OF y = f(x) at t=1.

y(t)=t^3-t
x(t)=1-t^2
dy/dx(vertical line) t-1 =?

Thanks in advance!
................................
Hi, Kerel,

You can do either of two things:

1. Eliminate the parameter to get  y as a function of x, then differentiate.

Here, you have:

y  = t^3 - t
x  = 1 - t^2

t^2 = 1 - x
t = sqrt(1 - x)

y = t(t^2 - 1)

y = sqrt(1 - x) (1 - x - 1)
y = sqrt(1 - x) ( - x)

y = - x sqrt(1 - x)

Now differentiate, using the product rule, etc.

dy              - 1
-- =  - [(x) -----------   + (1)sqrt(1 - x)]
dx           2 sqrt(1-x)

dy        + x
-- =   -----------  -  sqrt(1 - x)
dx     2 sqrt(1-x)

dy          x
-- =  ------------- - sqrt(1 - x)
dx    2 sqrt(1 - x)

dy     x - 2(1 - x)
-- =  -------------
dx    2 sqrt(1 - x)

dy     x - 2 + 2x
-- =  -------------
dx    2 sqrt(1 - x)

dy       3x - 2
-- =  -------------
dx    2 sqrt(1 - x)

Now if  t = 1,  x = 1 - t^2 = 1 - 1 = 0

dy     3(0) - 2
-- = -------------
dx   2 sqrt(1 - 0)

dy   -2
-- = --- = -1
dx    2

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OR

2. you use the rule: (A kind of chain rule.)

dy    dy/dt   3t^2 - 1
-- =  ----- = -----------
dx    dx/dt    - 2t

Now if t = 1, that is:
dy    3 - 1    2
-- =  ----- = --- = -1
dx     -2     -2

Whew! They came out the same. That's why you do things by two methods.

..............................
I'm not sure what your last line said.  What do you have to do about a vertical line?